7.1 Introduction

Chapter 7 — Gravitation

In the summer of 1666, plague had shut down Cambridge University, and a young Isaac Newton was sitting out the outbreak at his mother’s farmhouse in Woolsthorpe, England. According to a story Newton himself told friends late in his life, he was resting in the orchard when he saw an apple fall from a tree. The fall itself was ordinary — apples had been falling for as long as trees had grown. What was not ordinary was the question the twenty-three-year-old Newton asked next.

Why did the apple fall straight down, and not sideways or upward? And if the Earth’s pull reached the top of the tree, how far did that pull actually reach? Did it reach the top of a mountain? Did it reach the clouds? Did it, perhaps, reach all the way to the Moon?

That last thought was the leap. The Moon, Newton realised, was in a sense always “falling” toward the Earth — it just kept missing, because it was also moving sideways fast enough to curve around. If the same force that pulled the apple also held the Moon in its orbit, then one single law of nature might connect a falling fruit in an English orchard to the motion of every planet in the sky.

This chapter is the story of that single law — where it came from, what it says, and what it explains.

Figure to come

Fig. 7.0 – Split illustration: on the left, an apple falling from a tree toward the Earth’s surface; on the right, the Moon in a curved orbit around the Earth, with a dashed arrow showing the Moon’s “fall” direction toward Earth’s centre. A faint line connects the two, suggesting a single underlying force.

Curiosity Corner

Q1. Galileo showed that a heavy stone and a light stone dropped together hit the ground at the same time. Why does the mass of the falling object not change how fast it falls?

Q2. The Moon is constantly pulled toward the Earth by gravity. Why then does it not simply crash into us?

Q3. If you dig a deep mine and go far below the Earth’s surface, does gravity get stronger, weaker, or stay the same — and why?

Q4. A rocket needs a very specific minimum speed to escape the Earth forever. What decides that speed, and why is it the same for a small pebble and a large spacecraft?

Q5. Satellites orbit the Earth without any fuel keeping them up. What is holding them in orbit, and what decides how long they take to go around once?

By the end of this chapter, you will be able to answer each of these questions using a single law of gravitation that Newton first wrote down more than three centuries ago.

From the earliest moments of our lives, we learn one silent rule of the world around us: things fall. A ball tossed into the air comes back down. Rain drops from the clouds to the ground. Walking uphill tires us out far more quickly than walking downhill. Every one of these ordinary experiences is telling us the same thing — the Earth pulls material objects toward itself.

For a long time, though, no one could say clearly how it pulled them. In particular, a very old confusion needed to be cleared up first: does a heavier object fall faster than a lighter one?

Common sense says yes. A stone drops quickly, while a feather drifts down slowly. But the feather is not really being fooled by gravity — it is being slowed by air. What happens if we remove air from the picture? This is the question Galileo Galilei, an Italian physicist working in the late 1500s and early 1600s, set out to answer.

Galileo could not create a perfect vacuum, but he had a clever workaround. He rolled balls of different masses down smooth inclined planes and timed their motion carefully. By using inclined planes instead of a straight vertical drop, he slowed the fall down enough to measure it reliably. His results were remarkable: all bodies, regardless of their mass, sped up at the same rate as they moved down. In other words, the acceleration produced by gravity is the same for every object.

Real Incident / Discovery

In the late 1500s, Galileo Galilei carried out systematic experiments with bodies rolling down inclined planes at Padua and Pisa. Popular legend says he dropped two unequal weights from the Leaning Tower of Pisa to show they hit the ground together. Whether or not that public demonstration actually took place, his inclined-plane experiments were real, and they gave a value of the acceleration due to gravity very close to the accurate value we use today.

This was the first big idea: gravity produces a constant acceleration on all bodies near the Earth’s surface, no matter how heavy or how light they are.

While Galileo was studying falling bodies on Earth, a completely different puzzle had been on the human mind for thousands of years: the motion of objects in the sky. Ancient observers noticed that most stars stayed in fixed patterns year after year, but a few bright points wandered against that background of fixed stars. These wandering points are what we call the planets.

Explaining how the planets moved became one of the oldest scientific problems in history. Around 2000 years ago, the Greek astronomer Ptolemy proposed a model in which the Earth sat at the centre of the universe and everything else — the Sun, the Moon, the planets, and the stars — went around it. This is called the geocentric model (“geo” meaning Earth). In this picture, the only motion allowed for heavenly bodies was motion in a circle, because circles were thought to be the most “perfect” shape.

The trouble was that the planets did not move as if they were on simple circles. Sometimes they even appeared to go backwards for a while before moving forward again. To fit this awkward behaviour, Ptolemy had to invent complicated schemes: planets moved on small circles, whose centres moved on larger circles. Indian astronomers, some 400 years later, put forward similar circle-on-circle ideas.

A very different picture had also been suggested. Aryabhatta, in the 5th century AD, wrote in his treatise that the Sun — not the Earth — was at the centre, and the planets went around it. This is called the heliocentric model (“helio” meaning Sun). A thousand years later, in the 1500s, the Polish scholar Nicolas Copernicus revived and developed this idea into a full definitive model, in which the planets moved in circles around a fixed central Sun.

Side Note

Copernicus’s heliocentric model was strongly opposed by the church of his time. Galileo, who publicly supported it, was tried and put under house arrest for his beliefs — a reminder that even correct scientific ideas have sometimes had to fight for acceptance.

The next crucial step came from the patient work of one observer and one calculator. Around the same period as Galileo, a Danish nobleman named Tycho Brahe spent his entire life recording the positions of planets in the sky with the naked eye — no telescopes, just extremely careful measurements night after night. He built up a treasure chest of astronomical data unlike anything before.

Real Incident / Discovery

Tycho Brahe was famously particular about his data. Working in Denmark in the second half of the 1500s, he catalogued planetary positions with an accuracy no earlier astronomer had achieved. His observations, taken with the unaided eye, were precise enough that they would later be used to disprove all circle-based models of planetary motion.

After Tycho’s death, his assistant Johannes Kepler took over the mountain of observations. Kepler was not just an observer; he was a superb mathematician. By analysing Tycho’s data for years, he pulled three simple, elegant laws out of it — the laws that now carry his name.

These three laws of planetary motion described how planets move, but they did not explain why. That final explanation came later, from Isaac Newton. Newton realised that Kepler’s laws and Galileo’s falling apples were both consequences of one and the same idea — a universal law of gravitation that governs everything from a stone falling on Earth to a planet orbiting the Sun.

This chapter tells that story in physics form. We will begin with Kepler’s laws, move to Newton’s universal law of gravitation, and then use these ideas to understand how gravity behaves on and inside the Earth, how satellites stay in orbit, and how fast an object must be thrown so that it escapes the Earth’s pull altogether.

7.2 Kepler’s Laws

We ended Section 7.1 with Johannes Kepler sitting in front of a mountain of astronomical data left behind by Tycho Brahe. Over nearly two decades of patient calculation, Kepler squeezed three simple statements out of that data. These three statements — now called Kepler’s laws of planetary motion — describe how planets move around the Sun without yet saying why. They were the crucial bridge between careful observation and Newton’s universal law of gravitation.

Real Incident / Discovery

Kepler began by trying to fit the orbit of Mars using perfect circles, exactly as astronomers had done for two thousand years. The circles did not work. Tycho’s data was so precise that even a tiny mismatch of about 8 minutes of arc — a fraction of a degree — refused to disappear. Rather than dismiss this small error, Kepler took it seriously. That decision led him to abandon circles altogether and try ellipses instead, and it changed astronomy forever.

We now state and understand each of the three laws.

1. Law of orbits

Kepler’s first law says something that sounds simple but was revolutionary in the 1600s: planets do not move on circles. They move on ellipses.

Principle / Law

All planets move in elliptical orbits, with the Sun situated at one of the two foci of the ellipse.

An ellipse is a closed oval curve. To picture how it works, look at Fig. 7.1(b). Fix the two ends of a string to two points $F_1$ and $F_2$ using pins. Pull the string taut with a pencil tip and move the pencil around while keeping the string tight. The curve traced out is an ellipse.

Figure to come

Fig. 7.1(b) – Two pins at points F₁ and F₂ on a board, with a loop of string held taut by a pencil; the pencil is drawing an oval curve around both pins.

The two special points $F_1$ and $F_2$ are called the foci (singular: focus) of the ellipse. From the way we drew it, one property is obvious: for any point $T$ on the ellipse, the sum of the distances from $T$ to the two foci is a constant (it equals the length of the string).

Now look at Fig. 7.1(a), which shows a planet’s orbit around the Sun. If we join the two foci and extend the line, it meets the ellipse at two points, marked $P$ and $A$. The midpoint of $PA$ is called the centre of the ellipse, $O$. Half of the distance $PA$ — that is, $PO = AO$ — is called the semi-major axis of the ellipse. It plays the same role for an ellipse that the radius plays for a circle.

Figure to come

Fig. 7.1(a) – Ellipse with Sun marked at focus S, empty focus S′, planet’s closest point P (perihelion) on left, farthest point A (aphelion) on right; semi-major axis PO labelled, minor axis 2b vertical, major axis 2a horizontal.

The Sun sits at one focus, say $S$. The other focus $S'$ is just an empty point in space — nothing physical is there. The point $P$, where the planet is closest to the Sun, is called the perihelion. The point $A$, where it is farthest from the Sun, is called the aphelion.

A circle is actually a special case of an ellipse: if the two foci merge into a single point, the ellipse becomes a circle, and the semi-major axis becomes the radius. This is why Copernicus’s earlier picture — planets on circles around the Sun — was not entirely wrong. It was just a special, over-simplified case of Kepler’s more general truth.

2. Law of areas

Kepler noticed something curious in the data. A planet does not move at the same speed everywhere on its orbit. It speeds up when it is close to the Sun and slows down when it is far away. To capture this pattern precisely, Kepler proposed:

Principle / Law

The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

Look at Fig. 7.2. As the planet $P$ moves along its orbit, imagine a straight line drawn from the Sun $S$ to the planet. In a small time interval $\Delta t$, the planet moves a short distance and this line sweeps out a thin wedge of area, $\Delta A$. Kepler’s second law says that the rate at which area is swept is the same everywhere on the orbit — even though the planet’s speed itself changes.

Figure to come

Fig. 7.2 – Elliptical orbit with Sun at focus; planet at position P moving to P′ over time Δt; shaded triangular sliver of area ΔA between the two radius vectors and the arc; velocity vector v shown tangent to orbit at P; force F pointing from P toward Sun.

This is a strong statement, and it turns out to have a deeper reason behind it. Let us see where it comes from.

The law of areas is a direct consequence of the conservation of angular momentum, which is valid whenever the force on the planet is a central force. A central force is one that always points along the line joining the planet to a fixed centre — here, the Sun.

Definition

A central force acting on a particle is a force that is always directed along the line joining the particle to a fixed point (the “centre”).

Place the Sun at the origin. Let the planet’s position vector be $\vec{r}$ and its momentum be $\vec{p} = m\vec{v}$, where $m$ is the planet’s mass and $\vec{v}$ its velocity. In a small time $\Delta t$, the planet’s displacement is $\vec{v}\Delta t$. The area of the tiny triangular wedge swept by the position vector is half the magnitude of the cross product of $\vec{r}$ and this displacement:

\[\Delta \vec{A} = \tfrac{1}{2}\,(\vec{r} \times \vec{v}\,\Delta t) \tag{7.1}\]

Dividing both sides by $\Delta t$ and using $\vec{v} = \vec{p}/m$:

\[\frac{\Delta \vec{A}}{\Delta t} = \frac{1}{2m}\,(\vec{r} \times \vec{p}) = \frac{\vec{L}}{2m} \tag{7.2}\]

Here $\vec{L} = \vec{r} \times \vec{p}$ is the angular momentum of the planet about the Sun. For a central force, which is directed along $\vec{r}$, the torque about the centre is zero (because $\vec{r} \times \vec{F} = 0$ when $\vec{F}$ is parallel to $\vec{r}$). Zero torque means angular momentum is conserved — that is, $\vec{L}$ is a constant. From Eq. (7.2), if $\vec{L}$ is constant, then $\Delta \vec{A}/\Delta t$ is also constant. This is exactly Kepler’s second law.

Gravitation, as we shall see in the next section, is a central force. So the law of areas is not just an observed pattern — it is guaranteed by the physics.

3. Law of periods

Kepler’s third law connects the size of a planet’s orbit to how long it takes to complete one revolution.

Principle / Law

The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its elliptical orbit.

In symbols:

\[T^2 \propto a^3\]

where $T$ is the orbital period (time for one full revolution) and $a$ is the semi-major axis of the orbit. Equivalently, the quotient $T^2/a^3$ has the same value for every planet in the Solar System.

Table 7.1 shows the semi-major axes and orbital periods for the eight planets, along with the quotient $Q = T^2/a^3$. Notice how close the values of $Q$ are, even though the planets themselves are hugely different in size and distance.

Table 7.1 Data from measurement of planetary motions confirming Kepler’s Law of Periods (a ≡ semi-major axis in units of $10^{10}$ m; T ≡ time period of revolution in years; Q ≡ $T^2/a^3$ in units of $10^{-34}\ \text{y}^2\text{m}^{-3}$)

Planet	a	T	Q
Mercury	5.79	0.24	2.95
Venus	10.8	0.615	3.00
Earth	15.0	1	2.96
Mars	22.8	1.88	2.98
Jupiter	77.8	11.9	3.01
Saturn	143	29.5	2.98
Uranus	287	84	2.98
Neptune	450	165	2.99

The near-constancy of $Q$ across all eight planets is remarkable. It tells us that planets far from the Sun take much longer to orbit — not just because they travel a longer path, but because they also move more slowly. The law of periods will reappear later in this chapter when we derive it from Newton’s law of gravitation, and again when we study Earth’s satellites.

Side Note

Kepler’s laws were first written for the six planets known in his time. As new planets — Uranus and Neptune — were discovered later, and even for the many asteroids and dwarf planets discovered since, the same $T^2/a^3$ ratio holds. That is why Table 7.1 lists eight planets: the international astronomical community reclassified Pluto in 2006 as a “dwarf planet,” so it is no longer counted in the list of major planets.

Numerical 7.1

A hypothetical planet is discovered orbiting the Sun with a semi-major axis $4$ times that of the Earth. Using Kepler’s third law, find its orbital period in Earth years.

Numerical 7.2

A comet orbits the Sun in a highly elongated ellipse. At perihelion, its distance from the Sun is $0.6$ AU and its speed is $54\ \text{km s}^{-1}$. Using conservation of angular momentum, find its speed at aphelion, given that the aphelion distance is $35$ AU.

Now let us look at a solved example that ties together the law of areas and the law of orbits.

Solved Example 7.1

Consider a planet at its perihelion $P$ in Fig. 7.1(a), moving with speed $v_P$, where the Sun-planet distance is $r_P$. Relate $\{r_P, v_P\}$ to the corresponding quantities $\{r_A, v_A\}$ at the aphelion $A$. Will the planet take equal times to traverse the arcs $BAC$ and $CPB$?

Answer

At the perihelion $P$, the planet’s position vector $\vec{r}_P$ and velocity $\vec{v}_P$ are perpendicular to each other (this is a property of the endpoints of the major axis — the planet is momentarily moving purely sideways, not toward or away from the Sun). Therefore the magnitude of the angular momentum at $P$ is

\[L_P = m_p\, r_P\, v_P\]

Similarly, at the aphelion $A$, the vectors $\vec{r}_A$ and $\vec{v}_A$ are also perpendicular, so

\[L_A = m_p\, r_A\, v_A\]

By conservation of angular momentum (Kepler’s second law, expressed in physics terms):

\[m_p\, r_P\, v_P = m_p\, r_A\, v_A\]

which gives

\[\frac{v_P}{v_A} = \frac{r_A}{r_P}\]

Since the aphelion distance is greater than the perihelion distance ($r_A > r_P$), we conclude $v_P > v_A$ — the planet moves faster at perihelion than at aphelion. This is exactly what Kepler observed.

For the second part, notice from Fig. 7.1(a) that the area $SBAC$ (bounded by the ellipse and the radius vectors $SB$ and $SC$, and containing the aphelion) is larger than the area $SBPC$ (containing the perihelion). Kepler’s second law tells us that equal areas are swept in equal times. So sweeping a larger area needs a longer time. Hence the planet takes a longer time to traverse arc $BAC$ than arc $CPB$.

With Kepler’s three laws in hand, we now have a precise description of planetary motion. But we still do not know why planets obey these laws. That is Newton’s contribution, and it is the subject of the next section.

7.3 Universal Law of Gravitation

Kepler’s three laws described how planets move. Newton took the next, much larger step. He asked: what force makes them move that way? And even more boldly, he asked whether the same force that holds the planets in their orbits might also be the force pulling apples and stones toward the Earth.

The famous story of the apple in Newton’s garden captures the moment when this question first struck him. But an inspired guess is not physics — it needed a calculation. Newton wanted to actually check whether the Earth’s gravity, if it truly extended all the way to the Moon, could account for the Moon’s orbit.

Testing the idea with the Moon

The Moon goes around the Earth on a nearly circular orbit of radius $R_m$, completing one revolution in a time period $T$. Newton knew from Class 11 mechanics that an object moving in a circle is not moving in a straight line — it is being continuously pulled toward the centre. This inward pull produces a centripetal acceleration.

If the speed of the Moon in its orbit is $V$, then the centripetal acceleration $a_m$ pointing toward the Earth is

\[a_m = \frac{V^2}{R_m} = \frac{4\pi^2 R_m}{T^2} \tag{7.3}\]

The second form uses $V = 2\pi R_m / T$ — that is, one full circumference $2\pi R_m$ divided by the time $T$ for one revolution.

In Newton’s time, both $R_m$ and $T$ were already fairly well known: $T \approx 27.3$ days and $R_m \approx 3.84 \times 10^8$ m. Plugging these numbers into Eq. (7.3) gives a value of $a_m$ that is much smaller than the acceleration due to gravity $g \approx 9.8\ \text{m s}^{-2}$ felt on the Earth’s surface.

That was a clue, not a contradiction. Newton’s insight was that Earth’s gravity should get weaker at greater distances. The question was: how does it weaken?

The inverse-square guess

Newton tested the simplest reasonable guess: that the gravitational force from the Earth falls off as the inverse square of the distance from the Earth’s centre. If this is true, then the gravitational acceleration $a$ at distance $r$ should behave as $a \propto 1/r^2$.

Applied to our two cases:

\[a_m \propto \frac{1}{R_m^2}, \qquad g \propto \frac{1}{R_E^2}\]

where $R_E$ is the Earth’s radius. Taking the ratio removes the constant of proportionality:

\[\frac{g}{a_m} = \frac{R_m^2}{R_E^2} \approx 3600 \tag{7.4}\]

When Newton computed the same ratio directly from the value of $a_m$ obtained from Eq. (7.3) and the observed value of $g$, he got a number very close to 3600. The two calculations agreed. This was strong evidence that the Earth’s gravitational pull really does weaken as $1/r^2$.

Real Incident / Discovery

Newton’s Moon test — the calculation above — is one of the earliest examples in physics of testing a bold theoretical guess against observation. He did the calculation in the late 1660s but held back publication for nearly two decades, partly because he was uncertain about how to treat the Earth as a single point when the apple was only a few metres from its surface. The problem was solved once he developed calculus and could prove the “shell theorem” we meet later in this section.

Newton’s universal law of gravitation

The Moon test told Newton that the Earth pulls other objects with a force that decreases as the square of the distance. But he went further — he proposed that this rule was universal, applying not just to the Earth but to every pair of objects in the universe.

Principle / Law

Every body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

For two point masses $m_1$ and $m_2$ separated by a distance $r$, the magnitude of the gravitational force is

\[|\vec{F}| = G\,\frac{m_1 m_2}{r^2} \tag{7.5}\]

Here $G$ is a universal constant called the universal gravitational constant — universal because it has the same value everywhere in the universe, for every pair of masses, regardless of their nature. Its numerical value will be discussed in the next section. In SI units, $m_1$ and $m_2$ are in kilograms, $r$ is in metres, $|\vec{F}|$ is in newtons, and $G$ has units of $\text{N m}^2 \text{kg}^{-2}$.

Side Note

The quotation-style statement of the law above is essentially from Newton’s own famous treatise, Mathematical Principles of Natural Philosophy (usually just called the Principia), published in 1687. The Principia is often considered the single most important book in the history of physics.

Vector form of the law

Equation (7.5) gives only the magnitude of the force. To describe its direction as well, we need to write it as a vector equation.

Refer to Fig. 7.3. Let $m_1$ and $m_2$ be two point masses with position vectors $\vec{r}_1$ and $\vec{r}_2$ measured from some origin $O$. Define $\vec{r} = \vec{r}_2 - \vec{r}_1$ — the position vector from $m_1$ to $m_2$. Let $\hat{r}$ be the unit vector pointing from $m_1$ to $m_2$.

Figure to come

Fig. 7.3 – Origin O with two position vectors r₁ and r₂ pointing to masses m₁ and m₂; the vector r = r₂ − r₁ joins m₁ to m₂ with an arrow; small arrow labels a unit vector r̂ along r.

Then the gravitational force on mass $m_2$ due to mass $m_1$ is written as

\[\vec{F} = G\,\frac{m_1 m_2}{r^2}\,(-\hat{r}) = -G\,\frac{m_1 m_2}{r^2}\,\hat{r} = -G\,\frac{m_1 m_2}{|\vec{r}|^3}\,\vec{r}\]

The negative sign shows that the force on $m_2$ points from $m_2$ toward $m_1$ — the two masses attract each other, they do not repel. Gravity, unlike the electric force between charges, is always attractive.

By Newton’s third law, the force on $m_1$ due to $m_2$ is equal in magnitude and opposite in direction. In symbols,

\[\vec{F}_{12} = -\vec{F}_{21}\]

where $\vec{F}_{12}$ is the force on body 1 due to body 2 and $\vec{F}_{21}$ is the force on body 2 due to body 1.

Many masses — the superposition principle

Equation (7.5) applies to just two point masses. What if we have many? For example, a mass $m_1$ surrounded by three other masses $m_2$, $m_3$, and $m_4$ (see Fig. 7.4)?

Figure to come

Fig. 7.4 – Mass m₁ at centre with three surrounding masses m₂, m₃, m₄; three force vectors F₁₂, F₁₃, F₁₄ drawn on m₁, each pointing toward the respective attracting mass; unit vectors r̂₂₁, r̂₃₁, r̂₄₁ shown.

Nature turns out to be very kind here. The gravitational force on $m_1$ due to $m_2$ is not affected by the presence of $m_3$ or $m_4$. Each pair interacts as if the others were not there. To find the total force on $m_1$, we simply add up all the individual pairwise forces as vectors. This is the principle of superposition applied to gravity.

The total force on $m_1$ due to $m_2$, $m_3$, and $m_4$ is therefore

\[\vec{F}_1 = \frac{G m_2 m_1}{r_{21}^2}\,\hat{r}_{21} + \frac{G m_3 m_1}{r_{31}^2}\,\hat{r}_{31} + \frac{G m_4 m_1}{r_{41}^2}\,\hat{r}_{41}\]

where each $\hat{r}_{i1}$ is a unit vector from $m_1$ to $m_i$.

From point masses to extended bodies

There is a practical problem left. Newton’s law refers to point masses — objects with no size. But real bodies like the Earth, the Moon, or a stone are extended objects with finite size. Different parts of the Earth are at different distances from the apple. How do we handle this?

The correct method is to imagine the extended body as a huge collection of tiny point-mass pieces. Each piece pulls the outside object with a small force in its own direction, and we then add up (integrate) all these tiny force vectors. This addition is done using calculus.

Calculus is beyond our present scope, but Newton worked out two crucial results for the specific case of a hollow spherical shell of uniform density — and these results save us an enormous amount of trouble later:

Principle / Law

(Shell theorem — Case 1) The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside the shell is exactly the same as if the entire mass of the shell were concentrated at its centre.

Why should this be true? Qualitatively: the shell is symmetric around its centre. Each little patch of the shell pulls the outside mass in a direction slightly different from every other patch. All the sideways components cancel out neatly in pairs, and only the components pointing straight toward the centre survive. What is left behaves exactly as if the whole shell were shrunk to a single point at its centre.

Principle / Law

(Shell theorem — Case 2) The force of attraction due to a hollow spherical shell of uniform density on a point mass situated inside the shell is zero.

Again, symmetry gives us the answer. A point mass inside a hollow shell is pulled in every direction by different patches of the shell. Because of the perfect spherical symmetry, all these pulls cancel out exactly, giving zero net force — no matter where inside the shell the point mass is placed.

The shell theorem is why we can treat the Earth as if all its mass were concentrated at its centre when we compute the gravitational force it exerts on an object above its surface. Newton needed this result to justify his Moon test. It also gives the reason apples fall straight down toward the centre of the Earth, and why the Moon’s orbit responds to Earth’s mass as a whole rather than to some odd part of it.

Now let us look at a solved example that uses the superposition principle to compute the total gravitational force on a mass placed symmetrically among others.

Solved Example 7.2

Three equal masses of $m$ kg each are fixed at the vertices of an equilateral triangle $ABC$.

What is the force acting on a mass $2m$ placed at the centroid $G$ of the triangle?
What is the force if the mass at the vertex $A$ is doubled? Take $AG = BG = CG = 1$ m (see Fig. 7.5).

[Diagram: Fig. 7.5 – Equilateral triangle ABC with masses m at each vertex; a 2m mass placed at centroid G; three arrows from G pointing to A, B, C (or vice versa) showing the three gravitational forces; angle 30° marked between GC and positive x-axis, and between GB and negative x-axis.]

Answer

Set up coordinates so that vertex $A$ is directly above $G$ along the positive $y$-axis. Then $B$ is in the lower-left direction and $C$ in the lower-right direction, both making an angle of $30^\circ$ with the negative and positive $x$-axes respectively.

The gravitational force on the $2m$ mass at $G$ due to the mass at $A$ points from $G$ toward $A$, which is in the $+\hat{j}$ direction. With $AG = 1$ m:

\[\vec{F}_{GA} = \frac{G m (2m)}{1^2}\,\hat{j} = \frac{G m (2m)}{1}\,\hat{j}\]

The force due to $B$ is directed from $G$ toward $B$:

\[\vec{F}_{GB} = \frac{G m (2m)}{1}\,\left(-\hat{i}\cos 30^\circ - \hat{j}\sin 30^\circ\right)\]

The force due to $C$ is directed from $G$ toward $C$:

\[\vec{F}_{GC} = \frac{G m (2m)}{1}\,\left(+\hat{i}\cos 30^\circ - \hat{j}\sin 30^\circ\right)\]

By the principle of superposition, the net force is the vector sum:

\[\vec{F}_R = \vec{F}_{GA} + \vec{F}_{GB} + \vec{F}_{GC}\]

\[\vec{F}_R = 2Gm^2\,\hat{j} + 2Gm^2\,(-\hat{i}\cos 30^\circ - \hat{j}\sin 30^\circ) + 2Gm^2\,(\hat{i}\cos 30^\circ - \hat{j}\sin 30^\circ) = 0\]

The $\hat{i}$ components from $B$ and $C$ cancel, and the $\hat{j}$ contribution from $A$ is cancelled by the two $-\hat{j}\sin 30^\circ$ terms from $B$ and $C$ (which together give $-2Gm^2 \cdot \hat{j}$, since $2\sin 30^\circ = 1$). The net force is zero.

We could have guessed this from symmetry alone: three equal masses placed symmetrically around a common point pull that point equally in three symmetric directions, and the pulls cancel out.

Now suppose the mass at vertex $A$ is doubled from $m$ to $2m$. The forces from $B$ and $C$ are unchanged, but the force from $A$ doubles:

\[\vec{F}'_{GA} = \frac{G(2m)(2m)}{1}\,\hat{j} = 4Gm^2\,\hat{j}\]

The forces from $B$ and $C$ still cancel each other in the $\hat{i}$ direction and together contribute $-2Gm^2\,\hat{j}$ to the $\hat{j}$ direction (the same as before). So the net force is

\[\vec{F}'_R = \vec{F}'_{GA} + \vec{F}_{GB} + \vec{F}_{GC} = 4Gm^2\,\hat{j} - 2Gm^2\,\hat{j} = 2Gm^2\,\hat{j}\]

The net force is $2Gm^2$ directed from $G$ toward $A$ (the doubled mass) — which is exactly what physical intuition suggests.

Numerical 7.3

Two point masses of $3.0$ kg and $5.0$ kg are placed $0.40$ m apart in vacuum. Find the magnitude of the gravitational force between them. (Take $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$.)

Numerical 7.4

Two point masses of $8.0$ kg and $2.0$ kg are placed $1.0$ m apart. At what point on the line joining them will the net gravitational force on a small test mass be zero?

The universal law of gravitation is now fully in place. What remains is to actually measure the constant $G$ that appears in it — a delicate laboratory task that took more than a century after Newton to accomplish. That is the subject of the next section.

7.4 The Gravitational Constant

Newton’s universal law tells us that every pair of masses attracts every other pair with a force $F = G\,m_1 m_2 / r^2$. The law is precise and complete except for one thing — the value of $G$. Newton himself never measured $G$; in fact, he simply did not have the equipment. The masses in his equation had to be known, the distance between them had to be known, and above all, the extremely feeble force between two ordinary laboratory-sized objects had to be measured. In the 1600s, that was impossible.

More than a century passed before someone succeeded. In 1798, the English scientist Henry Cavendish carried out a beautifully sensitive experiment that finally yielded a value of $G$. The technique he used is still, in essence, the way $G$ is measured today.

The Cavendish experiment

The apparatus Cavendish used is shown schematically in Fig. 7.6. Its central idea is to convert a very weak gravitational force into an easily observable twist of a wire.

Figure to come

Fig. 7.6 – Horizontal bar AB with two small lead spheres attached at its ends, suspended at its centre by a fine vertical wire from a rigid support above; two large lead spheres $S_1$ and $S_2$ placed close to the small spheres on opposite sides of the bar (shown shaded); dotted circles indicate the alternate positions $S'_1$ and $S'_2$ when the big spheres are moved to the other side, causing the bar to rotate slightly.

A horizontal bar $AB$ has two small lead spheres fixed at its ends. The bar is suspended at its middle from a rigid support by a very fine wire, so it can rotate freely in the horizontal plane. Two large lead spheres $S_1$ and $S_2$ are then brought close to the small spheres, but on opposite sides of the bar, as shown.

Each big sphere gravitationally attracts the nearby small sphere. Because the two big spheres sit on opposite sides, they pull the two ends of the bar in opposite directions. The net force on the bar as a whole is zero, but the two equal and opposite pulls at the two ends together create a couple, or torque, that tries to rotate the bar.

The gravitational torque on the bar is easy to write. If $F$ is the gravitational force between a big sphere and its neighbouring small sphere, and $L$ is the length of the bar, then the torque about the wire is $F \times L$.

Turning the twist into a measurement

As the bar starts to rotate, the suspending wire begins to twist. A twisted wire pushes back — it exerts a restoring torque that tries to untwist itself. The more the wire is twisted, the stronger the restoring torque. In fact, for a fine wire twisted by a small angle $\theta$, the restoring torque is proportional to $\theta$:

\[\tau_{\text{restore}} = \tau \theta\]

Here $\tau$ (the Greek letter tau, without the $\theta$) is called the torsional constant or “restoring couple per unit angle of twist” of the wire. It has SI units of $\text{N m rad}^{-1}$.

The bar keeps rotating until the wire’s restoring torque exactly balances the gravitational torque. At that equilibrium angle $\theta$, the two are equal.

But how do we know $\tau$ for our particular wire? This is where the experiment is clever. The value of $\tau$ can be measured independently — by applying a known torque to the wire (for example, by attaching a small known mass at a known distance) and measuring the twist it produces. Once $\tau$ is known, it stays a property of the wire and can be reused for the gravitational measurement.

Real Incident / Discovery

Cavendish carried out his 1798 experiment inside a closed room, watching the deflection of the bar through a telescope from outside the room. He did this to avoid disturbing the delicate setup with air currents or with his own body heat. The gravitational torque he had to detect was so small that even ordinary temperature differences in the room would ruin the reading. His care paid off — his measurement was so accurate that it stood as the best value of $G$ for nearly a century.

Writing down the equation

There is one small subtlety. The gravitational law we wrote — $F = G m_1 m_2 / r^2$ — is really for two point masses. In Cavendish’s apparatus, both the small and the large spheres are extended objects. But we now know from Section 7.3 (the shell theorem) that a uniform sphere pulls an outside object exactly as if all its mass were concentrated at its centre. So we can safely apply the formula, with $r$ being the centre-to-centre distance.

Let $M$ and $m$ be the masses of a big sphere and its neighbouring small sphere, and let $d$ be the distance between their centres. Then

\[F = G\,\frac{M m}{d^2} \tag{7.6}\]

The gravitational torque on the bar is $F \times L$. Setting this equal to the wire’s restoring torque $\tau \theta$ at equilibrium:

\[G\,\frac{M m}{d^2}\,L = \tau \theta \tag{7.7}\]

Every quantity in this equation except $G$ is known: $M$ and $m$ from weighing the spheres, $d$ and $L$ by direct measurement, $\tau$ from the independent calibration, and $\theta$ from the deflection observed in the experiment. Solving Eq. (7.7) for $G$ gives Cavendish’s value.

Since 1798, the experiment has been refined many times using more sensitive equipment. The currently accepted value is

\[G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2} \tag{7.8}\]

This is one of the smallest fundamental constants in physics — the reason gravity is so weak on the scale of everyday objects. Yet it is the same $G$ that holds the planets in orbit around the Sun and holds galaxies together.

Quick Question

If $G$ is so small, why does gravity dominate our everyday life? We don’t feel electric forces from every nearby object, but we certainly feel our weight all the time.

Because we are always near a huge mass — the Earth itself, about $6 \times 10^{24}$ kg. Even with a tiny $G$, the product $G M_E m$ is large enough for every 1 kg mass to feel a force of about 9.8 N. Between two everyday objects, however, the same $G$ produces a force so small that we cannot feel it at all.

Side Note

The measurement of $G$ is deceptively hard — even today, different laboratories using different techniques give values of $G$ that differ from each other in the fifth decimal place, which is larger than the experimental uncertainties in either. This unresolved disagreement makes $G$ the least precisely known of all the fundamental constants of physics, despite being the first one to be measured in a laboratory.

Once $G$ is known, Newton’s law is a fully working tool. We can now apply it to answer questions that have been waiting in the background: how strong is Earth’s gravity at its surface? How does it change with altitude or depth? How fast must an object move to escape the Earth? Those are the questions the next sections take up.

7.5 Acceleration Due to Gravity of the Earth

We now have all the tools we need to compute the strength of Earth’s gravity at its own surface. Newton’s law tells us the force between two point masses. The shell theorem tells us how to treat a spherical body. And Cavendish’s experiment tells us the value of $G$. Putting these together, we can predict $g$ — the acceleration due to gravity on Earth — from first principles.

The Earth as a set of shells

Real planets are not point objects. The Earth has a radius of about $6400$ km. So we cannot use Newton’s law directly on “the Earth” as a single point. Instead, we imagine the Earth as being built from a very large number of thin, concentric spherical shells, one inside the other — the smallest at the very centre and the largest at the surface.

Figure to come

Fig. 7.5.1 – Cross-section of the Earth showing several nested concentric spherical shells drawn as concentric circles, with the outermost circle marked as Earth’s surface.

Consider a point object sitting somewhere outside the Earth — for example, an apple in the air, or a satellite in orbit. This point lies outside every one of the shells, because every shell is smaller than the Earth as a whole. Case 1 of the shell theorem (Section 7.3) then applies to each shell separately: each shell pulls the outside point as if the shell’s entire mass were concentrated at the shell’s centre.

But every shell has the same centre — the centre of the Earth. So when we add up the contributions of all the shells, we get a single result: the entire mass of the Earth pulls the outside object as if all that mass were concentrated at the Earth’s centre.

This is a hugely useful simplification. From the outside, no matter how complicated the internal layering of the Earth may be, gravitationally the Earth behaves like a point of the same total mass sitting at its centre.

Inside the Earth is different

For a point inside the Earth, the situation is more subtle. This is shown in Fig. 7.7, where a mass $m$ sits at a point $P$ located at a distance $r$ from the centre of the Earth (say, in a deep mine at a depth $d$ below the surface, so that $r = R_E - d$, where $R_E$ is the Earth’s radius).

Figure to come

Fig. 7.7 – Circle representing the Earth of radius R_E with centre O; a smaller dashed inner circle of radius r shown; point P on the inner circle with mass m; a depth d marked between P and the Earth’s surface; the mass of the inner sphere labelled M_r and total Earth mass labelled M_E.

Now the shells split into two groups:

Shells of radius greater than $r$ enclose the point $P$ in their interior. By Case 2 of the shell theorem, each such shell exerts zero net force on $P$. So all the material of the Earth farther from the centre than $P$ contributes nothing.
Shells of radius less than or equal to $r$ form a smaller sphere on whose surface the point $P$ lies. By Case 1, this smaller sphere pulls $P$ as if its whole mass were concentrated at the centre.

So the mass $m$ at $P$ is pulled only by the smaller sphere of radius $r$ that lies inside $P$. Let $M_r$ denote the mass of this inner sphere. The force on $m$ is then

\[F = \frac{G m\,M_r}{r^2} \tag{7.9}\]

Assuming uniform density

To take this one step further, let us assume for simplicity that the Earth is of uniform density $\rho$ throughout. Then the mass of a sphere depends only on its volume:

\[M_E = \frac{4\pi}{3} R_E^3\,\rho \qquad \text{(mass of the whole Earth)}\]

\[M_r = \frac{4\pi}{3} r^3\,\rho \qquad \text{(mass of the inner sphere of radius $r$)}\]

Substituting $M_r$ into Eq. (7.9):

\[F = G m \left(\frac{4\pi\rho}{3}\right) \frac{r^3}{r^2} = G m \left(\frac{M_E}{R_E^3}\right) r^3 \cdot \frac{1}{r^2}\]

Using $M_E = (4\pi/3) R_E^3 \rho$ to rewrite the density in terms of the Earth’s total mass, we obtain

\[F = \frac{G m\,M_E}{R_E^3}\,r \tag{7.10}\]

Notice that inside the Earth (with uniform density), the force is proportional to $r$ — not to $1/r^2$. It grows linearly as we move outward from the centre, becoming maximum at the surface, and then falls off as $1/r^2$ beyond the surface. We will return to this in the next section.

Acceleration at the surface

If the mass $m$ is right on the Earth’s surface, then $r = R_E$. Substituting this into Eq. (7.10):

\[F = \frac{G\,M_E\,m}{R_E^2} \tag{7.11}\]

The acceleration experienced by the mass $m$ is what we usually call $g$. From Newton’s second law, $F = m g$, so

\[g = \frac{F}{m} = \frac{G\,M_E}{R_E^2} \tag{7.12}\]

Look carefully at this formula. The mass of the falling object, $m$, has cancelled out completely. The value of $g$ depends only on Earth’s mass $M_E$, Earth’s radius $R_E$, and the universal constant $G$ — it does not depend on the mass of the object being pulled. This is exactly the fact Galileo demonstrated with his inclined-plane experiments: two objects of different masses, dropped from the same height, hit the ground together. Now we see it emerging as a direct consequence of Newton’s law of gravitation.

Curiosity Corner

Q. Galileo showed that a heavy stone and a light stone dropped together hit the ground at the same time. Why does the mass of the falling object not change how fast it falls? A. Because the mass of the falling body cancels out. Gravity pulls a body of mass $m$ with a force $G M_E m / R_E^2$, which is proportional to $m$, but by Newton’s second law the acceleration that force produces is the force divided by that same $m$. The result, $g = G M_E / R_E^2$, contains the Earth’s mass and radius but not the falling body’s mass at all. A heavy stone is pulled harder, yet it is exactly as much harder to accelerate, so every object falls with the same $g$.

“Cavendish weighed the Earth”

Equation (7.12) can be rearranged to solve for the Earth’s mass:

\[M_E = \frac{g\,R_E^2}{G}\]

$g$ is easily measured with a pendulum or by timing a falling body.
$R_E$ has been known since ancient times from geometrical astronomy (Eratosthenes measured it around 240 BCE).
$G$ was measured by Cavendish in his laboratory in 1798.

Once we plug these three numbers in, we get the mass of the entire Earth without ever setting foot on it. This is the reason for the popular saying about Cavendish’s experiment: “Cavendish weighed the Earth.” He didn’t weigh it directly, of course — but the moment he measured $G$, the mass of the Earth followed from a simple calculation.

Side Note

Using the modern values $g = 9.81\ \text{m s}^{-2}$, $R_E = 6.37 \times 10^6$ m, and $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$, Eq. (7.12) gives $M_E \approx 5.97 \times 10^{24}$ kg. This is one of the most important numbers in astronomy — it is used every time we compute the orbit of a satellite, the trajectory of a space mission, or the strength of Earth’s tides.

We have found $g$ at the Earth’s surface. But how does $g$ behave at other heights, or below the surface? That is the subject of the next section.

7.6 Acceleration Due to Gravity Below and Above the Surface of Earth

In the last section we found that at the Earth’s surface, the acceleration due to gravity is $g = G M_E / R_E^2$. This raises an immediate question: what if we don’t stay on the surface? What if we climb a tall mountain, or fly in an aeroplane, or go down deep into a mine? Does $g$ stay the same, get stronger, or get weaker?

The answer is not obvious, because two different effects fight against each other. Going up increases our distance from the centre of the Earth, so gravity should weaken. Going down brings us closer to the centre, so gravity should strengthen — but at the same time, part of the Earth is now above us, and by the shell theorem that part contributes nothing. We need to carefully work out each case.

Gravity above the surface

Consider a small mass $m$ placed at a height $h$ above the Earth’s surface, as shown in Fig. 7.8(a). Its distance from the centre of the Earth is now $R_E + h$, not $R_E$. Since it is still outside the whole Earth, the shell theorem lets us treat the entire mass of the Earth as concentrated at the centre.

Figure to come

Fig. 7.8(a) – Circle representing the Earth of radius R_E with centre marked; a small mass shown at height h above the surface, connected by a dashed line to the centre; label “Earth’s surface” on the circle.

Applying Newton’s law of gravitation, the magnitude of the force on the mass is

\[F(h) = \frac{G M_E m}{(R_E + h)^2} \tag{7.13}\]

The acceleration due to gravity at height $h$, which we denote by $g(h)$, is $F(h)/m$:

\[g(h) = \frac{F(h)}{m} = \frac{G M_E}{(R_E + h)^2} \tag{7.14}\]

Comparing with the surface value $g = G M_E / R_E^2$, we see $g(h) < g$ — as expected, gravity weakens with height. But by how much?

For heights that are small compared to the Earth’s radius (that is, $h \ll R_E$, which covers everything from tall buildings to aircraft altitudes and even low satellite orbits), we can approximate Eq. (7.14). Rewrite it as

\[g(h) = \frac{G M_E}{R_E^2 (1 + h/R_E)^2} = g\left(1 + \frac{h}{R_E}\right)^{-2}\]

For $h/R_E \ll 1$, the binomial expansion gives $(1 + x)^{-2} \approx 1 - 2x$ when $x$ is small. Applying this here:

\[g(h) \approx g\left(1 - \frac{2h}{R_E}\right) \tag{7.15}\]

Equation (7.15) is neat and easy to use. It tells us that for small heights, $g$ decreases by a factor of $(1 - 2h/R_E)$ — that is, roughly twice the fractional height $h/R_E$.

Real-World Application

Satellites in low Earth orbit (roughly 400 km above the surface, like the International Space Station) do not experience “zero gravity” in the way films often show. Using Eq. (7.15) with $h = 400$ km, the value of $g$ up there is only about $12\%$ smaller than at the ground. The reason astronauts feel weightless is not that gravity is gone — it is that both the astronaut and the spacecraft are in continuous free-fall around the Earth. This is exactly the point Newton first suggested with his cannonball thought experiment.

Gravity below the surface

Now consider going in the other direction — into a deep mine, at a depth $d$ below the Earth’s surface (Fig. 7.8b). The mass $m$ is now at a distance $R_E - d$ from the centre of the Earth.

Figure to come

Fig. 7.8(b) – Circle representing the Earth of radius R_E with centre marked; a dashed inner circle of radius (R_E − d); depth d shown from the outer surface down to the inner circle; mass m marked at that depth; labels M_E for total Earth mass, M_s for mass of the inner sphere.

Imagine splitting the Earth into two pieces:

A smaller inner sphere of radius $(R_E - d)$, containing the point where $m$ sits on its surface.
A spherical shell of thickness $d$, extending from radius $(R_E - d)$ up to $R_E$, containing the point where $m$ sits in its interior.

By Case 2 of the shell theorem, the outer shell of thickness $d$ exerts zero net force on $m$. So the mass in the outer shell contributes nothing, no matter how thick it is.

By Case 1, the smaller inner sphere of radius $(R_E - d)$ acts on $m$ as if all its mass were concentrated at its centre. Let $M_s$ be the mass of this smaller sphere. Then the force on $m$ at depth $d$ is

\[F(d) = \frac{G M_s\, m}{(R_E - d)^2} \tag{7.17}\]

Assuming the Earth has uniform density, the mass of a sphere is proportional to the cube of its radius (as we saw in Section 7.5):

\[\frac{M_s}{M_E} = \frac{(R_E - d)^3}{R_E^3} \tag{7.16}\]

Substituting $M_s = M_E (R_E - d)^3 / R_E^3$ into Eq. (7.17):

\[F(d) = \frac{G M_E m (R_E - d)}{R_E^3} \tag{7.18}\]

The acceleration due to gravity at depth $d$ is then

\[g(d) = \frac{F(d)}{m} = \frac{G M_E}{R_E^3}(R_E - d) = g\,\frac{R_E - d}{R_E} = g\left(1 - \frac{d}{R_E}\right) \tag{7.19}\]

Look carefully at Eqs. (7.15) and (7.19) side by side. Above the surface, $g$ decreases by a factor $(1 - 2h/R_E)$. Below the surface, $g$ decreases by a factor $(1 - d/R_E)$ — half as fast, per unit distance, but still decreasing. Whether we go up or down, $g$ gets smaller.

The peak at the surface

The remarkable conclusion is:

The acceleration due to gravity is maximum at the Earth’s surface, and decreases whether you go up or down.

Curiosity Corner

Q. If you dig a deep mine and go far below the Earth’s surface, does gravity get stronger, weaker, or stay the same — and why? A. It gets weaker. By the shell theorem, the shells of the Earth lying outside your position exert no net force on you at all, so only the sphere beneath your feet pulls you. Assuming uniform density, that inner sphere gives $g(d) = g\,(1 - d/R_E)$, which falls steadily as the depth $d$ increases and reaches zero at the centre. Gravity therefore has its greatest value at the surface, decreasing both as you go up and as you go down.

At the centre of the Earth ($d = R_E$), Eq. (7.19) gives $g(d) = 0$. This makes physical sense — at the very centre, mass is pulling equally in every direction, so the net gravitational force on an object placed there is zero.

Real Incident / Discovery

In the 1850s, the British Astronomer Royal George Biddell Airy attempted to measure gravity at the bottom of a coal mine at Harton Colliery in England. He used pendulums to compare the value of $g$ at the surface and at a depth of over 380 metres underground. The experiment was aimed at estimating the mean density of the Earth from the way $g$ changed with depth. Careful pendulum-based measurements like Airy’s showed clearly that gravity does change with depth, just as Eq. (7.19) predicts.

Quick Question

Real measurements sometimes show gravity increasing slightly a little way below the surface, not decreasing as Eq. (7.19) predicts. Why?

Equation (7.19) assumes the Earth has uniform density. The real Earth does not — the core is much denser than the mantle, and the crust is less dense than both. Near the surface, going down means we are leaving a low-density crust and getting closer to the dense core, which briefly increases $g$ before it starts to fall off toward zero at the centre.

Numerical 7.5

A satellite orbits the Earth at a height of $200$ km above the surface. Using $R_E = 6400$ km and $g = 9.8\ \text{m s}^{-2}$ at the surface, use the approximation in Eq. (7.15) to estimate the value of $g$ at the satellite’s altitude.

Numerical 7.6

A body weighs $500$ N at the Earth’s surface. Assuming the Earth is a sphere of uniform density with $R_E = 6400$ km, find its weight at a depth of $1600$ km below the surface.

We have now understood how gravity behaves on and near the Earth’s surface, both above and below. But so far we have been talking only about forces. In the next section we shift to a different but equally powerful language — the language of potential energy — which will let us tackle questions like escape speed and satellite motion.

7.7 Gravitational Potential Energy

So far we have been describing gravity in terms of force. There is a second, equally powerful way of looking at the same physics — in terms of energy. In many situations, especially involving escape speeds and satellite motion, the energy approach is much easier to work with than the force approach. In this section we set up the idea of gravitational potential energy carefully.

A quick recap: what “potential energy” means

You have already met potential energy in Class 11 mechanics. Very briefly: when a force does work on an object as the object moves from one position to another, the object’s energy state changes. If the force is a conservative force — one for which the work done depends only on the two endpoints, not on the path taken between them — we can associate an “energy stored in position” with each point in space. This is potential energy.

The definition is:

The change in potential energy of an object between two points equals the amount of work done on the object by the conservative force as the object moves between those points.

Gravity, we shall see, is a conservative force. So we can define a gravitational potential energy for it, and use energy conservation to solve problems that would be very awkward using forces alone.

Near the Earth’s surface: the familiar $mgh$

Let us start close to home. For a point very close to the Earth’s surface, the height above the surface is much smaller than the Earth’s radius. In this regime, the gravitational force is practically constant, of magnitude $mg$, directed straight down toward the centre of the Earth.

Consider two points along a vertical line: point 1 at a height $h_1$ above the surface, and point 2 directly above it at height $h_2$. Lift a mass $m$ from position 1 to position 2. The work $W_{12}$ done against gravity is force times displacement:

\[W_{12} = \text{Force} \times \text{displacement} = mg(h_2 - h_1) \tag{7.20}\]

Now let us define a potential energy function $W(h)$ at a height $h$ above the surface by the relation

\[W(h) = mgh + W_0 \tag{7.21}\]

where $W_0$ is some constant. If we take the difference of $W(h)$ at the two positions, the constant $W_0$ cancels:

\[W_{12} = W(h_2) - W(h_1) \tag{7.22}\]

So the work done in lifting the object is exactly the difference in its potential energy between the two points. This confirms that $W(h) = mgh + W_0$ is a valid potential energy for gravity near the surface.

If we set $h = 0$ in Eq. (7.21), we get $W(h = 0) = W_0$. So $W_0$ is the potential energy of a mass at the surface of the Earth. We can pick any value for $W_0$ we like — usually zero — and only the differences in potential energy between two points will have physical meaning.

Far from the Earth: gravity isn’t constant

The formula $W = mgh$ works nicely if you are lifting a book from the floor to a shelf. But it fails badly if you are launching a rocket to the Moon. Why? Because the assumption of constant gravitational force breaks down. As we saw in Section 7.6, gravity weakens with distance from the Earth.

At an arbitrary distance $r$ from the Earth’s centre (with $r > R_E$, so we are outside the Earth), the gravitational force on a particle of mass $m$ is

\[F = \frac{G M_E m}{r^2} \tag{7.23}\]

directed toward the centre of the Earth. Here $M_E$ is the Earth’s mass. Now let us lift the particle from an initial distance $r = r_1$ to a final distance $r = r_2$ (with $r_2 > r_1$), moving along a vertical path.

The work done against gravity is no longer just “force times distance” because the force varies with $r$. We must integrate. In place of Eq. (7.20), we get

\[W_{12} = \int_{r_1}^{r_2} \frac{G M_E m}{r^2}\,dr = -G M_E m \left(\frac{1}{r_2} - \frac{1}{r_1}\right) \tag{7.24}\]

Following the same idea as before, we can define a potential energy function $W(r)$ at distance $r$ by

\[W(r) = -\frac{G M_E m}{r} + W_1 \tag{7.25}\]

which is valid for $r > R$ (outside the Earth). Here $W_1$ is again an arbitrary constant. Taking the difference between two positions once more gives $W_{12} = W(r_2) - W(r_1)$, matching Eq. (7.24). So Eq. (7.25) is a valid gravitational potential energy.

The natural reference: infinity

Where should we set the reference? For the near-surface case, setting $W_0 = 0$ at the surface was natural. For gravity extending out into space, there is a much cleaner choice: set the potential energy to zero at infinity.

Substitute $r = \infty$ in Eq. (7.25): the first term becomes zero, and we get $W(\infty) = W_1$. So $W_1$ is the potential energy at infinity. Choosing the convention $W_1 = 0$ gives us

\[W(r) = -\frac{G M_E m}{r}\]

Two features of this formula deserve attention:

The potential energy is negative. This is not a mistake or an oddity. It reflects our chosen convention that PE is zero at infinity. Since gravity pulls the mass toward the Earth, dropping it in from infinity releases energy — so its PE at finite distance must be lower (i.e., more negative) than at infinity.
The potential energy at a point equals the work done in bringing a mass from infinity to that point (against gravity, done externally without change of kinetic energy).

The general two-particle formula

The result we have just derived is really about two interacting masses — the Earth and the object. It generalises immediately to any two point masses. For two particles of masses $m_1$ and $m_2$, separated by a distance $r$, the gravitational potential energy of the pair is

\[V = -\frac{G m_1 m_2}{r} \qquad \text{(taking $V = 0$ as $r \to \infty$)}\]

This is a mutual property of the pair — you cannot attach half of it to $m_1$ and half to $m_2$. It is the energy stored in the system by virtue of their gravitational interaction.

The concept of gravitational potential (energy per unit test mass) is closely related. The gravitational potential at a point in space, due to some mass distribution, is the gravitational potential energy that a unit mass would have if placed at that point. From the expression above, the potential due to a single mass $m_1$ at distance $r$ from it is $-Gm_1/r$ (per unit test mass).

Many particles: superposition

What if we have not just two masses but many? Suppose we have masses $m_1, m_2, m_3, \dots$ distributed at various positions. The total gravitational PE of this whole system is not “the PE of $m_1$” plus “the PE of $m_2$” separately — that would double-count. Instead, we sum the pairwise PE over every distinct pair of particles:

\[U_{\text{total}} = \sum_{\text{all pairs } i<j} \left(-\frac{G m_i m_j}{r_{ij}}\right)\]

where $r_{ij}$ is the distance between the $i$-th and $j$-th particles. This is the superposition principle applied to gravitational PE — the same principle we used earlier for forces.

The following solved example shows how to add up pairwise PEs for a small set of particles.

Solved Example 7.3

Find the gravitational potential energy of a system of four particles, each of mass $m$, placed at the vertices of a square of side $l$. Also obtain the gravitational potential at the centre of the square.

[Diagram: Fig. 7.9 – Square of side l with a mass m at each of the four corners; two diagonals drawn inside the square, showing the diagonal length $\sqrt{2}\,l$; the centre of the square marked.]

Answer

With four particles, there are $\binom{4}{2} = 6$ pairs. Look carefully at the geometry (Fig. 7.9):

Four pairs are formed by particles at the ends of the sides of the square — each pair is at distance $l$.
Two pairs are formed by particles at opposite corners of the square — each pair is at the diagonal distance $\sqrt{2}\,l$.

The gravitational PE of each pair is $-G m^2 / r$, where $r$ is the pair’s separation. Summing over all six pairs:

\[W(r) = -4\,\frac{G m^2}{l} - 2\,\frac{G m^2}{\sqrt{2}\,l}\]

\[= -\frac{2 G m^2}{l}\left(2 + \frac{1}{\sqrt{2}}\right) = -5.41\,\frac{G m^2}{l}\]

For the gravitational potential at the centre of the square, note that all four particles are at the same distance from the centre. Each side of the square has length $l$, so the half-diagonal (centre-to-corner distance) is $r = \sqrt{2}\,l / 2$. The gravitational potential due to a single point mass $m$ at distance $r$ is $-G m / r$. Adding the potentials due to all four masses (potentials are scalars, so we just add them):

\[U(r) = 4 \times \left(-\frac{G m}{\sqrt{2}\,l / 2}\right) = -4\sqrt{2}\,\frac{G m}{l}\]

Numerical 7.7

Three point masses, each of mass $2.0$ kg, are placed at the vertices of an equilateral triangle of side $0.50$ m. Find the total gravitational potential energy of the system. Take $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$.

Numerical 7.8

A body of mass $10$ kg is lifted from the Earth’s surface to a height equal to the Earth’s radius. Using $G$, $M_E$, and $R_E$ from the chapter, find the change in the body’s gravitational potential energy.

Gravitational potential energy is the key ingredient for the next question we tackle: how fast must an object move to leave the Earth for good? That is the topic of Section 7.8.

7.8 Escape Speed

Throw a stone straight up. It rises for a moment, slows down, and comes back to your hand. Throw it a little harder — it goes higher, but still comes back. Now imagine using a machine (a cannon, a rocket) that can throw an object with an arbitrarily large starting speed. The higher the launch speed, the higher the object rises before returning. A natural question forms in the mind:

Is there a critical launch speed such that the object never comes back?

This critical speed is called the escape speed. In this section we compute it, and see why it depends only on the properties of the planet you are launching from — not on the object being launched.

Setting up with energy conservation

The energy method (Section 7.7) is far easier here than the force method. Total mechanical energy — kinetic plus potential — is conserved for an object moving under gravity alone (with no air resistance, and no thrust after launch).

Suppose we throw an object of mass $m$ straight up from a point at a distance $(h + R_E)$ from the Earth’s centre — that is, at a height $h$ above the Earth’s surface. Its initial speed is $V_i$. We ask: what is the smallest value of $V_i$ for which the object never returns?

If the object were to “escape,” it would eventually reach an arbitrarily large distance — call this “infinity” — with some final speed $V_f \geq 0$. Using our convention that gravitational potential energy is zero at infinity, the total energy at infinity is purely kinetic (plus the reference constant $W_1$, which we conventionally take as zero):

\[E(\infty) = W_1 + \frac{1}{2} m V_f^2 \tag{7.26}\]

Initially, the object has kinetic energy $\tfrac{1}{2} m V_i^2$ and gravitational PE $-G m M_E / (h + R_E)$, so

\[E(h + R_E) = \frac{1}{2} m V_i^2 - \frac{G m M_E}{(h + R_E)} + W_1 \tag{7.27}\]

The condition for escape

By energy conservation, Eq. (7.26) and Eq. (7.27) must be equal. Setting them equal and rearranging:

\[\frac{m V_i^2}{2} - \frac{G m M_E}{(h + R_E)} = \frac{m V_f^2}{2} \tag{7.28}\]

Now look at the right side of Eq. (7.28). It is $\tfrac{1}{2} m V_f^2$, which is a positive quantity (or zero at best — the smallest possible KE at infinity is zero, when the object just barely reaches infinity with no speed left). The right side cannot be negative.

Therefore the left side cannot be negative either. That gives a condition:

\[\frac{m V_i^2}{2} - \frac{G m M_E}{(h + R_E)} \geq 0 \tag{7.29}\]

Equation (7.29) is the escape condition. It says: for the object to escape the Earth’s gravity, the initial kinetic energy must be at least enough to cancel out the (magnitude of the) initial gravitational potential energy.

The minimum value of $V_i$ occurs when the left side is exactly zero — the object reaches infinity with $V_f = 0$, using every last bit of its kinetic energy to climb out of the Earth’s gravitational “well.” From Eq. (7.29) with equality:

\[\frac{1}{2} m (V_i^2)_{\min} = \frac{G m M_E}{h + R_E} \tag{7.30}\]

Solving for $(V_i)_{\min}$:

\[(V_i)_{\min} = \sqrt{\frac{2 G M_E}{h + R_E}}\]

Launching from the Earth’s surface

For the special case where we launch from the surface itself, $h = 0$, and we get

\[(V_i)_{\min} = \sqrt{\frac{2 G M_E}{R_E}} \tag{7.31}\]

Using the useful relation $g = G M_E / R_E^2$ from Section 7.5, we can rewrite this in a form that involves only $g$ and $R_E$:

\[(V_i)_{\min} = \sqrt{2 g R_E} \tag{7.32}\]

Plugging in numerical values ($g \approx 9.8\ \text{m s}^{-2}$, $R_E \approx 6400$ km), we get

\[(V_i)_{\min} \approx 11.2\ \text{km s}^{-1}\]

This is the escape speed from the Earth’s surface. Any object launched vertically upward from the ground with a speed of at least $11.2$ km/s will never come back (assuming no air resistance and neglecting the pull of the Sun and other planets).

The term “escape velocity” is often used loosely — but strictly speaking, the direction of launch is not important (we could have launched at an angle, and as long as the object has this much speed and does not run into the ground, it will still escape). What matters is the speed, so “escape speed” is the more precise name.

Why $m$ dropped out — and what it means

Look carefully at Eq. (7.31). The mass $m$ of the object being launched has completely disappeared. The escape speed depends only on $G$, $M_E$, and $R_E$ — properties of the planet you are leaving, not of the thing you are launching.

This means the same critical speed applies to a marble, a bullet, a car, and a spacecraft. If you can throw a pebble at 11.2 km/s from the ground, it escapes the Earth just as surely as a rocket does. A much heavier spacecraft needs a much more powerful engine to accelerate it up to that same speed, but the target speed itself is the same.

Curiosity Corner

Q. A rocket needs a very specific minimum speed to escape the Earth forever. What decides that speed, and why is it the same for a small pebble and a large spacecraft? A. Escape speed is fixed by the mass and radius of the body being escaped from: $v_e = \sqrt{2 G M_E / R_E} = \sqrt{2 g R_E}$, about $11.2\ \text{km s}^{-1}$ for the Earth. It comes from demanding that the projectile’s total energy be at least zero, so that it can just reach infinity with no speed left over. The projectile’s own mass $m$ appears in both the kinetic and the potential energy terms and cancels when the condition is written down — which is why a pebble and a spacecraft need exactly the same launch speed.

Applying the same formula elsewhere

Equation (7.32) is written for the Earth, but the derivation used no property that is unique to the Earth. The same expression applies to any planet or moon — just replace $g$ with the value of gravitational acceleration on the surface of that body, and replace $R_E$ with the radius of that body.

For example, the Moon has a much smaller surface gravity and radius than the Earth. Both are smaller, so the escape speed from the Moon is smaller. It works out to about $2.3\ \text{km s}^{-1}$ — roughly one-fifth of Earth’s escape speed.

Real-World Application

The low escape speed of the Moon is the reason it has essentially no atmosphere. Gas molecules at the Moon’s surface temperature move with a wide range of speeds. A significant fraction of them travel faster than $2.3\ \text{km s}^{-1}$, which means those molecules simply fly off into space rather than being held back by the Moon’s weak gravity. Over billions of years, any atmosphere the Moon may once have had has leaked away. Earth’s higher escape speed — $11.2$ km/s — is fast enough compared to typical molecular speeds that our atmosphere is retained.

Real-World Application

The Voyager 1 and 2 spacecraft, launched by NASA in 1977, are the fastest human-made objects to leave the Solar System. They achieved not just Earth’s escape speed but also the higher escape speed from the Sun’s gravity at Earth’s orbit (about $42$ km/s), largely by using slingshot manoeuvres past Jupiter and Saturn. Voyager 1 has now crossed into interstellar space — a real demonstration that the same escape-speed logic works at every gravitational scale.

A solved example with two attracting bodies

Escape speed calculations become more interesting when there is more than one massive body pulling on the projectile. Here is a classic case.

Solved Example 7.4

Two uniform solid spheres of equal radii $R$, but masses $M$ and $4M$, are held fixed with their centres separated by a distance $6R$, as shown in Fig. 7.10. A projectile of mass $m$ is launched from the surface of the sphere of mass $M$, directly toward the centre of the second sphere. Find an expression for the minimum speed $v$ of the projectile so that it just reaches the surface of the second sphere.

[Diagram: Fig. 7.10 – Two circles side by side: on the left, a sphere of mass M with radius R, centre O; on the right, a sphere of mass 4M with radius R, centre C; the centres are 6R apart; a small projectile of mass m shown at the surface of M with velocity v pointing to the right; a point N marked between the two spheres (the “neutral point”) where the two gravitational pulls exactly cancel.]

Answer

The projectile is pulled left by the mass $M$ and right by the mass $4M$. Between the two spheres, there is a special point — the neutral point $N$ — where the two gravitational pulls exactly cancel. If we can just get the projectile to $N$ with any tiny remaining forward speed, the pull of the heavier sphere $4M$ from that point onward will do the rest of the work, dragging it inward toward the surface of the second sphere.

So the problem reduces to: give the projectile just enough speed at the surface of $M$ so that it reaches $N$ with zero speed.

Locating $N$: Let $ON = r$, so that the distance from $N$ to the centre of $4M$ is $(6R - r)$. Setting the two pulls on a small test mass equal:

\[\frac{G M m}{r^2} = \frac{4 G M m}{(6R - r)^2}\]

\[(6R - r)^2 = 4 r^2\]

\[6R - r = \pm 2r\]

\[r = 2R \quad \text{or} \quad r = -6R\]

The negative root does not concern us (it lies outside the physical region between the spheres). So $ON = r = 2R$.

Applying energy conservation: The mechanical energy of the projectile at the surface of $M$ is

\[E_i = \frac{1}{2} m v^2 - \frac{G M m}{R} - \frac{4 G M m}{5R}\]

The two negative terms are the gravitational PE due to $M$ (at distance $R$, the surface) and due to $4M$ (at distance $6R - R = 5R$).

At the neutral point $N$, the projectile’s speed is (just barely) zero, so its mechanical energy is purely potential:

\[E_N = -\frac{G M m}{2R} - \frac{4 G M m}{4R}\]

Here the two negative terms are the PE due to $M$ (at distance $r = 2R$) and due to $4M$ (at distance $6R - 2R = 4R$).

Setting $E_i = E_N$ (energy conservation):

\[\frac{1}{2} v^2 - \frac{G M}{R} - \frac{4 G M}{5R} = -\frac{G M}{2R} - \frac{G M}{R}\]

Rearranging and simplifying:

\[v^2 = \frac{2 G M}{R}\left(\frac{4}{5} - \frac{1}{2}\right)\]

\[v = \left(\frac{3 G M}{5 R}\right)^{1/2}\]

Notice that the projectile’s speed is zero at $N$, but non-zero when it reaches the surface of the heavier sphere $4M$ — as it falls in, it picks up kinetic energy from the gravitational field of the second sphere. Calculating that final impact speed is left as an exercise for the student.

Numerical 7.9

The mass of Mars is about $0.11$ times the mass of the Earth, and its radius is about $0.53$ times the radius of the Earth. Estimate the escape speed from the surface of Mars, given that the escape speed from Earth’s surface is $11.2\ \text{km s}^{-1}$.

Numerical 7.10

A satellite is launched from a point at a height of $R_E$ above the Earth’s surface (i.e., at a distance of $2R_E$ from the Earth’s centre). Find the minimum launch speed required for the satellite to escape the Earth’s gravitational field.

The escape speed idea leads naturally into the opposite question: what if we don’t want an object to escape, but instead want it to keep going around the Earth forever? That is exactly what a satellite does, and it is the subject of Section 7.9.

7.9 Earth Satellites

An Earth satellite is any object that revolves around the Earth in a stable orbit. Some satellites are natural — the Moon is the only natural satellite of the Earth. Others are artificial — built and launched by humans. In this section we work out the physics of satellite orbits and derive the relations that connect their orbital speed, radius, and time period.

Because a satellite is bound to the Earth by gravity in exactly the same way a planet is bound to the Sun, Kepler’s three laws apply to satellites too. Their orbits, like those of planets, may be circular or elliptical. The Moon’s orbit around the Earth is nearly circular, with a time period of about $27.3$ days — which, remarkably, is also roughly equal to the time the Moon takes to spin once on its own axis. That is why the Moon always shows us the same face.

Real Incident / Discovery

The age of artificial Earth satellites began on 4 October 1957, when the Soviet Union launched Sputnik 1 — a metal sphere the size of a beach ball, carrying a radio transmitter. As it circled the Earth once every $\sim 96$ minutes, its faint beeps could be picked up by radio operators across the world. India entered the space age in 1975 with the launch of Aryabhata, its first satellite, built by the Indian Space Research Organisation (ISRO). Since then, thousands of artificial satellites have been placed in orbit for telecommunication, weather forecasting, remote sensing, navigation (like the Indian NavIC system), and scientific research.

Deriving the orbital speed

Consider a satellite of mass $m$ moving in a circular orbit at a height $h$ above the Earth’s surface. Its distance from the Earth’s centre is $(R_E + h)$, where $R_E$ is the Earth’s radius. Let $V$ be its speed.

Figure to come

Fig. 7.9.1 – Earth shown as a circle with centre O; a satellite of mass m at height h above the surface moving on a circular orbit around the Earth; velocity vector V drawn tangent to the orbit; gravitational force F drawn from the satellite toward Earth’s centre.

For any object moving in a circle of radius $(R_E + h)$ at speed $V$, Class 11 mechanics tells us that a centripetal force is needed, pointing toward the centre:

\[F(\text{centripetal}) = \frac{m V^2}{(R_E + h)} \tag{7.33}\]

This centripetal force is not applied by any string or engine — the satellite is in free fall, and the gravitational pull of the Earth provides it. Using Newton’s law of gravitation for a satellite outside the Earth:

\[F(\text{gravitation}) = \frac{G m M_E}{(R_E + h)^2} \tag{7.34}\]

where $M_E$ is the mass of the Earth. Since the gravitational force is the centripetal force, we equate the right-hand sides of Eq. (7.33) and Eq. (7.34). The satellite mass $m$ cancels, giving

\[V^2 = \frac{G M_E}{(R_E + h)} \tag{7.35}\]

Notice again that the mass of the satellite does not appear on the right side. A small science satellite and a large communication satellite at the same orbital height must move with the same speed. What sets the orbital speed is only the Earth’s mass and the orbital radius.

Eq. (7.35) also shows that $V$ decreases as the orbital height $h$ increases. Higher orbits are slower orbits — think of the Moon, which is very far from Earth and moves at only about $1$ km/s, while the International Space Station (much closer) moves at nearly $8$ km/s.

For the special case of a satellite skimming just above the Earth’s surface, $h = 0$:

\[V^2 (h = 0) = \frac{G M_E}{R_E} = g R_E \tag{7.36}\]

where we used $g = G M_E / R_E^2$ from Section 7.5. Numerically this gives $V \approx 7.9$ km/s — the theoretical minimum orbital speed near Earth. (In practice, satellites are placed a few hundred kilometres up to avoid atmospheric drag, so their orbital speed is slightly less.)

Quick Question

The Moon is being pulled toward the Earth by gravity all the time. Why does it not simply fall in and crash?

It is falling — but sideways. At every instant, gravity pulls the Moon inward toward Earth’s centre, so its velocity vector bends slightly downward. But because the Moon is also moving horizontally at just the right speed, the amount it “falls” is exactly the amount the Earth’s surface curves away underneath. The Moon keeps missing the Earth as it falls, and traces a closed orbit. This is what an orbit is: continuous free-fall around a body.

Curiosity Corner

Q. The Moon is constantly pulled toward the Earth by gravity. Why then does it not simply crash into us? A. Because it is also moving sideways. The Moon is indeed always falling toward the Earth — gravity supplies exactly the centripetal force its circular path requires — but it has enough tangential speed that it keeps missing, curving around the Earth instead of striking it. Being pulled toward a centre does not mean moving toward that centre: the pull continually bends the velocity without ever giving it a component along the radius, which is precisely what a stable orbit is.

Time period of a satellite

In one full orbit, a satellite covers a circumference $2\pi(R_E + h)$ at speed $V$. Its time period $T$ (the time for one revolution) is therefore

\[T = \frac{2\pi(R_E + h)}{V}\]

Substituting for $V$ from Eq. (7.35):

\[T = \frac{2\pi(R_E + h)^{3/2}}{\sqrt{G M_E}} \tag{7.37}\]

Squaring both sides of Eq. (7.37):

\[T^2 = k\,(R_E + h)^3 \qquad \text{where}\quad k = \frac{4\pi^2}{G M_E} \tag{7.38}\]

Look at Eq. (7.38) carefully. It says $T^2 \propto (R_E + h)^3$ — the square of the time period is proportional to the cube of the orbital radius. This is exactly Kepler’s third law (Section 7.2) applied to Earth satellites, with the constant $k$ depending only on the mass of the central body (here, the Earth).

Curiosity Corner

Q. Satellites orbit the Earth without any fuel keeping them up. What is holding them in orbit, and what decides how long they take to go around once? A. Nothing holds a satellite up; gravity holds it in. The Earth’s gravitational pull supplies precisely the centripetal force needed for the circular path, so no fuel is required to maintain the orbit. Equating the two gives the orbital speed $v = \sqrt{G M_E / (R_E + h)}$, and the period follows as $T = 2\pi (R_E + h)^{3/2} / \sqrt{G M_E}$. The period therefore depends only on the size of the orbit and the Earth’s mass — not on the satellite’s own mass — so a higher orbit means a slower satellite and a longer time to go around once.

For a satellite orbiting very close to the Earth’s surface, $h$ is much smaller than $R_E$, and we can neglect $h$ in comparison to $R_E$. The period then reduces to a constant $T_0$:

\[T_0 = 2\pi \sqrt{\frac{R_E}{g}} \tag{7.39}\]

Plugging in $g \approx 9.8\ \text{m s}^{-2}$ and $R_E \approx 6.4 \times 10^6$ m:

\[T_0 = 2\pi \sqrt{\frac{6.4 \times 10^6}{9.8}}\ \text{s} \approx 85\ \text{minutes}\]

So the fastest possible Earth orbit lasts about $85$ minutes. Any satellite orbiting higher up takes longer.

Real-World Application

A special orbital height — called the geostationary orbit — is about $36{,}000$ km above the equator. Using Eq. (7.38), a satellite at this height has a time period of exactly $24$ hours. If it moves in the direction of Earth’s rotation, from the ground it appears to hang motionless in the sky. This is why communication and weather satellites (such as India’s INSAT and GSAT series) are placed in geostationary orbit — dish antennas on the ground can point at them permanently without tracking. The idea of using such an orbit for global communication was first suggested in 1945 by the British writer and scientist Arthur C. Clarke.

Sample calculations

Three worked examples in the original NCERT text apply the ideas of this section to real astronomical data. Here they are, one by one.

Solved Example 7.5

The planet Mars has two moons, Phobos and Deimos.

Phobos has a period of $7$ hours $39$ minutes and an orbital radius of $9.4 \times 10^3$ km. Calculate the mass of Mars.
Assume that the Earth and Mars move in circular orbits around the Sun, with the Martian orbit being $1.52$ times the orbital radius of the Earth. Find the length of the Martian year in days.

Answer

(i) Kepler’s law of periods, Eq. (7.38), applies to any body orbiting a central mass. For Phobos orbiting Mars, we simply replace $M_E$ with the mass of Mars, $M_m$:

\[T^2 = \frac{4\pi^2}{G M_m}\,R^3\]

Solving for $M_m$:

\[M_m = \frac{4\pi^2 R^3}{G T^2}\]

Plugging in $R = 9.4 \times 10^6$ m and $T = 7\text{h } 39\text{min} = (7 \times 60 + 39)\text{min} = 459 \text{ min} = 459 \times 60 \text{ s}$:

\[M_m = \frac{4 \times (3.14)^2 \times (9.4)^3 \times 10^{18}}{6.67 \times 10^{-11} \times (459 \times 60)^2}\]

\[M_m = \frac{4 \times (3.14)^2 \times (9.4)^3 \times 10^{18}}{6.67 \times (4.59 \times 6)^2 \times 10^{-5}}\]

\[M_m = 6.48 \times 10^{23}\ \text{kg}\]

(ii) Kepler’s third law relates the periods and orbital radii of two planets around the same Sun:

\[\frac{T_M^2}{T_E^2} = \frac{R_{MS}^3}{R_{ES}^3}\]

where $R_{MS}$ is the Mars–Sun distance and $R_{ES}$ is the Earth–Sun distance. Taking square roots after substituting $R_{MS} = 1.52\, R_{ES}$:

\[T_M = (1.52)^{3/2} \times 365 = 684\ \text{days}\]

So a Martian year is roughly $684$ Earth days long.

Side Note

The orbits of all planets except Mercury and Mars are very close to circular. For Earth, the ratio of the semi-minor to semi-major axis is $b/a = 0.99986$ — the orbit is an ellipse, but so close to a circle that the difference is hard to see on a scale drawing.

Solved Example 7.6

“Weighing the Earth” — You are given the following data: $g = 9.81\ \text{m s}^{-2}$, $R_E = 6.37 \times 10^6$ m, the distance to the Moon $R = 3.84 \times 10^8$ m, and the Moon’s orbital period is $27.3$ days. Obtain the mass of the Earth $M_E$ in two different ways.

Answer

Method 1 — from Eq. (7.12). The surface-gravity formula rearranges to give $M_E$ directly:

\[M_E = \frac{g R_E^2}{G} = \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = 5.97 \times 10^{24}\ \text{kg}\]

Method 2 — from Kepler’s third law. The Moon is a satellite of the Earth, so Eq. (7.38) applies:

\[T^2 = \frac{4\pi^2 R^3}{G M_E}\]

Solving for $M_E$:

\[M_E = \frac{4\pi^2 R^3}{G T^2}\]

Substituting $R = 3.84 \times 10^8$ m and $T = 27.3 \times 24 \times 60 \times 60$ s:

\[M_E = \frac{4 \times 3.14 \times 3.14 \times (3.84)^3 \times 10^{24}}{6.67 \times 10^{-11} \times (27.3 \times 24 \times 60 \times 60)^2}\]

\[M_E \approx 6.02 \times 10^{24}\ \text{kg}\]

Both methods yield almost the same answer — the difference between them is less than $1\%$, which is a nice cross-check.

Solved Example 7.7

Express the constant $k$ of Eq. (7.38) in days and kilometres. Given $k = 10^{-13}\ \text{s}^2 \text{m}^{-3}$. The Moon is at a distance of $3.84 \times 10^5$ km from the Earth. Obtain its time period of revolution in days.

Answer

We must convert $k$ from SI units ($\text{s}^2 \text{m}^{-3}$) to units of $\text{d}^2 \text{km}^{-3}$. Recall $1$ day $= 24 \times 60 \times 60$ s and $1$ m $= 1/1000$ km.

\[k = 10^{-13} \left[\frac{1}{(24 \times 60 \times 60)^2}\right] \text{d}^2 \left[\frac{1}{(1/1000)^3}\right] \text{km}^{-3}\]

\[k = 1.33 \times 10^{-14}\ \text{d}^2\, \text{km}^{-3}\]

Now using Eq. (7.38) with $R = R_E + h$ set equal to the Earth–Moon distance $3.84 \times 10^5$ km:

\[T^2 = (1.33 \times 10^{-14}) \times (3.84 \times 10^5)^3\]

\[T = 27.3\ \text{days}\]

which matches the known lunar period.

Elliptical orbits

Equation (7.38) was derived for a circular orbit. But it also holds — with one small modification — for elliptical orbits: simply replace $(R_E + h)$ by the semi-major axis of the ellipse. When this substitution is made, Eq. (7.38) becomes exactly Kepler’s third law in its general form (Section 7.2). The Earth then sits at one of the two foci of the elliptical orbit, just as the Sun sits at one focus of a planet’s orbit.

Numerical 7.11

A satellite is to be placed in a circular orbit at a height of $600$ km above the Earth’s surface. Find its orbital speed and time period. Take $g = 9.8\ \text{m s}^{-2}$ and $R_E = 6.4 \times 10^6$ m.

Numerical 7.12

A geostationary satellite orbits the Earth once every $24$ hours in the plane of the equator. Given $G M_E = 4.0 \times 10^{14}\ \text{m}^3 \text{s}^{-2}$, find the radius of its orbit and its distance above the Earth’s surface (take $R_E = 6.4 \times 10^6$ m).

Having understood the motion of a satellite, we now turn to its energy. This will let us answer questions like: how much energy does it take to boost a satellite from a lower orbit to a higher one? That is the topic of the final section of this chapter.

7.10 Energy of an Orbiting Satellite

A satellite in orbit has kinetic energy because it is moving, and gravitational potential energy because it sits at a finite distance from the Earth. In this final section of the chapter, we compute both, add them up to get the total mechanical energy, and see what that total tells us physically.

Kinetic energy of the satellite

For a satellite of mass $m$ moving in a circular orbit at height $h$ above the Earth’s surface, the orbital speed $v$ was found in Section 7.9 (Eq. 7.35):

\[v^2 = \frac{G M_E}{R_E + h}\]

The kinetic energy is therefore

\[K.E = \frac{1}{2} m v^2 = \frac{G m M_E}{2(R_E + h)} \tag{7.40}\]

Note that the KE is positive, as kinetic energy always is. It gets smaller as $h$ increases — satellites in higher orbits move more slowly.

Potential energy of the satellite

For gravitational potential energy, we use the general expression from Section 7.7 with the convention that PE is zero at infinity. At a distance $(R_E + h)$ from the Earth’s centre,

\[P.E = -\frac{G m M_E}{(R_E + h)} \tag{7.41}\]

The PE is negative, as we saw earlier, because gravity has pulled the satellite in from infinity and it has “fallen down” into the gravitational well.

Total mechanical energy — a striking relation

Adding the two:

\[E = K.E + P.E = \frac{G m M_E}{2(R_E + h)} + \left(-\frac{G m M_E}{(R_E + h)}\right)\]

\[E = -\frac{G m M_E}{2(R_E + h)} \tag{7.42}\]

Look at this result carefully. Comparing Eqs. (7.40), (7.41), and (7.42) side by side:

$K.E = +\dfrac{G m M_E}{2(R_E + h)}$
$P.E = -\dfrac{G m M_E}{(R_E + h)}$
Total $E = -\dfrac{G m M_E}{2(R_E + h)}$

The magnitude of the PE is exactly twice the KE, and the total energy equals the negative of the KE. This “half-magnitude” relation between KE and PE for a bound circular orbit is called the virial relation and shows up throughout gravitational physics.

Quick Question

Why is the total energy of a satellite negative? Doesn’t “negative energy” sound odd for something that is clearly moving and has mass?

The sign comes from our choice of reference: PE was set to zero at infinity. Since the satellite is trapped in the Earth’s gravitational well, it takes positive work to lift it out to infinity — so its energy relative to that reference is negative. Any bound orbit — meaning a closed orbit, one that keeps going around forever — has negative total energy.

What the sign of $E$ tells us

The sign of the total energy is a compact way of expressing the fate of an object under gravity.

$E < 0$ (negative total energy) → the object is bound; it moves on a closed orbit (circle or ellipse) around the Earth. It cannot escape to infinity. All ordinary satellites, planets, and moons fall in this category.
$E = 0$ (zero total energy) → the object has just enough energy to reach infinity with zero speed. This is the escape-speed condition from Section 7.8.
$E > 0$ (positive total energy) → the object has more than enough to escape; it will fly off to infinity with leftover kinetic energy.

Because real Earth satellites are always at some finite distance from the Earth, their total energy is always negative — never zero, never positive. This is consistent with the escape-speed discussion: at the moment a spacecraft is boosted to escape speed, its energy becomes zero, and any faster launch gives it positive energy that carries it away permanently.

Elliptical orbits

For a circular orbit, the satellite’s speed and its distance from the Earth stay constant, so KE and PE stay constant individually. What if the orbit is not circular but elliptical?

Then the situation changes point by point. As the satellite gets closer to the Earth (near perigee), it speeds up — KE rises, PE becomes more negative. As it swings out to apogee, it slows down — KE falls, PE becomes less negative. But their sum — the total mechanical energy — stays exactly the same throughout the orbit. This is just conservation of energy at work.

Even in the elliptical case, the total $E$ remains negative for any bound orbit. This is consistent with what we said earlier: if $E$ ever became zero or positive, the satellite would fly off, and there would be no orbit at all.

Real-World Application

Modern space missions carefully manage the total orbital energy to move a spacecraft between orbits. For example, when the Indian Space Research Organisation (ISRO) launched Chandrayaan-1 to the Moon in 2008, and again with Chandrayaan-3 in 2023, they did not fire the rocket continuously all the way. Instead, the spacecraft was placed in a low Earth orbit and then, using a sequence of short engine burns at carefully chosen points, its total energy was increased in steps. Each burn raised the orbit to a bigger ellipse, until finally the spacecraft was captured by the Moon’s gravity. This step-by-step raising is called a Hohmann transfer, and it is far more fuel-efficient than a single continuous burn.

A worked example on orbital energy changes

The following example applies Eq. (7.42) to the practical question of how much energy is needed to move a satellite from one orbit to another.

Solved Example 7.8

A $400$ kg satellite is in a circular orbit of radius $2R_E$ about the Earth. How much energy is required to transfer it to a circular orbit of radius $4R_E$? What are the changes in the kinetic and potential energies?

Answer

Using Eq. (7.42) with $(R_E + h)$ replaced by the orbital radius:

Initially, at orbital radius $2R_E$:

\[E_i = -\frac{G M_E\, m}{4 R_E}\]

Finally, at orbital radius $4R_E$:

\[E_f = -\frac{G M_E\, m}{8 R_E}\]

The change in the total energy — that is, the energy that must be supplied by the rocket engine to boost the satellite — is

\[\Delta E = E_f - E_i = \frac{G M_E\, m}{8 R_E} = \left(\frac{G M_E}{R_E^2}\right)\frac{m R_E}{8}\]

Recognising $G M_E / R_E^2 = g$ (Section 7.5), we can simplify:

\[\Delta E = \frac{g m R_E}{8} = \frac{9.81 \times 400 \times 6.37 \times 10^6}{8} = 3.13 \times 10^9\ \text{J}\]

Now for the individual components:

Change in kinetic energy. In a circular orbit, $K = |E|$ (from Eq. 7.42, $K = -E$), so as we go from $2R_E$ to $4R_E$ the satellite’s KE actually decreases. The change is

\[\Delta K = K_f - K_i = -3.13 \times 10^9\ \text{J}\]

The satellite moves slower in the higher orbit, exactly as Eq. (7.35) predicted.

Change in potential energy. The magnitude of the change in PE is twice the change in the total energy (this follows from KE = -E and PE = 2E for a circular orbit):

\[\Delta V = V_f - V_i = -6.25 \times 10^9\ \text{J}\]

The kinetic energy is reduced, and this reduction has the same magnitude as $\Delta E$. The potential energy changes by twice as much in magnitude.

Numerical 7.13

A satellite of mass $500$ kg is placed in a circular orbit at a height of $R_E$ above the Earth’s surface. Find the total mechanical energy of the satellite in orbit. Take $g = 9.8\ \text{m s}^{-2}$ and $R_E = 6.4 \times 10^6$ m.

Numerical 7.14

A geostationary satellite of mass $2000$ kg is in a circular orbit at an altitude of about $36{,}000$ km above the Earth’s equator. How much additional energy would need to be supplied to move it out of the Earth’s gravitational field completely? Use $G M_E = 4.0 \times 10^{14}\ \text{m}^3 \text{s}^{-2}$ and $R_E = 6.4 \times 10^6$ m.

This completes our study of gravitation. Starting from Galileo’s inclined plane and Tycho Brahe’s naked-eye observations, we followed the trail through Kepler’s laws, Newton’s universal law, and Cavendish’s laboratory experiment, all the way to the energy accountancy of a modern orbiting satellite. The single formula $F = G m_1 m_2 / r^2$ has explained everything from a falling apple to the launch of a spacecraft — a striking illustration of how one simple law of nature can weave together phenomena that at first look completely unrelated.

7.11 Summary

1. Newton’s law of universal gravitation. Every two point masses in the universe attract each other. The gravitational force between two particles of masses $m_1$ and $m_2$ separated by a distance $r$ has magnitude

\[F = G\,\frac{m_1 m_2}{r^2}\]

where $G$ is the universal gravitational constant, with the value $G = 6.672 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$. The force is always attractive and directed along the line joining the two particles.

2. Principle of superposition. To find the resultant gravitational force on a particle $m$ due to a number of other masses $M_1, M_2, \dots, M_n$, compute each pairwise force $\vec{F}_1, \vec{F}_2, \dots, \vec{F}_n$ from Newton’s law separately (each is uninfluenced by the presence of the others) and add them as vectors:

\[\vec{F}_R = \vec{F}_1 + \vec{F}_2 + \cdots + \vec{F}_n = \sum_{i=1}^{n} \vec{F}_i\]

3. Kepler’s laws of planetary motion. (a) Law of orbits — All planets move on elliptical orbits with the Sun at one of the two focal points. (b) Law of areas — The radius vector drawn from the Sun to a planet sweeps out equal areas in equal times. This law follows from the fact that the gravitational force is a central force, which conserves angular momentum. (c) Law of periods — The square of the orbital period is proportional to the cube of the semi-major axis of the elliptical orbit.

For a circular orbit of radius $R$ around the Sun (mass $M_s$),

\[T^2 = \left(\frac{4\pi^2}{G M_s}\right) R^3\]

For elliptical orbits, replace $R$ by the semi-major axis $a$. Most planets have nearly circular orbits.

4. Acceleration due to gravity. (a) At a height $h$ above the surface:

\[g(h) = \frac{G M_E}{(R_E + h)^2} \approx \frac{G M_E}{R_E^2}\left(1 - \frac{2h}{R_E}\right)\ \text{for } h \ll R_E\]

which can be written as $g(h) = g(0)\left(1 - 2h/R_E\right)$, where $g(0) = G M_E / R_E^2$ is the surface value.

At a depth $d$ below the surface (assuming uniform density):

\[g(d) = \frac{G M_E}{R_E^2}\left(1 - \frac{d}{R_E}\right) = g(0)\left(1 - \frac{d}{R_E}\right)\]

Gravity is maximum at the surface and decreases both above and below.

5. Gravitational potential energy. Gravity is a conservative force, so a potential energy function can be defined. For two particles separated by $r$,

\[V = -\frac{G m_1 m_2}{r}\]

with $V \to 0$ as $r \to \infty$. For a system of many particles, the total PE is the sum of pairwise PEs (superposition applied to energy).

6. Total mechanical energy. For a particle of mass $m$ moving with speed $v$ in the vicinity of a mass $M$,

\[E = \frac{1}{2} m v^2 - \frac{G M m}{r}\]

is the sum of kinetic and potential energies. In the absence of non-conservative forces, $E$ is a constant of motion.

7. Circular orbit of a small mass around a much larger mass. When $m$ moves in a circular orbit of radius $a$ around $M$ (with $M \gg m$), the KE, PE, and total energy are:

\[K = \frac{G M m}{2 a}, \qquad V = -\frac{G M m}{a}, \qquad E = -\frac{G M m}{2 a}\]

The total energy is negative for any bound system — that is, any closed orbit such as a circle or ellipse.

8. Escape speed. The minimum speed required for an object launched from the Earth’s surface to escape the Earth’s gravity is

\[v_e = \sqrt{\frac{2 G M_E}{R_E}} = \sqrt{2 g R_E}\]

with a value of about $11.2\ \text{km s}^{-1}$.

9. Shell theorem — outside a spherical body. If a particle is outside a uniform spherical shell, or outside any solid sphere with spherically symmetric mass distribution, the sphere attracts the particle as though its entire mass were concentrated at the centre of the sphere.

10. Shell theorem — inside a spherical body. If a particle is inside a uniform spherical shell, the gravitational force on it is zero. If a particle is inside a homogeneous solid sphere, the force acts toward the centre of the sphere. The pull on such a particle is exerted only by the spherical mass interior to it.

Table of physical quantities

Physical Quantity	Symbol	Dimensions	SI Unit	Remarks
Gravitational Constant	$G$	$[\text{M}^{-1} \text{L}^{3} \text{T}^{-2}]$	$\text{N m}^2 \text{kg}^{-2}$	$6.67 \times 10^{-11}$
Gravitational Potential Energy	$V(r)$	$[\text{M L}^{2} \text{T}^{-2}]$	J	$-\dfrac{G M m}{r}$ (scalar)
Gravitational Potential	$U(r)$	$[\text{L}^{2} \text{T}^{-2}]$	$\text{J kg}^{-1}$	$-\dfrac{G M}{r}$ (scalar)
Gravitational Intensity	$\vec{E}$ or $\vec{g}$	$[\text{L T}^{-2}]$	$\text{m s}^{-2}$	$\dfrac{G M}{r^2}\,\hat{r}$ (vector)

7.12 Points to Ponder

1. What is conserved under gravity? For an object moving under the gravitational influence of another body, the following are conserved: (a) angular momentum, and (b) total mechanical energy.

Linear momentum, however, is not conserved — the direction of motion changes constantly under the gravitational pull, so momentum, being a vector, keeps changing.

2. Kepler’s second law and central forces. The conservation of angular momentum leads to Kepler’s second law (the law of areas). This is not special to the inverse-square nature of gravity — it holds for any central force, whatever its strength dependence on distance.

3. Kepler’s third law and the constant $K_s$. In Kepler’s third law $T^2 = K_s R^3$, the constant $K_s$ has the same value for all planets moving in circular orbits around the Sun. The very same relation applies to satellites orbiting the Earth, with a different constant that depends on the Earth’s mass rather than the Sun’s.

4. Weightlessness of astronauts. An astronaut in a space satellite experiences weightlessness. This is not because the gravitational force is small at that location in space — Earth’s gravity there is still nearly as strong as at the surface. It is because both the astronaut and the satellite are in continuous “free fall” toward the Earth. Since they fall together at the same rate, the astronaut floats freely inside the satellite.

5. Choice of reference for potential energy. Gravitational potential energy is defined up to an additive constant:

\[V = -\frac{G m_1 m_2}{r} + \text{constant}\]

The simplest convention is to set this constant to zero, giving

\[V = -\frac{G m_1 m_2}{r}\]

which implies $V \to 0$ as $r \to \infty$. Choosing the location of the zero of gravitational energy is equivalent to choosing the arbitrary constant in the PE. Importantly, this choice does not alter the gravitational force between the particles — force is the negative gradient of PE, and adding a constant to PE gives no change in force.

6. Sign of the total mechanical energy. The total mechanical energy of an object is the sum of its KE (always positive) and PE. With PE taken as zero at infinity, the gravitational PE of a bound object is negative. Consequently, the total energy of any bound satellite is also negative. This negative total energy is the signature of a “captured” orbit, one from which the object cannot escape without additional energy.

7. The $mgh$ formula is an approximation. The familiar expression $mgh$ for gravitational potential energy is really the approximate change in the exact gravitational PE derived above, valid only when the height $h$ is small compared to the Earth’s radius. It is not the true PE; it is a convenient near-surface approximation.

8. Central force between finite bodies. The gravitational force between two point particles is central — it lies along the line joining them. But the force between two finite rigid bodies is not necessarily along the line joining their centres of mass. For a spherically symmetric body, however, the force on an external particle is exactly as if the entire mass were concentrated at the centre — and this force is therefore central.

9. No gravitational shielding. The gravitational force on a particle inside a spherical shell is zero. This might tempt one to think shells act as gravitational shields. They do not. Unlike a metallic shell — which shields electrical forces inside from external charges — a spherical shell does not prevent bodies outside the shell from exerting gravitational forces on a particle inside it. In other words, gravitational shielding is not possible: gravity cannot be blocked.

7.13 NCERT Questions

Answer the following:
1. You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
2. An astronaut inside a small spaceship orbiting around the Earth cannot detect gravity. If the space station orbiting around the Earth has a large size, can he hope to detect gravity?
3. If you compare the gravitational force on the Earth due to the Sun to that due to the Moon, you would find that the Sun’s pull is greater than the Moon’s pull. (You can check this yourself using the data in the succeeding exercises.) However, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun. Why?
Choose the correct alternative:
1. Acceleration due to gravity increases/decreases with increasing altitude.
2. Acceleration due to gravity increases/decreases with increasing depth (assume the Earth to be a sphere of uniform density).
3. Acceleration due to gravity is independent of mass of the Earth/mass of the body.
4. The formula $-G M m (1/r_2 - 1/r_1)$ is more/less accurate than the formula $mg(r_2 - r_1)$ for the difference of potential energy between two points $r_2$ and $r_1$ distance away from the centre of the Earth.
Suppose there existed a planet that went around the Sun twice as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Io, one of the satellites of Jupiter, has an orbital period of $1.769$ days and the radius of the orbit is $4.22 \times 10^8$ m. Show that the mass of Jupiter is about one-thousandth that of the Sun.
Let us assume that our galaxy consists of $2.5 \times 10^{11}$ stars each of one solar mass. How long will a star at a distance of $50{,}000$ ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be $10^5$ ly.
Choose the correct alternative:
1. If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
2. The energy required to launch an orbiting satellite out of Earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of Earth’s influence.
Does the escape speed of a body from the Earth depend on (a) the mass of the body, (b) the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?
A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Which of the following symptoms is likely to afflict an astronaut in space: (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem?
In the following two exercises, choose the correct answer from among the given ones. The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig. 7.11): (i) a, (ii) b, (iii) c, (iv) 0.

Figure to come

Fig. 7.11 – Hemispherical shell (bowl shape) with several arrows labelled a, b, c, d, e, f, g at various points showing possible directions of gravitational intensity; point P shown on the rim.

For the above problem, the direction of the gravitational intensity at an arbitrary point $P$ is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.
A rocket is fired from the Earth towards the Sun. At what distance from the Earth’s centre is the gravitational force on the rocket zero? Mass of the Sun $= 2 \times 10^{30}$ kg, mass of the Earth $= 6 \times 10^{24}$ kg. Neglect the effect of other planets etc. (Orbital radius $= 1.5 \times 10^{11}$ m.)
How will you “weigh the Sun” — that is, estimate its mass? The mean orbital radius of the Earth around the Sun is $1.5 \times 10^8$ km.
A Saturn year is $29.5$ times the Earth year. How far is Saturn from the Sun if the Earth is $1.50 \times 10^8$ km away from the Sun?
A body weighs $63$ N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?
Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the Earth if it weighed $250$ N on the surface?
A rocket is fired vertically with a speed of $5\ \text{km s}^{-1}$ from the Earth’s surface. How far from the Earth does the rocket go before returning to the Earth? Mass of the Earth $= 6.0 \times 10^{24}$ kg; mean radius of the Earth $= 6.4 \times 10^6$ m; $G = 6.67 \times 10^{-11}\ \text{N m}^2 \text{kg}^{-2}$.
The escape speed of a projectile on the Earth’s surface is $11.2\ \text{km s}^{-1}$. A body is projected out with thrice this speed. What is the speed of the body far away from the Earth? Ignore the presence of the Sun and other planets.
A satellite orbits the Earth at a height of $400$ km above the surface. How much energy must be expended to rocket the satellite out of the Earth’s gravitational influence? Mass of the satellite $= 200$ kg; mass of the Earth $= 6.0 \times 10^{24}$ kg; radius of the Earth $= 6.4 \times 10^6$ m; $G = 6.67 \times 10^{-11}\ \text{N m}^2 \text{kg}^{-2}$.
Two stars each of one solar mass ($= 2 \times 10^{30}$ kg) are approaching each other for a head-on collision. When they are a distance $10^9$ km apart, their speeds are negligible. What is the speed with which they collide? The radius of each star is $10^4$ km. Assume the stars to remain undistorted until they collide. (Use the known value of $G$.)
Two heavy spheres each of mass $100$ kg and radius $0.10$ m are placed $1.0$ m apart on a horizontal table. What is the gravitational force and potential at the mid-point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

7.14 Check Your Concepts

Two identical apples fall from the same tree — one from a lower branch, the other from a higher one. Show, using Newton’s law of gravitation, why both apples still accelerate at essentially the same value $g$ near the ground, even though the higher apple is technically farther from the Earth’s centre.
Kepler’s second law (the law of areas) is often stated as an observation about planets. Explain in your own words why this law is really a consequence of the fact that the gravitational force is a central force, and does not require the inverse-square dependence at all.
A student argues: “If the Moon is being pulled toward the Earth by gravity, then it must be slowly moving closer and closer to the Earth. Eventually it will fall in and crash.” Explain what is wrong with this reasoning and describe correctly what the Moon is doing.
The gravitational potential energy of a two-mass system is written as $V = -G m_1 m_2 / r$, which is negative. A student is confused: “How can any energy be negative? Aren’t kinetic and potential energies both supposed to be positive quantities?” Explain to the student why the negative sign appears here, and what it physically means.
Both the Moon and the International Space Station (ISS) orbit the Earth. The ISS is only about $400$ km above the Earth’s surface, while the Moon is more than $3.8 \times 10^5$ km away. Yet the ISS completes an orbit in about $90$ minutes while the Moon takes about $27$ days. Explain qualitatively, using Kepler’s third law, why the Moon takes so much longer.
Explain why the Moon has essentially no atmosphere but the Earth does. Frame your answer in terms of escape speed and typical molecular speeds.
Two satellites $A$ and $B$ of the same mass are in circular orbits around the Earth, with $A$ in a lower orbit than $B$. Compare the (i) kinetic energies, (ii) potential energies, and (iii) total mechanical energies of the two satellites. Which one has more total energy, and how would you interpret the result?
The formula $-G M m (1/r_2 - 1/r_1)$ for the change in gravitational potential energy reduces to $m g (r_2 - r_1)$ for objects near the Earth’s surface. Show clearly, using the binomial approximation, how the second expression arises from the first in the limit $r_2 - r_1 \ll R_E$.
Explain the difference in meaning between the terms “gravitational potential” and “gravitational potential energy.” How are the two related, and what are their SI units?
A spacecraft is placed in a geostationary orbit above the equator. Explain (i) why such an orbit is useful for communication satellites, (ii) why the orbit must be over the equator and not over the poles, and (iii) what determines the specific altitude at which the geostationary orbit lies.

7.15 Practice with Numericals

Two point masses of $6.0$ kg and $10.0$ kg are placed $0.50$ m apart. Find the magnitude of the gravitational force between them, and compute the gravitational potential energy of the pair. Take $G = 6.67 \times 10^{-11}\ \text{N m}^2 \text{kg}^{-2}$.
The mass of the planet Jupiter is $1.9 \times 10^{27}$ kg and its radius is $7.0 \times 10^7$ m. Find the value of the acceleration due to gravity on the surface of Jupiter. Take $G = 6.67 \times 10^{-11}\ \text{N m}^2 \text{kg}^{-2}$.
A planet has a mass twice that of the Earth and a radius twice that of the Earth. Find the ratio of the escape speed from the surface of the planet to the escape speed from the Earth’s surface.
The orbital radius of Neptune around the Sun is about $30$ times that of the Earth. Using Kepler’s third law, estimate the orbital period of Neptune in Earth years.
A satellite of mass $1000$ kg is placed in a circular orbit at a height of $500$ km above the Earth’s surface. Find (a) the orbital speed and (b) the total mechanical energy of the satellite in its orbit. Take $g = 9.8\ \text{m s}^{-2}$ and $R_E = 6.4 \times 10^6$ m.
Two identical stars, each of mass $M$, orbit their common centre of mass in a circular path of diameter $d$ (so each star is at a distance $d/2$ from the centre). Using Newton’s law of gravitation and the requirement that the gravitational pull between the two stars provides the centripetal force, derive an expression for the orbital period $T$ of the system in terms of $M$, $d$, and $G$.
A comet in a highly elongated orbit around the Sun has a perihelion distance of $0.5$ AU and travels there at $60\ \text{km s}^{-1}$. Its aphelion distance is $50$ AU. Using conservation of angular momentum, find the comet’s speed at aphelion.
A body is projected vertically upward from the Earth’s surface with a speed equal to half the escape speed. Using energy conservation, find the maximum height above the Earth’s surface that the body reaches. Take $R_E = 6.4 \times 10^6$ m.

Physical Quantity	Symbol	Dimensions	SI Unit	Remarks
Gravitational Constant	\(G\)	\([\text{M}^{-1} \text{L}^{3} \text{T}^{-2}]\)	\(\text{N m}^2 \text{kg}^{-2}\)	\(6.67 \times 10^{-11}\)
Gravitational Potential Energy	\(V(r)\)	\([\text{M L}^{2} \text{T}^{-2}]\)	J	\(-\dfrac{G M m}{r}\) (scalar)
Gravitational Potential	\(U(r)\)	\([\text{L}^{2} \text{T}^{-2}]\)	\(\text{J kg}^{-1}\)	\(-\dfrac{G M}{r}\) (scalar)
Gravitational Intensity	\(\vec{E}\) or \(\vec{g}\)	\([\text{L T}^{-2}]\)	\(\text{m s}^{-2}\)	\(\dfrac{G M}{r^2}\,\hat{r}\) (vector)

Chapter 7 — Gravitation

7.2 Kepler’s Laws

1. Law of orbits

2. Law of areas

3. Law of periods

7.3 Universal Law of Gravitation

Testing the idea with the Moon

The inverse-square guess

Newton’s universal law of gravitation

Vector form of the law

Many masses — the superposition principle

From point masses to extended bodies

7.4 The Gravitational Constant

The Cavendish experiment

Turning the twist into a measurement

Writing down the equation

7.5 Acceleration Due to Gravity of the Earth

The Earth as a set of shells

Inside the Earth is different

Assuming uniform density

Acceleration at the surface

“Cavendish weighed the Earth”

7.6 Acceleration Due to Gravity Below and Above the Surface of Earth

Gravity above the surface

Gravity below the surface

The peak at the surface

7.7 Gravitational Potential Energy

A quick recap: what “potential energy” means

Near the Earth’s surface: the familiar \(mgh\)

Far from the Earth: gravity isn’t constant

The natural reference: infinity

The general two-particle formula

Many particles: superposition

7.8 Escape Speed

Setting up with energy conservation

The condition for escape

Launching from the Earth’s surface

Why \(m\) dropped out — and what it means

Applying the same formula elsewhere

A solved example with two attracting bodies

7.9 Earth Satellites

Deriving the orbital speed

Time period of a satellite

Sample calculations

Elliptical orbits

7.10 Energy of an Orbiting Satellite

Kinetic energy of the satellite

Potential energy of the satellite

Total mechanical energy — a striking relation

What the sign of \(E\) tells us

Elliptical orbits

A worked example on orbital energy changes

7.11 Summary

Table of physical quantities

7.12 Points to Ponder

7.13 NCERT Questions

7.14 Check Your Concepts

7.15 Practice with Numericals