Chapter 5: Gravitation
In the early 1800s, astronomers had a problem they could not explain. The planet Uranus, discovered only a few decades earlier, refused to follow the path predicted for it. Year after year it drifted slightly ahead of, and then behind, the position calculated from Newton’s law of gravitation. Something was wrong — either the law itself was failing at the edge of the Solar System, or something unseen was tugging at Uranus.
Two mathematicians, Urbain Le Verrier in France and John Couch Adams in England, working independently, refused to abandon the law. Instead, they trusted it completely. If an unknown planet existed further out, its gravitational pull would disturb Uranus in exactly the way observed. Working backwards from the disturbance, they calculated where that hidden planet must be — using nothing but pen, paper, and the law of gravitation.
In 1846, the astronomer Johann Galle at the Berlin Observatory pointed his telescope at the patch of sky Le Verrier had specified. Neptune was there, almost exactly where the mathematics said it would be.
A planet had been discovered not by looking, but by calculating. That is the power of the law you are about to study — a single rule of attraction that governs falling apples, orbiting satellites, and worlds no one has yet seen.
Diagram
Fig. 5.1 – Uranus’s orbit shown drifting off its predicted path, with an unseen outer planet (Neptune) pulling on it, and a telescope aimed at the predicted position.
By the end of this chapter, you will be able to answer each of these questions for yourself.
5.1 Introduction
The Universal Tendency to Fall
Every material object around us shows the same natural tendency — it gets pulled towards the Earth. A coconut breaks free from a tree and drops to the ground. Raindrops form high up in the clouds and come straight down. A ball thrown upwards slows, stops, and returns.
These look like very different events, but Physics treats them as one and the same phenomenon: the Earth attracts every object towards itself.
An important observation is that all bodies are attracted towards the Earth with the same constant acceleration. Whether the object is heavy or light, it gains speed at the same rate as it falls (if we ignore the resistance of air).
This fact was recognised by the Italian physicist Galileo. He is said to have demonstrated it by releasing two balls of different masses from the top of the leaning tower of Pisa; both balls reached the ground at the same time.
From the Sky to the Ground: A Long History
The motion of the sky has been studied for a very long time. The Indian astronomer and mathematician Aryabhatta (476–550 A.D.) studied the motion of the Moon, the Earth and the other planets in the 5th century A.D. In his book Aryabhatiya, he concluded that the Earth rotates about its own axis and moves in a circular orbit around the Sun, and that the Moon revolves in a circular orbit around the Earth.
Almost a thousand years after Aryabhatta, Tycho Brahe (1546–1601) and Johannes Kepler (1571–1630) studied planetary motion through careful observations. Tycho Brahe patiently recorded a huge amount of data on the positions of the planets over many years.
Kepler analysed this meticulously recorded data and established three laws of planetary motion. He showed that the motion of the planets follows these laws. But Kepler could only say how the planets move — he could not say why they move that way.
The reason why planets obey these laws was provided by Newton. He explained that gravitation is the phenomenon responsible for keeping the planets in their orbits around the Sun. The Moon also revolves around the Earth due to gravitation.
Gravitation does even more than this. It compels dispersed matter to come together and coalesce. This is why the Earth, the Sun and all the material macroscopic objects of the universe exist at all — scattered matter was drawn together by mutual gravitational attraction.
What Kind of Force Is Gravitation?
Every massive object in the universe experiences gravitational force. It is the force of mutual attraction between any two objects by virtue of their masses — that is, an object possesses this ability to attract simply because it has mass, and for no other reason.
This force has four striking properties that you should fix firmly in mind, because every later section of this chapter depends on them.
It is always attractive. Unlike the electrostatic force, which can pull or push depending on the signs of the charges, gravitation only ever pulls. There is no “gravitational repulsion.”
It has infinite range. The force becomes weaker and weaker as the two objects move apart, but it never becomes exactly zero at any finite separation. Strictly, it vanishes only at infinite distance. This is why the Sun can hold a planet as far away as Neptune in its orbit.
It does not depend on the intervening medium. If you place a wall, a sheet of lead, or a vacuum between two masses, the gravitational force between them stays exactly the same. There is no way to “shield” or “block” gravity, unlike an electric field, which can be screened by a conductor.
It is much weaker than the other fundamental forces. Gravitational force is \(10^{-39}\) times weaker than the strong nuclear force.
Notice the comparison in Fig. 5.2: individually, gravity is negligible between two small objects, but over astronomical masses it becomes the dominant force of the universe.
Diagram
Fig. 5.2 – A bar chart comparing the relative strengths of the four fundamental forces (strong nuclear, electromagnetic, weak nuclear, gravitational) on a logarithmic scale, with gravitation shown as by far the smallest.
In the sections that follow, we will first look at how the planets move (Kepler’s laws), then at why they move that way (Newton’s universal law of gravitation), and finally at what this law tells us about acceleration due to gravity, potential energy, escape velocity and satellites.
5.2 Kepler’s Laws
Kepler’s laws of planetary motion describe the orbits of the planets around the Sun. Kepler published the first two laws in 1609 and the third law in 1619.
These laws did not come from a theory. They are the result of the analysis of the data collected by Tycho Brahe through years of patient observations of planetary motion. Kepler searched through those numbers until he found the geometrical rules hidden inside them.
Keep this in mind as you read: Kepler’s laws are descriptions of what the planets actually do. The explanation of why they do it comes later, from Newton.
Understanding the Ellipse
Before stating the first law, you must be clear about what an ellipse is, because the entire law is built on it.
Diagram
Fig. 5.3 – Two drawing pins A and B a distance d apart, a loop of thread around them, and a pencil pulling the thread taut to trace an ellipse.
Notice what the thread guarantees: since its length never changes, the sum of the distances from the pencil tip to A and to B stays the same at every point of the curve. That is exactly the definition of an ellipse, and A and B are its two foci.
1. Law of Orbit
The orbit of a planet around the Sun is shown in Fig. 5.4.
Diagram
Fig. 5.4 – An ellipse traced by a planet with the Sun at the focus S; foci S and S′, points P (perihelion) and A (aphelion), centre O, major axis PA = 2a, minor axis MN = 2b.
Here, S and S′ are the foci of the ellipse, the Sun being at S. There is nothing at all at the other focus S′ — it is only a geometrical point.
The important terms marked in Fig. 5.4 are:
- P is the closest point along the orbit from S, and is called the perihelion.
- A is the farthest point from S, and is called the aphelion.
- PA is the major axis \(= 2a\).
- PO and AO are the semimajor axes \(= a\).
- MN is the minor axis \(= 2b\).
- MO and ON are the semiminor axes \(= b\).
2. Law of Areas
Kepler observed that planets do not move around the Sun with uniform speed. They move faster when they are nearer to the Sun, while they move slower when they are farther from the Sun. This law explains that behaviour.
Fig. 5.5 shows the orbit of a planet P. The shaded areas are the areas swept by SP, the line joining the planet and the Sun, in fixed intervals of time. According to the second law, these two shaded areas are equal, even though one of them is a long thin sliver far from the Sun and the other is a short wide wedge close to the Sun.
Diagram
Fig. 5.5 – The orbit of a planet P moving around the Sun S, showing two shaded regions of equal area swept in equal times, with \(\vec{r}\), \(\vec{p}\) and \(\vec{F}\) marked.
Think about what this forces the planet to do. Near the Sun, the line SP is short, so to sweep out the same area in the same time, the planet must travel a longer arc — that is, it must move faster. Far from the Sun, SP is long, so a small arc already sweeps a large area, and the planet can afford to crawl.
Why the Law of Areas Is True
The law of areas can be understood as an outcome of conservation of angular momentum. It is valid for any central force.
A central force on an object is a force which is always directed along the line joining the position of the object and a fixed point, usually taken to be the origin of the coordinate system.
The force of gravity due to the Sun on a planet is always along the line joining the Sun and the planet (Fig. 5.5). It is thus a central force. This one fact is enough to prove the second law.
Suppose the Sun is at the origin. The position of the planet is denoted by \(\vec{r}\), and the perpendicular component of its momentum is denoted by \(\vec{p}\) (the component \(\perp \vec{r}\)).
The area swept by the planet of mass \(m\) in a given interval \(\Delta t\) is \(\overrightarrow{\Delta A}\), which is given by
\[\overrightarrow{\Delta A} = \frac{1}{2}\left(\vec{r} \times \vec{v}\,\Delta t\right) \qquad \text{--- (5.1)}\]
Here \(\vec{r}\) is the position vector of the planet from the Sun (SI unit: metre, m) and \(\vec{v}\) is its velocity (SI unit: m/s). In a small time \(\Delta t\), the planet moves through a small displacement \(\vec{v}\,\Delta t\). The region swept out is a thin triangle with the two sides \(\vec{r}\) and \(\vec{v}\,\Delta t\), and the area of a triangle formed by two vectors is one-half of their cross product — which is exactly what Eq. (5.1) says.
As, for small \(\Delta t\), \(\vec{v}\) is perpendicular to \(\vec{r}\), this is the area of the triangle. Dividing by \(\Delta t\),
\[\therefore \frac{\overrightarrow{\Delta A}}{\Delta t} = \frac{1}{2}\left(\vec{r} \times \vec{v}\right) \qquad \text{--- (5.2)}\]
The quantity \(\dfrac{\overrightarrow{\Delta A}}{\Delta t}\) is the areal velocity — the area swept per unit time. If we can show that this stays constant, the law of areas is proved.
Linear momentum (\(\vec{p}\)) is the product of mass and velocity.
\[\vec{p} = m\vec{v} \qquad \text{--- (5.3)}\]
Putting \(\vec{v} = \vec{p}/m\) in the above equation, we get
\[\frac{\overrightarrow{\Delta A}}{\Delta t} = \frac{1}{2}\left(\vec{r} \times \frac{\vec{p}}{m}\right) \qquad \text{--- (5.4)}\]
Angular momentum \(\vec{L}\) is the rotational equivalent of linear momentum and is defined as
\[\therefore \vec{L} = \vec{r} \times \vec{p} \qquad \text{--- (5.5)}\]
Its SI unit is kg m² s⁻¹.
For a central force, the angular momentum is conserved — that is, \(\vec{L}\) does not change with time. (The reason is that a central force produces no torque about the force centre: the force and the position vector lie along the same line, so their cross product, and hence the torque, is zero. With zero torque, angular momentum cannot change.)
Substituting Eq. (5.5) into Eq. (5.4),
\[\therefore \frac{\overrightarrow{\Delta A}}{\Delta t} = \frac{\vec{L}}{2m} = \text{constant} \qquad \text{--- (5.6)}\]
Since \(m\) is the constant mass of the planet and \(\vec{L}\) is conserved, the area swept per unit time is constant. This proves the law of areas. It is a consequence of the gravitational force being a central force.
3. Law of Periods
If \(r\) is the length of the semimajor axis, then this law states that
\[T^2 \propto r^3\]
\[\text{or} \quad \frac{T^2}{r^3} = \text{constant} \qquad \text{--- (5.7)}\]
Here \(T\) is the period of revolution of the planet (the time for one complete orbit) and \(r\) is the semimajor axis of its orbit (SI unit: metre, m).
The physical content of this law is that the outer planets are slow in two ways at once: they have a longer path to cover and they move more slowly along it. That is why the period grows faster than the size of the orbit — as \(r^{3/2}\), not as \(r\).
Kepler’s laws were based on regular observations of the motion of planets. Kepler did not know why the planets obey these laws — that is, he had not derived these laws. He had only found them in the data.
Table 5.1 gives data from measurements of planetary motions which confirm Kepler’s law of periods. Look down the last column: the ratio \(T^2/r^3\) is very nearly the same number for every planet, from Mercury to Pluto, even though their orbits differ enormously in size.
Table 5.1: Kepler’s third law
| Planet | Semi-major axis in units of 10¹⁰ m | Period in years | T²/r³ in units of 10⁻³⁴ y² m⁻³ |
|---|---|---|---|
| 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 |
| Pluto | 590 | 248 | 2.99 |
5.3 Universal Law of Gravitation
When objects are released near the surface of the Earth, they always fall down to the ground. In other words, the Earth attracts objects towards itself. Galileo (1564–1642) pointed out that heavy and light objects, when released from the same height, fall towards the Earth at the same speed — that is, they have the same acceleration.
Newton went beyond this. He was not content with the Earth and the objects falling on it. He proposed that the force of attraction between masses is universal — that the same force which pulls a stone down also holds the Moon in its orbit and holds the planets around the Sun.
Newton stated the universal law of gravitation, which led to an explanation of terrestrial gravitation. It also explains Kepler’s laws and provides the reason behind the observed motion of the planets around the Sun. Recall that Kepler could only describe the motion; Newton’s law is what finally explains it.
Newton’s Test with the Moon
In 1665, Newton studied the motion of the Moon around the Earth. His idea was simple but bold: if the force that makes an apple fall is the same force that holds the Moon, then the Moon must also be “falling” — and the numbers must agree.
It was known that the Moon completes one revolution about the Earth in 27.3 days. The distance from the Earth to the Moon is \(3.85 \times 10^5\) km. The motion of the Moon is in almost a circular orbit around the Earth with constant angular speed \(\omega\), as shown in Fig. 5.6.
Diagram
Fig. 5.6 – The Moon in a circular orbit of radius r around the Earth, with the centripetal force directed towards the centre of the Earth.
As it is a circular motion, the Moon must be constantly acted upon by a force directed towards the Earth, which is at the centre of the circle. This force is the centripetal force, and is given by
\[F = mr\omega^2 \qquad \text{--- (5.8)}\]
where \(m\) is the mass of the Moon (SI unit: kg), \(r\) is the distance between the centres of the Moon and the Earth (SI unit: m), and \(\omega\) is the angular speed (SI unit: rad/s).
Also we have \(F = ma\) from Newton’s laws of motion. Equating the two expressions for the same force,
\[\therefore ma = mr\omega^2\]
\[\therefore a = r\omega^2 \qquad \text{--- (5.9)}\]
Notice that the mass \(m\) has cancelled out. The acceleration of the Moon does not depend on the Moon’s own mass — a hint of Galileo’s result appearing again, this time in the sky.
As angular velocity in terms of time period is given as
\[\omega = \frac{2\pi}{T}\]
we get
\[a = r\left(\frac{2\pi}{T}\right)^2 \qquad \text{--- (5.10)}\]
Substituting values of \(r\) and \(T\) — note that the distance is converted from km to m by multiplying by \(10^3\), and the period from days to seconds by multiplying by \(24 \times 60 \times 60\) — we get
\[a = \frac{3.85 \times 10^5 \times 10^3 \, 4\pi^2 \, \text{m}}{(27.3 \times 24 \times 60 \times 60)^2 \, \text{s}^2}\]
\[\therefore a = 0.0027 \ \text{m/s}^2\]
This is the acceleration of the Moon, which is directed towards the centre of the Earth, i.e., the centre of the orbit in which the Moon revolves.
Comparing the Fall of the Moon with the Fall of an Apple
What could be the force which produces this acceleration?
Newton assumed that the laws of nature are the same for Earthly objects and for celestial bodies. As this acceleration (\(0.0027 \ \text{m/s}^2\)) is much smaller than the acceleration felt by bodies near the surface of the Earth while falling on the Earth (\(9.8 \ \text{m/s}^2\)), he concluded that the acceleration felt by an object due to the gravitational force of the Earth must be decreasing with distance of the body from the Earth.
(Remember that the value of acceleration due to Earth’s gravity at the surface is \(9.8 \ \text{m/s}^2\).)
Now compare the two accelerations. We have,
\[\frac{a_{object}}{a_{moon}} = \frac{9.8 \ \text{m/s}^2}{0.0027 \ \text{m/s}^2} \approx 3600\]
Also, comparing the two distances from the Earth’s centre — the Moon at \(3.85 \times 10^5\) km, and an object at the surface, one Earth radius (6378 km) from the centre:
\[\frac{\text{distance of moon from the Earth's centre}}{\text{distance of object from the Earth's centre}} = \frac{3.85 \times 10^5 \ \text{km}}{6378 \ \text{km}} \approx 60\]
Now look carefully at the two numbers Newton had in front of him: the accelerations are in the ratio 3600, and the distances are in the ratio 60. And \(60^2 = 3600\). This is not a coincidence — it is the inverse-square law revealing itself in real data.
Thus from the above two equations we get
\[\therefore \frac{a_{object}}{a_{moon}} = \left[\frac{\text{distance of moon}}{\text{distance of object}}\right]^2 \qquad \text{--- (5.11)}\]
Newton therefore concluded that the acceleration of an object towards the Earth is inversely proportional to the square of the distance of the object from the centre of the Earth.
\[\therefore a \propto \frac{1}{r^2}\]
Building the Law Step by Step
As, \(F = ma\), and \(a \propto 1/r^2\), therefore the force exerted by the Earth on an object of mass \(m\) at a distance \(r\) from it is
\[F \propto \frac{m}{r^2}\]
Similarly, an object also exerts a force on the Earth, which is
\[F_E \propto \frac{M}{r^2}\]
where \(M\) is the mass of the Earth.
According to Newton’s third law of motion, the force on a body due to the Earth has to be equal to the force on the Earth due to the object. But the first expression contains only \(m\), and the second only \(M\). Both must describe the same force. Hence the force \(F\) is also proportional to the mass of the Earth. Hence Newton concluded that the gravitational force between the Earth and an object of mass \(m\) is
\[F \propto \frac{Mm}{r^2}\]
He then generalized it to the gravitational force between any two objects and stated his universal law of gravitation as follows.
This law is applicable to all material objects in the universe. Hence it is known as the universal law of gravitation.
The Law in Symbols
If two bodies of masses \(m_1\) and \(m_2\) are separated by a distance \(r\), then the gravitational force of attraction between them can be written as
\[F \propto \frac{m_1 m_2}{r^2}\]
\[\text{or,} \quad F = G\frac{m_1 m_2}{r^2} \qquad \text{--- (5.12)}\]
Here \(m_1\) and \(m_2\) are the two masses (SI unit: kg), \(r\) is the distance between them (SI unit: m), \(F\) is the force of attraction (SI unit: newton, N), and \(G\) is a constant known as the universal gravitational constant. Its value in SI units is given by
\[G = 6.67 \times 10^{-11} \ \text{N m}^2/\text{kg}^2\]
and its dimensions are
\[[G] = [\text{L}^3\text{M}^{-1}\text{T}^{-2}]\]
The word “universal” in the name of \(G\) is important: it has the same value for any two objects anywhere in the universe — two marbles on a table, or two galaxies. That is a very strong claim, and it is why the measurement of \(G\) (Section 5.4) mattered so much.
The gravitational force is an attractive force and it acts along the line joining the two bodies. The forces exerted by two bodies on each other have the same magnitude but have opposite directions; they form an action–reaction pair.
The Vector Form of the Law
The magnitude of the force is not the whole story — gravitation is a vector, and in problems with several bodies we must add the forces as vectors. So we now write Eq. (5.12) in vector form.
Figure 5.7 shows two point masses \(m_1\) and \(m_2\) with position vectors \(\vec{r}_1\) and \(\vec{r}_2\) respectively from the origin O.
Diagram
Fig. 5.7 – Gravitational force between masses \(m_1\) and \(m_2\), showing position vectors \(\vec{r}_1\), \(\vec{r}_2\) from origin O and the relative position vector \(\vec{r}_{21}\).
The position vector of \(m_2\) with respect to \(m_1\) is then given by
\[\vec{r}_{21} = \vec{r}_2 - \vec{r}_1\]
Similarly
\[\vec{r}_{12} = \vec{r}_1 - \vec{r}_2 = -\vec{r}_{21}\]
and if \(r = \left|\vec{r}_{12}\right| = \left|\vec{r}_{21}\right|\), the formula for the force on \(m_2\) due to \(m_1\) can be expressed in vector form as
\[\vec{F}_{21} = G\frac{m_1 m_2}{r^2}(-\hat{r}_{21}) \qquad \text{--- (5.13)}\]
where \(\hat{r}_{21}\) is the unit vector from \(m_1\) to \(m_2\). The minus sign is the mathematical statement that gravitation is attractive: the force on \(m_2\) points opposite to the direction pointing away from \(m_1\), i.e. it points back towards \(m_1\). The force \(\vec{F}_{21}\) is directed from \(m_2\) to \(m_1\).
Similarly, the force experienced by \(m_1\) due to \(m_2\) is \(\vec{F}_{12}\)
\[\vec{F}_{12} = G\frac{m_1 m_2}{r^2}(-\hat{r}_{12}) \qquad \text{--- (5.14)}\]
\[\therefore \vec{F}_{12} = -\vec{F}_{21} \qquad \text{--- (5.15)}\]
Equation (5.15) is simply Newton’s third law written in the language of gravitation: the two forces are equal in magnitude and opposite in direction.
Force Due to Many Masses: The Principle of Superposition
This law refers to two point masses. For a collection of point masses, the force on any one of them is the vector sum of the gravitational forces exerted by all the other point masses.
As shown in Fig. 5.8, the resultant force on point mass \(m_1\) is the vector sum of forces \(\vec{F}_{12}\), \(\vec{F}_{13}\) and \(\vec{F}_{14}\) due to point masses \(m_2\), \(m_3\) and \(m_4\) respectively.
Diagram
Fig. 5.8 – Gravitational force on \(m_1\) due to a collection of masses \(m_2\), \(m_3\) and \(m_4\), showing the three force vectors \(\vec{F}_{12}\), \(\vec{F}_{13}\), \(\vec{F}_{14}\) acting at \(m_1\).
Masses \(m_2\), \(m_3\) and \(m_4\) are also attracted towards mass \(m_1\), and there is also mutual attraction between masses \(m_2\), \(m_3\) and \(m_4\), but these forces are not shown in the figure.
For \(n\) particles, the force on the \(i^{\text{th}}\) mass is
\[\vec{F}_i = \sum_{\substack{j=1 \\ j \neq i}}^{n} \vec{F}_{ij}\]
where \(\vec{F}_{ij}\) is the force on the \(i^{\text{th}}\) particle due to the \(j^{\text{th}}\) particle. The condition \(j \neq i\) is there because a particle does not exert a gravitational force on itself.
Gravitational Force Due to an Extended Object
Real objects like the Earth are not point masses. The gravitational force between an extended object like the Earth and a point mass A can be obtained by taking the vector sum of the forces on the point mass A due to each of the point masses which make up the extended object.
In general this is a difficult calculation. But we can consider the following two special cases, for which we can get a simple result. We will state the result here and show how it can be understood qualitatively.
(1) The gravitational force of attraction due to a hollow, thin spherical shell of uniform density, on a point mass situated inside it, is zero.
This can be qualitatively understood as follows. First let us consider the case when the point mass A is at the centre of the hollow thin shell, as in Fig. 5.9. In this case, as every point on the shell is equidistant from A, all points exert forces of equal magnitude on A, but the directions of these forces are different.
Diagram
Fig. 5.9 – A hollow thin spherical shell with a point mass A inside it; pairs of diametrically opposite elements of the shell shown exerting equal and opposite forces on A.
Now consider the forces on A due to two diametrically opposite points on the shell. The forces on A due to them will be of equal magnitude but will be in opposite directions, and will cancel each other. Thus forces due to all pairs of points diametrically opposite to each other will cancel, and there will be no net force on A due to the shell.
When the point object is situated elsewhere inside the shell, the situation is not so symmetric. Gravitational force varies directly with mass and inversely with the square of the distance. Some part of the shell may be closer to point A, but its mass is less. The remaining part will then have larger mass, but its centre of mass is away from A. These two effects — nearer but lighter on one side, farther but heavier on the other — balance each other exactly. However, mathematically it can be shown that the net gravitational force on A is still zero, so long as it is inside the shell.
In fact, the gravitational force at any point inside any hollow closed object of any shape is zero.
(2) The gravitational force of attraction between a hollow spherical shell or solid sphere of uniform density and a point mass situated outside is just as if the entire mass of the shell or sphere is concentrated at the centre of the shell or sphere.
Gravitational force caused by different regions of the shell can be resolved into components along the line joining the point mass to the centre, and along a direction perpendicular to this line. The components perpendicular to this line cancel each other, and the resultant force remains along the line joining the point to the centre. By mathematical calculations it can be shown to be equal to the force that would have been exerted if the entire mass of the shell was present at the centre of the shell.
It is obvious that case (2) is applicable for any uniform sphere (solid or hollow), so long as the point is outside the sphere.
5.4 Measurement of the Gravitational Constant (G)
The magnitude of the gravitational constant \(G\) can be found by measuring the force of gravitational attraction between two bodies of masses \(m_1\) and \(m_2\) separated by a certain distance ‘\(L\)’. Once \(F\), \(m_1\), \(m_2\) and the separation are known, \(G\) follows directly from Eq. (5.12).
The difficulty is entirely practical. Gravitation is by far the weakest of the fundamental forces, so the attraction between two laboratory-sized objects is fantastically small — far too small to be read on any ordinary balance, and easily swamped by air currents or a passing footstep. The measurement therefore needs an instrument that can detect a force smaller than the weight of a grain of sand. This can be measured by using the Cavendish balance.
The Construction of the Cavendish Balance
The Cavendish balance consists of a light rigid rod. It is supported at the centre by a fine vertical metallic fibre about 100 cm long.
Two small spheres \(s_1\) and \(s_2\) of lead, having equal mass \(m\) and diameter about 5 cm, are mounted at the ends of the rod, and a small mirror M is fastened to the metallic fibre, as shown in Fig. 5.11.
Diagram
Fig. 5.11 – The Cavendish balance: a light rod carrying two small lead spheres \(s_1\), \(s_2\) suspended from a fine fibre with a mirror M, and two large lead spheres \(L_1\), \(L_2\) brought close to them on opposite sides; a light beam reflected from M onto a scale.
The mirror can be used to reflect a beam of light onto a scale and thereby measure the angle through which the wire will be twisted.
The mirror is the heart of the method. If the fibre twists through a tiny angle, the mirror turns through that same tiny angle — but the reflected beam of light turns through twice that angle, and by the time it travels several metres to the scale, the spot has moved a large, easily readable distance. The instrument thus converts an invisible rotation into a visible displacement.
Setting the Experiment in Motion
Two large lead spheres \(L_1\) and \(L_2\) of equal mass \(M\) and diameter of about 20 cm are brought close to the small spheres on opposite sides, as shown in Fig. 5.11.
The big spheres attract the nearby small spheres by equal and opposite forces. Let \(\vec{F}\) be the force of attraction between a big sphere and the small sphere near to it.
Because the two forces are equal in magnitude but act in opposite directions at the two ends of the rod, they do not push the rod as a whole in any direction — the net force on the bar is zero. What they do produce is a pure torque, a twisting effect about the fibre. Hence a torque will be generated without exerting any net force on the bar.
Due to this torque the bar turns, and the suspension wire gets twisted, till the restoring torque due to the elastic property of the wire becomes equal to the gravitational torque. At that moment the rod comes to rest at a definite angle, and the experiment can be read.
The Working Equation
The gravitational force between the spherical balls is the same as if their masses are concentrated at their centres. (This is exactly the second special case established in Section 5.3, and it is what allows two large spheres to be treated as point masses.)
If \(r\) is the initial distance of separation between the centres of the big and the neighbouring small sphere, then the magnitude of the force between them is
\[F = G\frac{mM}{r^2}\]
where \(m\) is the mass of a small sphere and \(M\) the mass of a big sphere (SI unit: kg), and \(r\) is the centre-to-centre separation (SI unit: m).
If the length of the rod is L, then the magnitude of the torque arising out of these forces is
\[\tau = FL = G\frac{mM}{r^2}L \qquad \text{--- (5.16)}\]
Torque is force multiplied by the perpendicular distance from the axis; here the two forces at the two ends together twist the rod about the fibre.
At equilibrium, it is equal and opposite to the restoring torque of the twisted fibre.
\[\therefore G\frac{mM}{r^2}L = K\theta \qquad \text{--- (5.17)}\]
where \(K\) is the restoring torque per unit angle (SI unit: N m/rad) and \(\theta\) is the angle of twist (measured in radian).
The right-hand side says that the fibre resists twisting in proportion to how far it has been twisted — the same idea as a spring, but for rotation.
Finding K, and Finally G
The constant \(K\) belongs to the fibre alone and must be found separately. By applying a known torque \(\tau_1\) and measuring the corresponding angle of twist \(\alpha\), the restoring torque per unit twist can be determined as
\[K = \tau_1/\alpha\]
Thus, in the actual experiment, measuring \(\theta\) and knowing the values of \(\tau\), \(m\), \(M\) and \(r\), the value of \(G\) can be calculated from Eq. (5.17).
Rearranged, Eq. (5.17) gives
\[G = \frac{K\theta \, r^2}{mML}\]
Every quantity on the right-hand side is directly measurable in the laboratory, so \(G\) is obtained without any assumption about the Earth at all.
The gravitational constant measured in this way is found to be
\[G = 6.67 \times 10^{-11} \ \text{N m}^2/\text{kg}^2\]
5.5 Acceleration due to Gravity
We have seen in Section 5.3 that the magnitude of the gravitational force on a point object of mass \(m\) due to another point object of mass \(M\) at a distance \(r\) from it is given by the equation
\[F = G\frac{mM}{r^2}\]
This formula can be used to calculate the gravitational force on an object due to the Earth. But there is one obstacle: we know that the Earth is an extended object, not a point.
In many practical applications the Earth can be assumed to be a uniform sphere. As seen in Section 5.3, its entire mass can then be assumed to be concentrated at its centre, provided the object lies outside the Earth. This single result is what makes the whole of this section possible.
Thus, if the mass of the Earth is \(M\), that of the point object is \(m\), and the distance of the point object from the centre of the Earth is \(r\), then the force of attraction between them is given by
\[F = G\frac{Mm}{r^2}\]
Diagram
Fig. 5.12 – The Earth shown as a sphere with its entire mass M marked at the centre, and a point object of mass m at a distance r from the centre, with the force of attraction directed towards the centre.
Notice carefully, as shown in Fig. 5.12, that \(r\) is measured from the centre of the Earth and not from its surface. This is one of the commonest sources of error in numerical problems.
From Force to Acceleration
If the point object is not acted upon by any other force, it will be accelerated towards the centre of the Earth under the action of this force. Its acceleration can be calculated by using Newton’s second law, \(F = ma\).
Dividing the gravitational force by the mass of the object gives its acceleration:
Acceleration due to the gravity of the Earth
\[= G\frac{Mm}{r^2} \times \frac{1}{m}\]
\[= \frac{GM}{r^2} \qquad \text{--- (5.18)}\]
This is known as the acceleration due to gravity of the Earth and is denoted by \(g\). Its SI unit is m/s².
Look at what has happened in that one line of algebra. The mass \(m\) of the object appears in the force, but it cancels out completely in the acceleration. The heavier the object, the harder the Earth pulls it — but the harder it is also to accelerate. The two effects cancel exactly.
If the object is close to the surface of the Earth, \(r \cong R\), the radius of the Earth, and then
\[g_{\text{Earth's surface}} = \frac{GM}{R^2} \qquad \text{--- (5.19)}\]
Equation (5.19) is one of the most useful relations in this chapter. Read backwards, it says \(GM = gR^2\) — a substitution you will meet again and again in later sections, because it lets you replace the unknown product \(GM\) by two quantities that are easy to measure.
5.6 Variation in the Acceleration due to Gravity with Altitude, Depth, Latitude and Shape
In the previous section we treated \(g\) as a single fixed number, \(9.8 \ \text{m/s}^2\). That is only an approximation. The value of \(g\) is not the same everywhere on or near the Earth. It changes if you climb a mountain, if you go down a mine, if you travel from the equator towards the pole, and it is affected by the fact that the Earth is not a perfect sphere.
We now examine these four effects one by one.
Variation in g with Altitude
Consider a body of mass \(m\) on the surface of the Earth. The acceleration due to gravity on the Earth’s surface is given by,
\[g = \frac{GM}{R^2}\]
When the body is at height \(h\) above the surface of the Earth as shown in Fig. 5.13, its distance from the centre of the Earth becomes \((R + h)\) instead of \(R\). Since gravity depends on the distance from the centre, the acceleration due to gravity changes to
Diagram
Fig. 5.13 – Acceleration due to gravity at height h above the Earth’s surface: a sphere of radius R with centre O, and a mass m at a height h above the surface.
\[g_h = \frac{GM}{(R+h)^2}\]
To express this in terms of the familiar surface value \(g\), we divide by \(g\) so that the unknown quantities \(G\) and \(M\) cancel:
\[\therefore \frac{g_h}{g} = \frac{\dfrac{GM}{(R+h)^2}}{\dfrac{GM}{R^2}}\]
\[\therefore \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\]
\[\therefore g_h = \frac{g\,R^2}{(R+h)^2} \qquad \text{--- (5.20)}\]
Here \(g_h\) is the acceleration due to gravity at height \(h\) (SI unit: m/s²), \(R\) is the radius of the Earth (m), and \(h\) is the height above the surface (m).
This equation shows that the acceleration due to gravity goes on decreasing with increase in the altitude of a body from the surface of the Earth. The denominator grows as you go up, so \(g_h\) shrinks — and it does so as an inverse square, not linearly.
The Useful Approximation for Small Heights
Equation (5.20) is exact, but clumsy for everyday heights. We can simplify it. We can rewrite Eq. (5.20) as
\[g_h = \frac{g\,R^2}{R^2\left(1 + \dfrac{h}{R}\right)^2}\]
\[\therefore g_h = g\left(1 + \frac{h}{R}\right)^{-2}\]
For small altitude \(h\), i.e., for \(\dfrac{h}{R} \ll 1\), we expand this binomially and keep only the first two terms:
\[\therefore g_h \simeq g\left(1 - \frac{2h}{R}\right) \qquad \text{--- (5.21)}\]
(by neglecting higher power terms of \(\dfrac{h}{R}\) as \(\dfrac{h}{R} \ll 1\))
This expression can be used to calculate the value of \(g\) at height \(h\) above the surface of the Earth as long as \(h \ll R\).
Variation in g with Depth
The Earth can be imagined to be a sphere made of a large number of concentric uniform spherical shells. The total mass of the Earth is the combined mass of all the shells. When an object is on the surface of the Earth, it experiences the gravitational force as if the entire mass of the Earth is concentrated at its centre.
The acceleration due to gravity according to Eq. (5.19) is
\[g = \frac{GM}{R^2}\]
To handle depth, it is convenient to write the mass of the Earth in terms of its density. Assuming that the density of the Earth is uniform, it is given by
\[\rho = \frac{\text{Mass }(M)}{\text{Volume }(V)}\]
Since the volume of a sphere of radius \(R\) is \(\frac{4}{3}\pi R^3\),
\[\therefore M = \frac{4}{3}\pi R^3 \rho\]
Substituting this into \(g = GM/R^2\),
\[\therefore g = \frac{G \times \dfrac{4}{3}\pi \text{R}^3 \rho}{\text{R}^2}\]
\[\therefore g = \frac{4}{3}\pi \text{R}\rho \text{G} \qquad \text{--- (5.22)}\]
Here \(\rho\) is the (assumed uniform) density of the Earth, in kg/m³. Notice that \(g\) now appears to be directly proportional to the radius — this looks like the opposite of the inverse-square law, but there is no contradiction: as \(R\) grows, so does the mass enclosed, and the two effects together leave \(g \propto R\).
Consider a body at a point P at the depth \(d\) below the surface of the Earth as shown in Fig. 5.14.
Diagram
Fig. 5.14 – Acceleration due to gravity at depth d below the Earth’s surface: a point P at depth d, with the inner sphere of radius (R − d) drawn and the outer shaded shell shown.
Here the force on a body at P due to the material outside the inner sphere — shown by the shaded region — can be shown to cancel out due to symmetry. This is exactly the shell result established in Section 5.3: a point mass inside a uniform hollow shell feels no net gravitational force from that shell. The material above P forms just such a shell, and so contributes nothing.
The net force on P is only due to the material inside the inner sphere of radius \(\text{OP} = R - d\). Acceleration due to gravity because of this sphere is
\[g_d = \frac{GM'}{(R-d)^2}\]
where \(M'\) = volume of the inner sphere × density
\[\therefore M' = \frac{4}{3}\pi (R-d)^3 \times \rho\]
\[\therefore g_d = \frac{G \times \dfrac{4}{3}\pi (R-d)^3 \rho}{(R-d)^2}\]
\[\therefore g_d = G \times \frac{4}{3}\pi (R-d)\rho \qquad \text{--- (5.23)}\]
Dividing Eq. (5.23) by Eq. (5.22), the constants \(G\), \(\frac{4}{3}\pi\) and \(\rho\) all cancel, leaving a strikingly simple result:
\[\therefore \frac{g_d}{g} = \frac{R-d}{R}\]
\[\therefore \frac{g_d}{g} = 1 - \frac{d}{R}\]
\[\therefore g_d = g\left[1 - \frac{d}{R}\right] \qquad \text{--- (5.24)}\]
This equation gives the acceleration due to gravity at depth \(d\) below the Earth’s surface. It shows that the acceleration due to gravity decreases with depth.
Special case:
At the centre of the Earth, where \(d = R\), Eq. (5.24) gives \(g_d = 0\).
Hence, a body of mass \(m\), if taken to the centre of the Earth, will not experience the force of gravity due to the Earth. This can also be understood to be due to symmetry. The case is similar to the force of gravity on an object placed at the centre of a spherical shell as seen in Section 5.3 — at the centre, every bit of the Earth is matched by an equal bit diametrically opposite, and all the pulls cancel.
The Complete Picture: How g Varies with Distance from the Centre
Thus, the value of acceleration due to gravity is maximum at the surface of the Earth. The value goes on decreasing with
- increase in depth below the Earth’s surface (varies linearly with \((R - d) = r\))
- increase in height above the Earth’s surface (varies inversely with \((R + h)^2 = r^2\))
Graphically, the variation of acceleration due to gravity according to depth and height can be expressed as follows. We have plotted the value of \(g\) as a function of \(r\), the distance from the centre of the Earth, in Fig. 5.15.
For \(r < R\), i.e. below the surface of the Earth, we use Eq. (5.24), according to which
\[g_d = g\left(1 - \frac{d}{R}\right)\]
Writing \(R - d = r\), the distance from the centre of the Earth, we get the value of \(g\) as a function of \(r\),
\[g(r) = g\frac{r}{R}\]
which is the equation of a straight line with slope \(g/R\) and passing through the origin.
Diagram
Fig. 5.15 – Variation of g due to depth and altitude from the Earth’s surface: a straight line rising from the origin to a peak at r = R, then falling as an inverse-square curve for r > R.
For \(r > R\) we have to use Eq. (5.20). Writing \(R + h = r\) we have
\[g(r) = \frac{g\,\text{R}^2}{r^2}\]
which is plotted in Fig. 5.15.
Notice in Fig. 5.15 that the two curves meet at the surface, \(r = R\), where \(g\) has its peak value. Inside, the graph is a straight line; outside, it is a decaying curve. The graph never becomes zero on the right — it only approaches zero as \(r\) becomes very large, which is the infinite-range property of gravitation we noted in Section 5.1.
Variation in g with Latitude and Rotation of the Earth
So far we have assumed the Earth to be at rest. It is not — it spins, and this spin steals a part of gravity.
Latitude is an angle made by the radius vector of any point from the centre of the Earth with the equatorial plane. Obviously it ranges from \(0^\circ\) at the equator to \(90^\circ\) at the poles.
Diagram
Fig. 5.16 – Variation of g with latitude: the Earth rotating with angular velocity ω, a point P at latitude θ moving in a dotted circle of radius r about the point O′ on the polar axis.
The Earth rotates about its polar axis from west to east with uniform angular velocity \(\omega\). Hence every point on the surface of the Earth (except the poles) moves in a circle parallel to the equator.
The motion of a mass \(m\) at point P on the Earth is shown by the dotted circle with centre at O′ in Fig. 5.16. Let the latitude of P be \(\theta\) and the radius of the circle be \(r\).
\[\text{PO}' = r\]
\[\angle \text{EOP} = \theta, \quad \text{E being a point on the equator}\]
\[\therefore \angle \text{OPO}' = \theta\]
In \(\Delta\) OPO′,
\[\cos\theta = \frac{\text{PO}'}{\text{PO}} = \frac{r}{R}\]
\[\therefore r = R\cos\theta\]
This is an important geometrical step: the circle in which P travels is not of radius \(R\), but of the smaller radius \(R\cos\theta\). As you move towards the pole, this circle shrinks to nothing.
Since P is moving in a circle, it needs a centripetal acceleration. The centripetal acceleration for the mass \(m\), directed along PO′ (that is, towards the axis of rotation, not towards the centre of the Earth), is
\[a = r\omega^2\]
\[a = R\omega^2\cos\theta\]
The component of this centripetal acceleration along PO, i.e., towards the centre of the Earth, is
\[a_r = a\cos\theta\]
\[\therefore a_r = R\omega^2\cos\theta \cdot \cos\theta\]
\[a_r = R\omega^2\cos^2\theta\]
Part of the gravitational force of attraction on P acting towards PO is utilized in providing this component of centripetal acceleration. In other words, gravity has two jobs at once: to hold you down, and to keep you turning with the Earth. What is left over after the turning is paid for is what you actually feel as your weight.
Thus the effective force of gravitational attraction on \(m\) at P can be written as
\[mg' = mg - mR\omega^2\cos^2\theta\]
\(g'\) being the effective acceleration due to gravity at P, i.e., at latitude \(\theta\). This is thus given by
\[g' = g - R\omega^2\cos^2\theta \qquad \text{--- (5.25)}\]
As the value of \(\theta\) increases, \(\cos\theta\) decreases. Therefore \(g'\) will increase as we move away from the equator towards any pole, due to the rotation of the Earth.
Special case I: At the equator, \(\theta = 0\)
\[\cos\theta = 1\]
\[g' = g - R\omega^2\]
The effective acceleration due to gravity is minimum at the equator, as here it is reduced by the maximum amount. The reduction here is \(g - g' = R\omega^2\).
\(R = 6.4 \times 10^6\) m — Radius of the Earth, and
\(\omega\) = Angular velocity of rotation of the Earth. Since the Earth completes one rotation in 24 hours,
\[\omega = \frac{2\pi}{T}\]
\[\therefore \omega = \frac{2\pi}{24 \times 60 \times 60}\]
\[\therefore \omega = 7.275 \times 10^{-5}\ \text{s}^{-1}\]
\[\therefore g - g' = R\omega^2\]
\[\therefore g - g' = 0.03386 \ \text{m/s}^2\]
Case II: At the poles, \(\theta = 90^\circ\)
\[\cos\theta = 0\]
\[\therefore g' = g - R\omega^2\cos\theta\]
\[= g - 0\]
\[= g\]
There is no reduction in acceleration due to gravity at the poles, due to the rotation of the Earth, as the poles are lying on the axis of rotation and do not revolve. A person standing exactly at the pole simply spins on the spot, and needs no centripetal force at all.
Variation of \(g\) with latitude at sea level is given in the following table.
Table 5.2: Variation of g with latitude
| Latitude (°) | g (m/s²) |
|---|---|
| 0 | 9.7804 |
| 10 | 9.7819 |
| 20 | 9.7864 |
| 30 | 9.7933 |
| 40 | 9.8017 |
| 50 | 9.8107 |
| 60 | 9.8192 |
| 70 | 9.8261 |
| 80 | 9.8306 |
| 90 | 9.8322 |
Effect of the Shape of the Earth
Quite often we assume the Earth to be a sphere. However, it is actually an ellipsoid, bulged at the equator. Hence the equatorial radius of the Earth (6378 km) is greater than the polar radius (6356 km).
Thus, on the equator, there are two effects acting together to reduce the force of gravity: the greater radius (which reduces \(g\) through the inverse-square dependence) and the rotation (which subtracts \(R\omega^2\)). Both work in the same direction.
As a result, the acceleration due to gravity on the equator is \(g_E = 9.7804 \ \text{m/s}^2\) and on the poles it is \(g = 9.8322 \ \text{m/s}^2\). These are precisely the first and last entries of Table 5.2.
Weight and Its Variation
Weight of an object is the force with which the Earth attracts that object.
As the value of \(g\) changes with altitude, depth and latitude, the weight also changes. Weight of an object is minimum at the equator. Similarly, the weight of an object reduces with increasing height above the Earth’s surface and with increasing depth below its surface.
Mass, on the other hand, remains exactly the same throughout. A 60 kg student is 60 kg at the pole, at the equator, on a mountain and in a mine — but their weight in newtons differs slightly in each case.
5.7 Gravitational Potential and Potential Energy
In earlier standards, you have studied potential energy as the energy possessed by an object on account of its position or configuration.
The word configuration corresponds to the distribution of the particles in the object. Two objects made of exactly the same particles can have different configurations if those particles are arranged, or spaced, differently — a coiled spring and a stretched spring are the same spring in two different configurations.
More specifically, potential energy is the work done against conservative force (or forces) in achieving a certain position or configuration of a given system. It always depends upon the relative positions of the particles in that system.
Read that definition carefully, because two words in it decide everything that follows. The first is against — potential energy is the work you have to do to fight the force. The second is system — potential energy never belongs to a single lonely object.
The Principle of Minimum Potential Energy
There is a universal principle that governs how systems settle down.
Obviously, in order to change the configuration, you will have to do work. A system left to itself will never spontaneously climb to a higher potential energy; it must be pushed there by an external agent.
Examples
(I) The spring. A spring in its natural state possesses minimum potential energy. Whenever we stretch it or compress it, we perform work against the conservative force — in this case, the elastic restoring force.
Due to this work, the relative distances between the particles of the system change (the configuration changes) and its potential energy increases. The spring finally regains its original configuration of minimum potential energy on removal of the applied force. Left alone, it snaps back — because that is the state of least potential energy available to it.
(II) The object and the Earth. When an object is lying on the Earth, the system of that object and the Earth has minimum potential energy. This is the gravitational potential energy of the system, as these two are bound by the gravitational force.
While lifting the object to some height (a new position), we do work against the conservative gravitational force in order to achieve the new position.
Diagram
Fig. 5.17 – Left: a spring in its natural, stretched and compressed states, showing minimum potential energy in the natural state. Right: an object at rest on the ground and the same object held at a height h, with the lifting force shown acting against gravity.
In its new position, the object is at rest due to balanced forces. If you are holding the object, the force of static friction between the object and your fingers balances the gravitational force. If it is kept on a surface, the normal reaction force given by the surface balances the gravitational force.
However, now the object has a capacity to acquire kinetic energy, when given an opportunity — that is, when it is allowed to fall. We call this increase in the capacity the potential energy gained by the system.
As we raise it more and more, this capacity, and hence the potential energy of the system, increases. It falls back on the Earth to achieve the configuration of minimum potential energy on dropping it from the new position.
The General Relation
Thus, in general, we can write:
In symbols,
\[\therefore \vec{F} \cdot d\vec{x} = dU\]
Here \(dU\) is the change in potential energy while displacing the object through \(d\vec{x}\), \(\vec{F}\) being the force acting on the object. The SI unit of potential energy is the joule (J), and the dot product tells us that only the component of the force along the displacement contributes.
Potential Energy Belongs to the System
It should be remembered that potential energy is always of the system as a whole.
For an object on the Earth, it is the potential energy of the system of the object and the Earth — and not of that object alone. When we casually say “the potential energy of the ball,” we are using a shorthand for “the potential energy of the ball–Earth system.”
There is no meaning to the potential energy of an isolated object in intergalactic (gravity-free) space, in the absence of any conservative force acting upon it. Take away the Earth, and the ball’s “potential energy” simply ceases to mean anything, because there is no longer any conservative force to do work against and no second body whose relative position matters.
This section is the foundation for what follows. In Section 5.7.1 we will use exactly this idea — work done against the gravitational force — to derive an exact expression for gravitational potential energy that is valid at any distance, not merely near the Earth’s surface.
5.7.1 Expression for Gravitational Potential Energy
Work done against gravitational force \(\vec{F}_g\), in displacing an object through a small displacement \(d\vec{r}\), appears as increase in the potential energy of the system.
\[\therefore dU = -\vec{F}_g \cdot d\vec{r}\]
The negative sign appears because \(dU\) is the work done by us (the external agent) against the gravitational force \(\vec{F}_g\). We must pull outward while gravity pulls inward, so the work we do is the negative of the work gravity does.
For displacement of the object from an initial position \(\vec{r}_i\) to the final position \(\vec{r}_f\), the change in potential energy \(\Delta U\) can be obtained by integrating \(dU\). A single small step gives \(dU\); adding up all the small steps between the two positions means integrating.
\[\therefore \Delta U = \int_{r_i}^{r_f} \left(dU\right) = \int_{r_i}^{r_f} \left(-\vec{F}_g \cdot d\vec{r}\right)\]
Putting in the Gravitational Force
Gravitational force of the Earth is
\[\vec{F}_g = -\frac{GMm}{r^2}\hat{r}\]
where \(\hat{r}\) is the unit vector in the direction of \(\vec{r}\). The negative sign appears here because \(\vec{r}\) is drawn from the centre of the Earth to the object, while \(\vec{F}_g\) is directed towards the centre of the Earth — the two point in opposite directions.
Here \(M\) is the mass of the Earth and \(m\) the mass of the object (SI unit: kg), \(r\) is the distance of the object from the centre of the Earth (SI unit: m), and \(\vec{F}_g\) is measured in newton (N).
\(\therefore\) For the ‘Earth and mass’ system,
\[\Delta U = \int_{r_i}^{r_f} dU = \int_{r_i}^{r_f} -\left(-\frac{GMm}{r^2}\hat{r}\right) \cdot d\vec{r}\]
\[= GMm\int_{r_i}^{r_f}\left(\frac{dr}{r^2}\right) \quad \text{as } d\vec{r} \text{ is along } \hat{r}\]
The two negative signs have cancelled, and because the displacement is radially outward, \(\hat{r} \cdot d\vec{r} = dr\), so the dot product disappears and we are left with an ordinary integral.
Using \(\displaystyle\int \frac{dr}{r^2} = -\frac{1}{r}\),
\[= GMm\left(-\frac{1}{r}\right)_{r_i}^{r_f}\]
\[= GMm\left(\frac{1}{r_i} - \frac{1}{r_f}\right) \qquad \text{--- (5.26)}\]
Equation (5.26) is exact. It works whether the object is lifted by one metre or carried to the Moon.
Where Should Zero Potential Energy Be?
Change in potential energy corresponds to the work done against conservative forces. The absolute value of potential energy is not defined — only differences in potential energy have physical meaning. We are therefore free to choose a reference point where we agree to call the potential energy zero.
It is logical as well as convenient to choose the point of zero potential energy to be the point of zero force. For the gravitational force, such a point is taken at \(r = \infty\), because the force \(GMm/r^2\) dies away to zero only at infinite separation.
This point should be chosen as the initial point, so that initially the potential energy is zero.
\[\therefore U(r_i) = 0 \text{ at } r_i = \infty\]
The final point \(r_f\) is obviously the point where we need to determine the potential energy of the system.
\[\therefore r_f = r\]
Substituting these into Eq. (5.26),
\[\therefore U(\text{r})_{grav.} = GMm\left(\frac{1}{r_i} - \frac{1}{r_f}\right)\]
\[= GMm\left(\frac{1}{\infty} - \frac{1}{r}\right)\]
Since \(\dfrac{1}{\infty} = 0\),
\[= -\frac{GMm}{r} \qquad \text{--- (5.27)}\]
This is the gravitational potential energy of the system of an object of mass \(m\) and the Earth of mass \(M\) having separation \(r\) (between their centres of mass).
Diagram
Fig. 5.18 – A graph of gravitational potential energy U against distance r from the centre of the Earth, showing a curve rising from a large negative value near the surface and approaching zero asymptotically as r increases.
Making Sense of the Negative Sign
Students often find Eq. (5.27) troubling: how can energy be negative?
The negative sign does not mean the system has “less than nothing.” It is a direct consequence of our choice of zero at infinity, as Fig. 5.18 shows. Bringing the object in from infinity to a distance \(r\) means gravity does positive work on it, so the system’s energy drops below the value it had at infinity — which we called zero. Hence it becomes negative.
5.7.2 Connection of Potential Energy Formula with mgh
In earlier classes you always wrote the gravitational potential energy of a lifted body as \(mgh\). In Section 5.7.1 we derived a completely different-looking expression, \(-GMm/r\). Both cannot be independent truths — one must follow from the other. We now show that \(mgh\) is simply the special case of the exact formula, valid only near the Earth’s surface.
If the object is on the surface of the Earth, \(r = R\)
\[\therefore U_1 = -\frac{GMm}{R}\]
If the object is lifted to a height \(h\) above the surface of the Earth, its distance from the centre becomes \(R + h\), and the potential energy becomes
\[U_2 = -\frac{GMm}{R+h}\]
The increase in the potential energy is given by
\[\Delta U = U_2 - U_1\]
\[= GMm\left(-\frac{1}{R+h} - \left[-\frac{1}{R}\right]\right)\]
Taking the two fractions to a common denominator, \(\dfrac{1}{R} - \dfrac{1}{R+h} = \dfrac{(R+h) - R}{R(R+h)} = \dfrac{h}{R(R+h)}\), we get
\[= GMm\left(\frac{h}{R(R+h)}\right)\]
\[= \frac{GMmh}{R(R+h)}\]
Bringing g into the Picture
The expression above contains \(G\) and \(M\), which are awkward to carry around. We eliminate them using the surface relation established in Section 5.5.
If \(g\) is the acceleration due to the Earth on the surface of the Earth, then from Eq. (5.19), \(GM = gR^2\). Substituting this,
\[\therefore \Delta U = mgh\left(\frac{R}{R+h}\right) \qquad \text{--- (5.28)}\]
Eq. (5.28) gives the work to be done (or the energy to be supplied) to raise an object of mass \(m\) to a height \(h\) above the surface of the Earth. Here \(m\) is in kg, \(g\) in m/s², and \(h\) and \(R\) in m, so that \(\Delta U\) comes out in joule (J).
Notice the structure of this result. It is the familiar \(mgh\), multiplied by a correction factor \(\dfrac{R}{R+h}\), which is always slightly less than 1. The exact answer is therefore always a little smaller than \(mgh\) — which makes sense, since \(g\) itself weakens as you rise.
The Familiar Special Case
If \(h \ll R\), we can use \(R + h \cong R\), so the correction factor becomes 1. Only in this case
\[\Delta U = mgh \qquad \text{--- (5.29)}\]
Thus, \(mgh\) is the increase in the gravitational potential energy of the Earth–mass system if an object of mass \(m\) is lifted to a height \(h\), provided \(h\) is negligible compared to the radius of the Earth (up to a few kilometres).
5.7.3 Concept of Potential
From Eq. (5.27), the gravitational potential energy of the system of the Earth and any mass \(m\) at a distance \(r\) from the centre of the Earth is given by
\[U = -\frac{GMm}{r}\]
Now separate the mass of the object from everything else in this expression:
\[= \left(-\frac{GM}{r}\right)m\]
\[= \left[\left(V_E\right)_r\right]m \qquad \text{--- (5.30)}\]
Look closely at what has been isolated. The factor
\[-\frac{GM}{r} = \left(V_E\right)_r\]
depends only upon the mass of the Earth and the location — the distance \(r\). It contains no reference whatsoever to the object placed there.
Thus, it is the same for any mass \(m\) bound to the Earth. A 1 kg stone and a 1000 kg satellite at the same distance from the Earth’s centre have very different potential energies, but they sit at exactly the same potential.
Conveniently, this quantity is defined as the gravitational potential of the Earth at distance \(r\) from its centre.
In terms of potential, we can write the potential energy of the Earth–mass system as
Gravitational potential energy, \(U\) = Gravitational potential \(V_r\) × mass \(m\)
or, equivalently, gravitational potential is gravitational potential energy per unit mass, i.e.,
\[V_r = \frac{U}{m}\]
The concept of potential can be defined on similar lines for any conservative force field.
Diagram
Fig. 5.19 – Concentric spherical surfaces around the Earth, each labelled with a constant value of gravitational potential, becoming less negative as the radius increases.
Since \(V_r\) depends only on \(r\), all points at the same distance from the centre of the Earth have the same potential, as shown by the spheres in Fig. 5.19. Moving along such a surface requires no work against gravity at all.
Gravitational Potential Difference
Gravitational potential difference between any two points in a gravitational field can be written as
\[V_2 - V_1 = \left(\frac{U_2 - U_1}{m}\right) = \frac{dW}{m} \qquad \text{--- (5.31)}\]
= Work done (or change in potential energy) per unit mass
In words: the potential difference between two points is the work that must be done against the gravitational force to carry a unit mass from the first point to the second. Its SI unit, again, is J/kg.
Potential for Any Two Masses
There is nothing special about the Earth in this argument. In general, for a system of any two masses \(m_1\) and \(m_2\), separated by \(r\), we can write
Gravitational potential energy,
\[U = -\frac{Gm_1m_2}{r} = \left(V_1\right)m_2 = \left(V_2\right)m_1 \qquad \text{--- (5.32)}\]
Here \(V_1\) and \(V_2\) are the gravitational potentials at \(r\) due to \(m_1\) and \(m_2\) respectively.
Equation (5.32) says something rather elegant: you may compute the potential energy of the pair either by taking the potential set up by \(m_1\) and multiplying it by \(m_2\), or by taking the potential set up by \(m_2\) and multiplying it by \(m_1\). Both give the same answer — as they must, because the potential energy belongs to the system as a whole, and not to either mass individually.
5.7.4 Escape Velocity
When any object is thrown vertically up, it falls back to the Earth after reaching a certain height. The higher the speed with which the object is thrown up, the greater will be the height it reaches.
If we keep on increasing the velocity, a stage will come when the object will reach heights so large that it will escape the gravitational field of the Earth and will not fall back on the Earth. This initial velocity is called the escape velocity.
Obviously, as the gravitational force due to the Earth becomes zero only at infinite distance, the object has to reach infinite distance in order to escape. This is the exact condition we must translate into an equation.
Setting Up the Energy Equation
Let us consider the kinetic and potential energies of an object thrown vertically upwards with escape velocity \(v_e\), when it is at the surface of the Earth and when it reaches infinite distance.
On the surface of the Earth,
\[K.E. = \frac{1}{2}m\text{v}_e^2\]
\[P.E. = -\frac{GMm}{R}\]
Here \(m\) is the mass of the object (kg), \(M\) the mass of the Earth (kg), \(R\) the radius of the Earth (m), and the energies are in joule (J). The potential energy is the one obtained in Section 5.7.1, evaluated at \(r = R\).
Total energy = \(P.E. + K.E.\)
\[= \frac{1}{2}m\text{v}_e^2 - \frac{GMm}{R} \qquad \text{--- (5.33)}\]
The kinetic energy of the object will go on decreasing with time as it is pulled back by the Earth’s gravitational force. It will become zero when it reaches infinity — and “just barely escaping” means arriving at infinity with nothing left over.
Thus at infinite distance from the Earth
\[K.E. = 0\]
Also,
\[P.E. = -\frac{GMm}{\infty} = 0\]
\[\therefore \text{Total energy} = P.E. + K.E. = 0\]
Solving for the Escape Velocity
As energy is conserved (gravitation is a conservative force, so no energy is lost on the way), the total energy at the surface must equal the total energy at infinity, which we have just shown to be zero.
\[\frac{1}{2}m\text{v}_e^2 - \frac{GMm}{R} = 0\]
\[\text{or,} \quad \text{v}_e = \sqrt{\frac{2GM}{R}} \qquad \text{--- (5.34)}\]
Using the numerical values of \(G\), \(M\) and \(R\), the escape velocity is 11.2 km/s.
Notice at once what has vanished from Eq. (5.34): the mass \(m\) of the object. It cancelled out when we divided through. A pebble, a cricket ball and a rocket all need exactly the same escape speed of 11.2 km/s. What differs is the energy needed to give them that speed, which does depend on mass.
Notice also what the formula contains: only \(G\), and the mass and radius of the body being escaped from. Escape velocity is a property of the planet, not of the projectile.
Since \(GM = gR^2\) from Section 5.5, Eq. (5.34) can also be written in the very useful form
\[\text{v}_e = \sqrt{2gR}\]
This version is handy in numerical problems where \(g\) and \(R\) are given but \(M\) is not.
5.8 Earth Satellites
The objects which revolve around the Earth are called Earth satellites.
The Moon is the only natural satellite of the Earth. It revolves in almost a circular orbit around the Earth with a period of revolution of nearly 27.3 days — the same figure Newton used in Section 5.3 when he compared the fall of the Moon with the fall of a stone.
Artificial satellites have been launched by several countries including India. These satellites have different periods of revolution according to their practical use, such as navigation, surveillance, communication, looking into space, and monitoring the weather.
The period is not chosen freely. As we shall prove in Section 5.8.3, the period of a satellite is fixed entirely by the radius of its orbit. So the moment a designer decides what the satellite is for, the height of its orbit — and therefore its period — is decided as well. The two main families of satellites below show this clearly.
Diagram
Fig. 5.20 – The Earth with a geostationary satellite in a large circular orbit in the equatorial plane, and a polar satellite in a small circular orbit passing over both poles in a north–south direction.
Communication Satellites
These are geostationary satellites. They revolve around the Earth in the equatorial plane. They have the same sense of rotation as that of the Earth and the same period of rotation as that of the Earth, i.e., one day or 24 hours.
Due to this, they appear stationary from the Earth’s surface. Hence they are called geostationary satellites or geosynchronous satellites.
The word “stationary” needs care. Such a satellite is not standing still — it is racing around the Earth at thousands of metres per second. But because the ground beneath it is turning at exactly the same angular rate, the satellite always stays above the same point on the equator. Relative to an observer on the Earth, it never moves.
Geostationary satellites are used for communication, television transmission, telephones and radiowave signal transmission — for example, the INSAT group of satellites launched by India.
Polar Satellites
These satellites are placed in lower polar orbits. They are at a low altitude of 500 km to 800 km.
Polar satellites are used for weather forecasting and meteorological purposes. They are also used for astronomical observations and the study of solar radiations.
The period of revolution of a polar satellite is nearly 85 minutes, so it can orbit the Earth 16 times per day. (The reason for this much shorter period is that the orbit is much smaller — a point that Eq. (5.36) will make quantitative in Section 5.8.3.)
They go around the poles of the Earth in a north–south direction, while the Earth rotates in an east–west direction about its own axis. This combination is the clever part of the design. The satellite keeps sweeping over the poles along the same fixed plane, but because the Earth is turning underneath it, a fresh strip of the surface passes below on every orbit.
The polar satellites have cameras fixed on them. The camera can view small stripes of the Earth in one orbit. In an entire day the whole Earth can be viewed strip by strip. Polar and equatorial regions at close distances can be viewed by these satellites.
5.8.1 Projection of Satellite
For the projection of an artificial satellite, it is necessary for the satellite to have a certain velocity and a minimum two-stage rocket. A single-stage rocket cannot achieve this.
Why not? The reasoning is worth following carefully, because it explains the peculiar shape of every launch you have seen on television.
When the fuel in the first stage of the rocket is ignited on the surface of the Earth, it raises the satellite vertically. The velocity of projection of the satellite normal to the surface of the Earth is the vertical velocity.
If this vertical velocity is less than the escape velocity (\(v_e\)), the satellite returns to the Earth’s surface. It simply goes up, slows down, stops and falls back — exactly like a stone thrown upward.
While, if the vertical velocity is greater than or equal to the escape velocity, the satellite will escape from the Earth’s gravitational influence and go to infinity. It never comes back at all.
So a purely vertical launch offers only two outcomes — fall back, or fly away for ever. Neither of them is an orbit. Hence launching of a satellite in an orbit round the Earth cannot take place by use of a single-stage rocket. It requires a minimum two-stage rocket.
The Two Stages
With the help of the first stage of the rocket, the satellite can be taken to a desired height above the surface of the Earth.
Then the launcher is rotated in the horizontal direction, i.e. through \(90^\circ\), using remote control, and the first stage of the rocket is detached.
Then, with the help of the second stage of the rocket, a specific horizontal velocity (\(v_h\)) is given to the satellite so that it can revolve in a circular path round the Earth.
The exact horizontal velocity of projection that must be given to a satellite at a certain height so that it can revolve in a circular orbit round the Earth is called the critical velocity or orbital velocity (\(v_c\)).
The Five Possible Orbits
A satellite follows different paths depending upon the horizontal velocity provided to it. Four different possible cases are shown in Fig. 5.21.
Diagram
Fig. 5.21 – Various possible orbits from a point of projection P depending on the value of \(v_h\): an ellipse with P as apogee, a circle, an ellipse with P as perigee, a parabola, and a hyperbola.
Notice in Fig. 5.21 that all five paths start from the same point of projection and differ only in the horizontal speed given there. The whole future of the satellite is decided in that one instant.
Case (I) \(v_h < v_c\):
If the tangential velocity of projection \(v_h\) is less than the critical velocity, the orbit of the satellite is an ellipse with the point of projection as apogee (farthest from the Earth) and the Earth at one of the foci.
The satellite has been given too little sideways speed to hold its height, so it falls inward — but it still misses the Earth and swings back out, tracing an ellipse.
During this elliptical path, if the satellite passes through the Earth’s atmosphere, it experiences a nonconservative force of air resistance. As a result it loses energy and spirals down to the Earth. (Note the word nonconservative: unlike gravity, air resistance permanently removes mechanical energy from the system, so the orbit cannot repeat itself.)
Case (II) \(v_h = v_c\):
If the horizontal velocity is exactly equal to the critical velocity, the satellite moves in a stable circular orbit round the Earth. This is the case aimed for in almost every launch.
Case (III) \(v_c < v_h < v_e\):
If the horizontal velocity is greater than the critical velocity and less than the escape velocity at that height, the satellite again moves in an elliptical orbit round the Earth, with the point of projection as perigee (point closest to the Earth).
Compare this with Case (I). The path is again an ellipse, but the roles have reversed: too much speed now carries the satellite outward from the launch point, so the launch point becomes the nearest point instead of the farthest.
Case (IV) \(v_h = v_e\):
If the horizontal speed of projection is equal to the escape speed at that height, the satellite travels along a parabolic path and never returns to the point of projection. Its speed will be zero at infinity — this is the borderline case, in which it just barely escapes with nothing to spare.
Case (V) \(v_h > v_e\):
If the horizontal velocity is greater than the escape velocity, the satellite escapes from the gravitational influence of the Earth, traversing a hyperbolic path. It escapes with speed still left over at infinity.
Expression for Critical Speed
Consider a satellite of mass \(m\) revolving round the Earth at height \(h\) above its surface.
Let \(M\) be the mass of the Earth and \(R\) be its radius. If the satellite is moving in a circular orbit of radius \((R + h) = r\), its speed must be the magnitude of the critical velocity \(v_c\).
Now we ask the key question: what keeps the satellite in its circular path? Any body moving in a circle needs a centripetal force directed towards the centre. For a satellite there is only one force available, and that is gravity.
The centripetal force necessary for the circular motion of the satellite is provided by the gravitational force exerted by the Earth on the satellite.
\[\therefore \text{Centripetal force} = \text{Gravitational force}\]
\[\frac{m\text{v}_c^2}{r} = \frac{GMm}{r^2}\]
The mass \(m\) of the satellite appears on both sides, and one power of \(r\) cancels:
\[\therefore \text{v}_c^2 = \frac{GM}{r}\]
\[\therefore \text{v}_c = \sqrt{\frac{GM}{r}}\]
\[\therefore \text{v}_c = \sqrt{\frac{GM}{(R+h)}} = \sqrt{g_h(R+h)} \qquad \text{--- (5.35)}\]
(The last form follows from Section 5.6: at height \(h\), the acceleration due to gravity is \(g_h = GM/(R+h)^2\), so \(GM = g_h(R+h)^2\).)
Here \(v_c\) is in m/s, \(M\) is the mass of the Earth in kg, \(R\) and \(h\) are in m, and \(g_h\) is the acceleration due to gravity at that height, in m/s².
This is the expression for critical speed in the orbit of radius \((R + h)\).
What the Formula Tells Us
It is clear that the critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is revolving, or the gravitational acceleration at that altitude.
The critical speed of a satellite decreases with increase in the height of the satellite. This may seem odd at first — a higher satellite moves slower, not faster. But the reason is plain from Eq. (5.35): further away, gravity is weaker, so less centripetal force is available, and only a gentler speed can be sustained in a circle.
Special Case: Satellite Close to the Earth’s Surface
When the satellite is revolving close to the surface of the Earth, the height is very small as compared to the radius of the Earth. Hence the height can be neglected and the radius of the orbit is nearly equal to \(R\) (i.e. \(R \gg h\), \(R + h \approx R\))
\[\therefore \text{Critical speed } \text{v}_c = \sqrt{\frac{GM}{R}}\]
As \(G\) is related to acceleration due to gravity by the relation,
\[g = \frac{GM}{R^2}\]
\[\therefore GM = gR^2\]
\(\therefore\) Critical speed in terms of acceleration due to gravity can be obtained as
\[\text{v}_c = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}\]
\[= 7.92 \ \text{km/s}\]
Obviously, this is the maximum possible critical speed, because any real orbit must lie at some height above the surface, and we have just seen that critical speed falls as height rises. This is at least 25 times the speed of the fastest passenger aeroplanes.
5.8.2 Weightlessness in a Satellite
According to Newton’s second law of motion, \(F = ma\), where \(F\) is the net force acting on an object having acceleration \(a\).
To understand why an astronaut floats inside an orbiting satellite, we do not begin in space at all. We begin with something far more familiar — a lift.
Let us consider the example of a lift or elevator from an inertial frame of reference.
The Two Forces on a Passenger in a Lift
Whether the lift is at rest or in motion, a passenger in it experiences only two forces:
- Gravitational force \(mg\), directed vertically downwards (towards the centre of the Earth), and
- Normal reaction force \(N\), directed vertically upwards, exerted by the floor of the lift.
As these forces are oppositely directed, the net force in the downward direction will be \(F = mg - N\).
Diagram
Fig. 5.22 – A passenger standing on a weighing machine inside a lift, showing the downward gravitational force mg and the upward normal reaction N, drawn for the cases of zero acceleration, upward acceleration and downward acceleration.
Now comes the central idea of this section, and it is one that students often miss.
Though the weight of a body (the passenger, in this case) is the gravitational force acting upon it, we experience or feel our weight only due to the normal reaction force \(N\) exerted by the floor. This, in turn, is equal and opposite to the relative force between the body and the lift.
If you are standing on a weighing machine in a lift, the force recorded by the weighing machine is nothing but the normal reaction \(N\).
Case I: Lift Having Zero Acceleration
This happens when the lift is at rest, or is moving upwards or downwards with constant velocity.
The net force
\[F = 0 = mg - N \quad \therefore \quad mg = N\]
Hence in this case we feel our normal weight \(mg\).
Note carefully that moving at constant speed — even fast — changes nothing. It is acceleration, not velocity, that alters the reading.
Case II: Lift Having Net Upward Acceleration \(a_u\)
This happens when the lift just starts moving upwards, or is about to stop at a lower floor during its downward motion (remember, while stopping during downward motion, the acceleration must be upwards).
As the net acceleration is upwards, the upward force must be greater.
\[\therefore F = ma_u = N - mg \quad \therefore \quad N = mg + ma_u\]
i.e., \(N > mg\), hence, we feel heavier.
It should also be remembered that this is not an apparent feeling. The weighing machine really records a reading greater than \(mg\). The extra push from the floor is physically real — it is what accelerates you upwards.
Case III (a): Lift Having Net Downward Acceleration \(a_d\)
This happens when the lift just starts moving downwards, or is about to stop at a higher floor during its upward motion (remember, while stopping during upward motion, the acceleration must be downwards).
As the net acceleration is downwards, the downward force must be greater.
\[\therefore F = ma_d = mg - N \quad \therefore \quad N = mg - ma_d\]
i.e., \(N < mg\), hence, we feel lighter.
It should be remembered that this is not an apparent feeling. The weighing machine really records a reading less than \(mg\).
Case III (b): State of Free Fall
This will be possible if the cables of the lift are cut. In this case, the downward acceleration \(a_d = g\) — the lift and everything in it fall together, freely.
If the downward acceleration becomes equal to the gravitational acceleration \(g\), we get,
\[N = mg - ma_d = 0\]
Thus, there will not be any feeling of weight. This is the state of total weightlessness, and the weighing machine will record zero.
Read that last sentence again. Weightlessness does not mean the absence of gravity. Gravity is acting on the falling passenger with its full strength — it is exactly what is producing the acceleration \(g\). What has vanished is the support, and with it the sensation of weight.
Weightlessness in an Orbiting Satellite
In the case of a revolving satellite, the satellite is performing a circular motion. The acceleration for this motion is centripetal, which is provided by the gravitational acceleration \(g\) at the location of the satellite. (Recall Section 5.8.1: the entire centripetal force comes from gravity, and nothing else.)
In this case, \(a_d = g\) — the same condition as in the lift with its cables cut. So the satellite, along with the astronaut, is in the state of free fall.
Diagram
Fig. 5.23 – A satellite in circular orbit around the Earth with an astronaut floating inside it, showing the gravitational acceleration g directed towards the Earth’s centre and the tangential velocity of the satellite perpendicular to it.
Obviously, the apparent weight will be zero, giving the feeling of total weightlessness. Perhaps you might have seen in some videos that astronauts are floating inside the satellite. It is really difficult for them to change their position, because there is nothing to push against — nothing presses on them, and they press on nothing.
The astronaut and the satellite are falling together, at the same rate, so neither exerts any force on the other. That is precisely the condition \(N = 0\).
Why Does the Satellite Not Fall Down?
In spite of free fall, why is the satellite not falling on the Earth?
The reason is that the revolving satellite is having a tangential velocity, which manages to keep it moving in a circular orbit at that height.
The satellite is indeed falling — continuously. But because it is also moving sideways very fast, the surface of the Earth curves away beneath it at exactly the rate at which the satellite falls towards it. The result is that it never gets any closer. It falls for ever, all the way around the Earth.
5.8.3 Time Period of a Satellite
The time taken by a satellite to complete one revolution round the Earth is its time period.
Consider a satellite of mass \(m\) projected to a height \(h\) and provided a horizontal velocity equal to the critical velocity. The satellite revolves in a circular orbit of radius \((R + h) = r\).
Deriving the Period
The distance traced by the satellite in one revolution is equal to the circumference of the circular orbit, within the periodic time \(T\). Since the motion is uniform circular motion, speed is simply distance divided by time:
\[\therefore \text{Critical speed} = \frac{\text{Circumference of the orbit}}{\text{Time period}}\]
\[\text{v}_c = \frac{2\pi r}{T}\]
But we have, from Eq. (5.35),
\[\text{v}_c = \sqrt{\frac{GM}{r}}\]
We now have two independent expressions for the same speed. Equating them is what produces the period.
\[\therefore \sqrt{\frac{GM}{r}} = \frac{2\pi r}{T}\]
Squaring both sides,
\[\text{or,} \quad \frac{GM}{r} = \frac{4\pi^2 r^2}{T^2}\]
Rearranging for \(T^2\),
\[\therefore T^2 = \frac{4\pi^2 r^3}{GM}\]
Here \(r = (R + h)\) is the radius of the orbit in metre, \(M\) is the mass of the Earth in kg, \(G\) is the universal gravitational constant, and \(T\) is in second.
A Familiar Law Reappears
As \(\pi^2\), \(G\) and \(M\) are constant,
\[T^2 \propto r^3\]
i.e., the square of the period of revolution of a satellite is directly proportional to the cube of the radius of the orbit.
Pause on this result. It is exactly Kepler’s third law from Section 5.2 — the law that Kepler found by staring at Tycho Brahe’s data and could not explain. Here it has emerged in three lines of algebra from Newton’s law of gravitation, and it applies not just to planets around the Sun but to satellites around the Earth. This is what it means for a law to be explained rather than merely observed.
Taking the square root,
\[T = 2\pi\sqrt{\frac{r^3}{GM}}\]
\[\therefore T = 2\pi\sqrt{\frac{(R+h)^3}{GM}} \qquad \text{--- (5.36)}\]
This is an expression for the period of a satellite revolving in a circular orbit round the Earth.
The period of a satellite does not depend on its mass. It depends on the mass of the Earth, the radius of the Earth, and the height of the satellite. If the height of projection is increased, the period of the satellite increases.
The Period in Terms of g
The period of the satellite can also be obtained in terms of acceleration due to gravity — a form that is often more convenient, since \(g\) is easier to look up than \(GM\).
As \(GM = g_h (R+h)^2\) (from Section 5.6),
\[\therefore T = 2\pi\sqrt{\frac{(R+h)^3}{g_h(R+h)^2}}\]
Cancelling \((R+h)^2\) inside the root,
\[\therefore T = 2\pi\sqrt{\frac{R+h}{g_h}}\]
\[\therefore T = 2\pi\sqrt{\frac{r}{g_h}} \qquad \text{--- (5.37)}\]
where \(g_h\) is the acceleration due to gravity at the height of the satellite (m/s²).
Special case:
When the satellite revolves close to the surface of the Earth, \(R + h \approx R\) and \(g_h \approx g\). Hence the minimum period of revolution is
\[(T)_{min} = 2\pi\sqrt{\frac{R}{g}} \qquad \text{--- (5.38)}\]
This is a minimum for the same reason that the critical speed close to the surface was a maximum: no orbit can be smaller than one skimming the Earth’s surface, and smaller orbits mean shorter periods.
5.8.4 Binding Energy of an Orbiting Satellite
The minimum energy required by a satellite to escape from the Earth’s gravitational influence is the binding energy of the satellite.
The name is well chosen. The satellite is bound to the Earth, in the same sense that an electron is bound to a nucleus, and the binding energy is the price of setting it free.
Expression for Binding Energy of a Satellite Revolving in a Circular Orbit Round the Earth
Consider a satellite of mass \(m\) revolving at height \(h\) above the surface of the Earth in a circular orbit. It possesses potential energy as well as kinetic energy — and to know whether it is bound, we must add both.
Let \(M\) be the mass of the Earth, \(R\) be the radius of the Earth, \(v_c\) be the critical velocity of the satellite, and \(r = (R + h)\) be the radius of the orbit.
Kinetic energy. The satellite is moving with the critical speed, so from Eq. (5.35), \(v_c^2 = GM/r\). Substituting this into \(\frac{1}{2}mv_c^2\):
\[\therefore \text{Kinetic energy of satellite} = \frac{1}{2}m\text{v}_c^2\]
\[= \frac{1}{2}\frac{GMm}{r} \qquad \text{--- (5.39)}\]
Potential energy. The gravitational potential at a distance \(r\) from the centre of the Earth is \(-\dfrac{GM}{r}\) (Section 5.7.3).
\(\therefore\) Potential energy of satellite = Gravitational potential × mass of satellite
\[= -\frac{GMm}{r} \qquad \text{--- (5.40)}\]
Total energy. The total energy of the satellite is given as
\[T.E. = K.E. + P.E.\]
\[= \frac{1}{2}\frac{GMm}{r} - \frac{GMm}{r}\]
\[= -\frac{1}{2}\frac{GMm}{r} \qquad \text{--- (5.41)}\]
Here \(m\) is the mass of the satellite (kg), \(M\) the mass of the Earth (kg), \(r = (R+h)\) the radius of the orbit (m), and the energies are in joule (J).
Why the Sign Matters
Total energy of a circularly orbiting satellite is negative. The negative sign indicates that the satellite is bound to the Earth, due to the gravitational force of attraction.
Look at the structure of Eq. (5.41). The potential energy is negative and twice as large in magnitude as the kinetic energy is positive. The negative term wins, and the sum comes out negative. This is not an accident of arithmetic — it is the mathematical signature of a bound orbit.
For the satellite to be free from the Earth’s gravitational influence, its total energy should become non-negative (zero or positive). Recall from Section 5.7.4 that a body which just escapes to infinity has exactly zero total energy.
Hence the minimum energy to be supplied to unbind the satellite is
\[+\frac{1}{2}\frac{GMm}{r}\]
This is the binding energy of a satellite. It is exactly the amount needed to lift the total energy from \(-\frac{1}{2}\dfrac{GMm}{r}\) up to zero.
5.9 Summary
Gravitation and Its Nature
- Every material object is attracted towards the Earth. All bodies fall with the same acceleration, independent of their mass — a fact recognised by Galileo.
- Gravitational force is the force of mutual attraction between any two objects by virtue of their masses.
- It is always attractive, has infinite range, does not depend on the intervening medium, and is much weaker than the other fundamental forces. It is \(10^{-39}\) times weaker than the strong nuclear force.
Kepler’s Laws
- Law of orbit: All planets move in elliptical orbits around the Sun, with the Sun at one of the foci of the ellipse. The closest point is the perihelion, the farthest is the aphelion; the semimajor axis is \(a\) and the semiminor axis is \(b\).
- Law of areas: The line that joins a planet and the Sun sweeps equal areas in equal intervals of time. Hence planets move faster near the Sun and slower far from it.
- The law of areas follows from the conservation of angular momentum, and is valid for any central force — a force always directed along the line joining the object to a fixed point.
\[\overrightarrow{\Delta A} = \frac{1}{2}\left(\vec{r} \times \vec{v}\,\Delta t\right), \qquad \vec{p} = m\vec{v}, \qquad \vec{L} = \vec{r} \times \vec{p}\]
\[\frac{\overrightarrow{\Delta A}}{\Delta t} = \frac{\vec{L}}{2m} = \text{constant}\]
- Law of periods: The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of its ellipse.
\[T^2 \propto r^3 \qquad \text{or} \qquad \frac{T^2}{r^3} = \text{constant}\]
Newton’s Universal Law of Gravitation
- Every particle of matter attracts every other particle of matter with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
\[F = G\frac{m_1 m_2}{r^2}\]
- \(G\) is the universal gravitational constant: \(G = 6.67 \times 10^{-11}\) N m²/kg², with dimensions \([G] = [\text{L}^3\text{M}^{-1}\text{T}^{-2}]\).
- In vector form, the force on \(m_2\) due to \(m_1\) is
\[\vec{F}_{21} = G\frac{m_1 m_2}{r^2}(-\hat{r}_{21}), \qquad \vec{F}_{12} = G\frac{m_1 m_2}{r^2}(-\hat{r}_{12}), \qquad \vec{F}_{12} = -\vec{F}_{21}\]
- Principle of superposition: the force on the \(i^{\text{th}}\) mass due to many masses is the vector sum
\[\vec{F}_i = \sum_{\substack{j=1 \\ j \neq i}}^{n} \vec{F}_{ij}\]
- Shell results: (1) The gravitational force on a point mass placed anywhere inside a uniform hollow spherical shell is zero. (2) A uniform shell or solid sphere attracts a point mass lying outside it as if its entire mass were concentrated at its centre.
Measurement of G
- \(G\) is measured with the Cavendish balance, a torsion balance in which large lead spheres twist a fine fibre carrying small lead spheres. At equilibrium, gravitational torque = restoring torque:
\[\tau = FL = G\frac{mM}{r^2}L, \qquad G\frac{mM}{r^2}L = K\theta\]
where \(K\) is the restoring torque per unit angle (\(K = \tau_1/\alpha\)) and \(\theta\) is the angle of twist.
Acceleration due to Gravity
\[g = \frac{GM}{r^2}, \qquad g_{\text{Earth's surface}} = \frac{GM}{R^2}\]
- \(g\) depends only on the properties of the Earth and the location, not on the mass of the falling object.
- The useful substitution \(GM = gR^2\) is used repeatedly throughout the chapter.
Variation of g
- With altitude:
\[g_h = \frac{g\,R^2}{(R+h)^2}, \qquad \text{and for } \frac{h}{R} \ll 1, \quad g_h \simeq g\left(1 - \frac{2h}{R}\right)\]
- With depth (uniform density \(\rho\), using \(g = \frac{4}{3}\pi R\rho G\)):
\[g_d = g\left(1 - \frac{d}{R}\right)\]
At the centre of the Earth (\(d = R\)), \(g_d = 0\).
- \(g\) is maximum at the surface; it varies linearly with \((R - d) = r\) below the surface and inversely with \((R + h)^2 = r^2\) above it.
- With latitude and rotation:
\[g' = g - R\omega^2\cos^2\theta\]
At the equator (\(\theta = 0\)), \(g' = g - R\omega^2\), and the reduction \(g - g' = 0.03386\) m/s², using \(\omega = 7.275 \times 10^{-5}\) s⁻¹ and \(R = 6.4 \times 10^6\) m. At the poles (\(\theta = 90^\circ\)), \(g' = g\) — no reduction. 5. With shape: the Earth is an ellipsoid bulged at the equator; equatorial radius 6378 km, polar radius 6356 km. Hence \(g_E = 9.7804\) m/s² at the equator and \(g = 9.8322\) m/s² at the poles. 6. Weight \(w = mg\) therefore changes with altitude, depth and latitude; it is minimum at the equator. Mass never changes.
Gravitational Potential Energy and Potential
- Potential energy is the work done against a conservative force in achieving a given position or configuration of a system. Every system tends to minimise its potential energy.
\[\vec{F} \cdot d\vec{x} = dU\]
- Gravitational potential energy of the Earth–object system, taking \(U = 0\) at \(r = \infty\):
\[\Delta U = GMm\left(\frac{1}{r_i} - \frac{1}{r_f}\right), \qquad U(r) = -\frac{GMm}{r}\]
- Connection with \(mgh\):
\[\Delta U = mgh\left(\frac{R}{R+h}\right), \qquad \text{and only if } h \ll R, \quad \Delta U = mgh\]
- Gravitational potential is potential energy per unit mass (SI unit: J/kg):
\[V_r = -\frac{GM}{r} = \frac{U}{m}, \qquad V_2 - V_1 = \frac{U_2 - U_1}{m} = \frac{dW}{m}\]
- For any two masses: \(U = -\dfrac{Gm_1m_2}{r} = (V_1)m_2 = (V_2)m_1\).
Escape Velocity
- The minimum velocity with which a body must be thrown vertically upwards from the Earth’s surface so that it escapes the Earth’s gravitational field.
\[\frac{1}{2}m\text{v}_e^2 - \frac{GMm}{R} = 0 \qquad \Rightarrow \qquad \text{v}_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}\]
- For the Earth, \(v_e = 11.2\) km/s. It is independent of the mass of the escaping body.
Earth Satellites
- Geostationary (communication) satellites revolve in the equatorial plane, in the same sense as the Earth and with the same period of 24 hours, so they appear stationary (e.g. INSAT).
- Polar satellites orbit at altitudes of 500 km to 800 km with a period of nearly 85 minutes, circling the Earth 16 times a day and scanning it strip by strip; used for weather forecasting, astronomical observation and study of solar radiation.
Projection and Critical Velocity
- A minimum two-stage rocket is needed: the first stage lifts the satellite vertically to the required height; the second stage supplies the horizontal (tangential) velocity.
- Depending on the horizontal speed \(v_h\) of projection:
- \(v_h < v_c\): ellipse with the point of projection as apogee
- \(v_h = v_c\): stable circular orbit
- \(v_c < v_h < v_e\): ellipse with the point of projection as perigee
- \(v_h = v_e\): parabolic path, speed zero at infinity
- \(v_h > v_e\): hyperbolic path, escapes
- Critical (orbital) velocity, from centripetal force = gravitational force:
\[\text{v}_c = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{(R+h)}} = \sqrt{g_h(R+h)}\]
- It is independent of the mass of the satellite and decreases as the height increases. Close to the surface, \(\text{v}_c = \sqrt{gR} = 7.92\) km/s, the maximum possible critical speed.
- For a planet of radius \(R\) and mean density \(\rho\): \(\text{v}_c = 2R\sqrt{\dfrac{G\pi\rho}{3}}\).
Weightlessness
- We feel our weight only through the normal reaction \(N\); a weighing machine records \(N\), not \(mg\).
- Lift at rest or constant velocity: \(N = mg\). Upward acceleration: \(N = mg + ma_u\) (feel heavier). Downward acceleration: \(N = mg - ma_d\) (feel lighter). Free fall (\(a_d = g\)): \(N = 0\) — total weightlessness.
- In an orbiting satellite the centripetal acceleration is supplied by \(g\) itself, so the satellite and astronaut are in free fall and \(N = 0\). Gravity is not absent.
- The satellite does not fall to the Earth because it possesses a tangential velocity that keeps it moving in its circular orbit.
Time Period of a Satellite
\[T^2 = \frac{4\pi^2 r^3}{GM} \quad \Rightarrow \quad T^2 \propto r^3\]
\[T = 2\pi\sqrt{\frac{(R+h)^3}{GM}}, \qquad T = 2\pi\sqrt{\frac{r}{g_h}}, \qquad (T)_{min} = 2\pi\sqrt{\frac{R}{g}}\]
- The period does not depend on the mass of the satellite; it increases with the height of projection.
- For a satellite close to a planet of density \(\rho\): \(T = \sqrt{\dfrac{3\pi}{G\rho}}\) — independent of the planet’s radius.
- A polar satellite close to the Earth’s surface has \(T \approx 85\) minutes.
Energy and Binding Energy of an Orbiting Satellite
\[K.E. = \frac{1}{2}\frac{GMm}{r}, \qquad P.E. = -\frac{GMm}{r}, \qquad T.E. = -\frac{1}{2}\frac{GMm}{r}\]
- The total energy is negative, which shows that the satellite is bound to the Earth.
- Binding energy is the minimum energy needed to make the total energy zero:
\[B.E. = +\frac{1}{2}\frac{GMm}{r}\]
- Hence \(B.E. = K.E. = |T.E.|\) for a satellite in a circular orbit.
5.10 Check Your Concepts
Two students are arguing. One says a heavy stone must fall faster than a light one because the Earth pulls it harder. The other disagrees. Using the expression for acceleration due to gravity, explain who is right and where exactly the first student’s reasoning breaks down.
Gravitation is by far the weakest of the four fundamental forces, yet it governs the structure of the entire universe, while the far stronger electrostatic force does not. Explain why.
Kepler’s law of areas is a consequence of the gravitational force being a central force. Would the law of areas still hold if the force between the Sun and a planet varied as \(1/r^3\) instead of \(1/r^2\)? Justify your answer.
A planet moves faster at perihelion than at aphelion. Which conservation law explains this, and why does the planet’s speed change even though no external torque acts on it?
Why must the distance \(r\) in Newton’s law of gravitation be measured from the centre of the Earth rather than from its surface? Which result of this chapter permits this?
A hollow spherical shell of uniform density is given. Explain why a point mass placed anywhere inside it experiences no net gravitational force, even when it is not at the centre.
Why is a long, fine metallic fibre — rather than a short, stiff wire — used to suspend the rod in the Cavendish balance? What would happen to the measured angle of twist if a stiffer wire were used?
The value of \(g\) decreases both as you go up above the Earth’s surface and as you go down below it, yet the two variations are described by different formulas. State both formulas and explain why the dependence is inverse-square in one case and linear in the other.
Explain why the acceleration due to gravity is least at the equator. Identify the two separate causes and state whether they act in the same direction or oppose each other.
A body is taken to the centre of the Earth. What happens to (i) its mass, (ii) its weight, (iii) the acceleration due to gravity acting on it? Explain each answer.
Gravitational potential energy is negative for all finite separations. What is the physical significance of this negative sign, and what would a positive total energy imply for a satellite?
Why is \(mgh\) an acceptable expression for the change in gravitational potential energy when lifting a book onto a shelf, but a badly wrong one when lifting a satellite into orbit?
Distinguish clearly between gravitational potential and gravitational potential energy. State the SI unit of each.
The escape velocity of a body from the Earth is 11.2 km/s and is independent of the mass of the body. Explain why the mass does not appear in the result, and why the escape velocity is nevertheless finite even though the body must travel an infinite distance.
The Moon has no atmosphere while the Earth has one. Explain this using the concept of escape velocity.
A satellite cannot be placed in an orbit around the Earth using a single-stage rocket launched vertically upwards, no matter what speed is given to it. Explain why.
A satellite is projected horizontally at a certain height with a speed slightly less than the critical velocity at that height. Describe the resulting orbit and state whether the point of projection is the apogee or the perigee.
Two satellites of different masses are placed in the same circular orbit around the Earth. Compare their (i) orbital speeds, (ii) periods of revolution, (iii) binding energies. Justify each answer.
A satellite in a higher orbit moves more slowly and hence has less kinetic energy, yet more energy must be supplied to place it there. Resolve this apparent contradiction using the expressions for kinetic, potential and total energy of an orbiting satellite.
An astronaut inside an orbiting spacecraft floats freely and a weighing machine placed under her reads zero. Does this mean the gravitational force on her is zero? Explain the state of weightlessness carefully, distinguishing true weight from apparent weight.
In spite of being in a state of continuous free fall, an orbiting satellite never lands on the Earth. Explain why.
A geostationary satellite must be placed at one particular height above the equator, and nowhere else. Explain what physically forces this choice.
A person standing on a weighing machine inside a lift observes the reading increase for a few seconds and then return to normal. Describe the motion of the lift during that interval, and write the expression for the normal reaction in that phase.
Show that the period of a satellite revolving very close to the surface of a planet depends only on the density of the planet, and not on its radius. What does this imply for two planets of very different size but equal density?
What are the dimensions of the universal gravitational constant?
What do you mean by geostationary satellite?
On which factors does the escape speed of a body from the surface of the Earth depend?
As we go from one planet to another planet, how will the mass and weight of a body change?
What is the periodic time of a geostationary satellite?
State Newton’s law of gravitation and express it in vector form.
What do you mean by gravitational constant? State its SI units.
State the conditions for various possible orbits of a satellite depending upon the tangential speed of projection.
State any four applications of a communication satellite.
Show that the acceleration due to gravity at height \(h\) above the Earth’s surface is \(g_h = g\left(\dfrac{R}{R+h}\right)^2\)
Draw a labelled diagram to show different trajectories of a satellite depending upon the tangential projection speed.
Derive an expression for the binding energy of a body at rest on the Earth’s surface.
Draw a graph showing the variation of gravitational acceleration due to the depth and altitude from the Earth’s surface.
At which place on the Earth’s surface is the gravitational acceleration maximum? Why?
Define the binding energy of a satellite. Obtain an expression for the binding energy of a satellite revolving around the Earth at a certain altitude.
Obtain the formula for acceleration due to gravity at the depth ‘d’ below the Earth’s surface.
State Kepler’s three laws of planetary motion.
State the formula for acceleration due to gravity at depth ‘d’ and altitude ‘h’. Hence show that their ratio is equal to \(\left(\dfrac{R-d}{R-2h}\right)\) by assuming that the altitude is very small as compared to the radius of the Earth.
What is critical velocity? Obtain an expression for the critical velocity of an orbiting satellite. On what factors does it depend?
Define escape speed. Derive an expression for the escape speed of an object from the surface of the Earth.
Describe how an artificial satellite using a two-stage rocket is launched in an orbit around the Earth.
5.11 Practice with Numericals
Two spheres, each of mass 80 kg, are placed with their centres 0.40 m apart. Calculate the gravitational force of attraction between them. Take \(G = 6.67 \times 10^{-11}\) N m²/kg².
The gravitational force between two bodies is 8 N. Calculate the new force if (i) the distance between them is made three times the original distance, and (ii) one of the masses is doubled while the distance is halved.
A planet revolves around a star in a circular orbit of radius 9 times the radius of the Earth’s orbit around the Sun. Assuming the star has the same mass as the Sun, calculate the period of revolution of the planet in Earth years.
The semimajor axis of the orbit of a planet is \(4.5 \times 10^{12}\) m and its period of revolution is 165 years. Calculate the value of \(T^2/r^3\) for this planet in y² m⁻³, and compare it with the value for the Earth, whose semimajor axis is \(15.0 \times 10^{10}\) m and period is 1 year.
Calculate the mass of a planet whose radius is \(3.4 \times 10^6\) m and on whose surface the acceleration due to gravity is 3.7 m/s². Take \(G = 6.67 \times 10^{-11}\) N m²/kg².
The mass of a planet is 5 times that of the Earth and its radius is 2 times that of the Earth. Calculate the acceleration due to gravity on the surface of this planet. Take \(g = 9.8\) m/s² on the Earth.
At what height above the surface of the Earth will the acceleration due to gravity be 25% of its value at the surface? Take \(R = 6400\) km and use the exact expression.
Calculate the value of the acceleration due to gravity at a depth of 1600 km below the surface of the Earth. Take \(R = 6400\) km and \(g = 9.8\) m/s².
At a certain depth \(d\) below the Earth’s surface, the acceleration due to gravity is the same as it is at a height of 32 km above the surface. Assuming the height to be small compared with the radius of the Earth, find \(d\). Take \(R = 6400\) km.
Calculate the reduction in the acceleration due to gravity at a place of latitude \(60^\circ\) due to the rotation of the Earth. Take \(R = 6.4 \times 10^6\) m and \(\omega = 7.275 \times 10^{-5}\) s⁻¹.
A body of mass 20 kg is taken from the surface of the Earth to a height of 6400 km above it. Calculate the increase in the gravitational potential energy of the Earth–body system. Take \(M = 6 \times 10^{24}\) kg, \(R = 6400\) km and \(G = 6.67 \times 10^{-11}\) N m²/kg².
Calculate the gravitational potential at a point \(1.6 \times 10^7\) m from the centre of the Earth. Hence find the potential energy of a 50 kg body placed at that point. Take \(M = 6 \times 10^{24}\) kg and \(G = 6.67 \times 10^{-11}\) N m²/kg².
Two point masses of 5 kg and 20 kg are placed 2.0 m apart. Find the gravitational potential at a point on the line joining them, 0.5 m from the 5 kg mass. Take \(G = 6.67 \times 10^{-11}\) N m²/kg².
Calculate the escape velocity from the surface of a planet whose mass is 9 times that of the Earth and whose radius is 4 times that of the Earth. Take the escape velocity from the Earth as 11.2 km/s.
Calculate the escape velocity from the surface of the Moon, given that the mass of the Moon is \(7.3 \times 10^{22}\) kg and its radius is \(1.74 \times 10^6\) m. Take \(G = 6.67 \times 10^{-11}\) N m²/kg².
A body is projected vertically upwards from the Earth’s surface with a speed of 8 km/s. Using conservation of energy, find the maximum height it reaches above the Earth’s surface. Take \(R = 6400\) km and \(g = 9.8\) m/s².
Calculate the critical velocity and the period of revolution of a satellite orbiting at a height of 2000 km above the surface of the Earth. Take \(R = 6400\) km and \(g = 9.8\) m/s².
A satellite revolves around the Earth in a circular orbit of radius \(9.6 \times 10^6\) m. Calculate its orbital speed and its period of revolution. Take \(M = 6 \times 10^{24}\) kg and \(G = 6.67 \times 10^{-11}\) N m²/kg².
Two satellites of a planet have periods of revolution of 2 days and 16 days. If the radius of the orbit of the first satellite is \(3 \times 10^4\) km, calculate the radius of the orbit of the second satellite.
A planet has a mean density of \(5.5 \times 10^3\) kg/m³. Calculate the period of revolution of a satellite orbiting very close to its surface. Take \(G = 6.67 \times 10^{-11}\) N m²/kg².
An artificial satellite of mass 500 kg revolves around the Earth in a circular orbit of radius \(7.0 \times 10^6\) m. Calculate its kinetic energy, potential energy, total energy and binding energy. Take \(M = 6 \times 10^{24}\) kg and \(G = 6.67 \times 10^{-11}\) N m²/kg².
The binding energy of a satellite revolving around the Earth in a circular orbit is \(5 \times 10^9\) J. State its kinetic energy, potential energy and total energy in that orbit.
A person of mass 70 kg stands on a weighing machine in a lift. Take \(g = 9.8\) m/s². Find the reading of the machine, in newton, when the lift (i) is at rest, (ii) moves upwards with an acceleration of 1.5 m/s², (iii) moves downwards with an acceleration of 1.5 m/s², and (iv) falls freely.
A body weighs 60 N on the surface of the Earth. Calculate its weight at a height equal to half the radius of the Earth above the surface.
At what distance below the surface of the Earth does the acceleration due to gravity decrease by 10% of its value at the surface, given that the radius of the Earth is 6400 km.
If the Earth were made of wood, the mass of the wooden Earth would have been 10% as much as it is now (without change in its diameter). Calculate the escape speed from the surface of this Earth.
Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth. Given: \(G = 6.67 \times 10^{-11}\) Nm²/kg², \(R = 6400\) km, \(M = 6 \times 10^{24}\) kg
Two satellites A and B are revolving around a planet. Their periods of revolution are 1 hour and 8 hours respectively. The radius of the orbit of satellite B is \(4 \times 10^4\) km. Find the radius of the orbit of satellite A.
Find the gravitational force between the Sun and the Earth. Given: Mass of the Sun \(= 1.99 \times 10^{30}\) kg, Mass of the Earth \(= 5.98 \times 10^{24}\) kg, the average distance between the Earth and the Sun \(= 1.5 \times 10^{11}\) m.
Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (\(M = 5.98 \times 10^{24}\) kg, \(R = 6400\) km).
Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth’s surface. \(M_E = 5.98 \times 10^{24}\) kg, \(R = 6.4 \times 10^6\) m.
Calculate the value of acceleration due to gravity on the surface of Mars if the radius of Mars \(= 3.4 \times 10^3\) km and its mass is \(6.4 \times 10^{23}\) kg.
A planet has mass \(6.4 \times 10^{24}\) kg and radius \(3.4 \times 10^6\) m. Calculate the energy required to remove an object of mass 800 kg from the surface of the planet to infinity.
Calculate the value of the universal gravitational constant from the given data. Mass of the Earth \(= 6 \times 10^{24}\) kg, Radius of the Earth \(= 6400\) km and the acceleration due to gravity on the surface \(= 9.8\) m/s².
A body weighs 5.6 kg wt on the surface of the Earth. How much will be its weight on a planet whose mass is 7 times the mass of the Earth and radius twice that of the Earth’s radius.
What is the gravitational potential due to the Earth at a point which is at a height of \(2R_E\) above the surface of the Earth? Mass of the Earth is \(6 \times 10^{24}\) kg, radius of the Earth \(= 6400\) km and \(G = 6.67 \times 10^{-11}\) Nm² kg⁻².
5.12 Objective Questions
- The value of acceleration due to gravity is maximum at
- the equator of the Earth.
- the centre of the Earth.
- the pole of the Earth.
- slightly above the surface of the Earth.
- The weight of a particle at the centre of the Earth is
- infinite.
- zero.
- same as that at other places.
- greater than at the poles.
- The gravitational potential due to the Earth is minimum at
- the centre of the Earth.
- the surface of the Earth.
- a point inside the Earth but not at its centre.
- infinite distance.
- The binding energy of a satellite revolving around a planet in a circular orbit is \(3 \times 10^9\) J. Its kinetic energy is
- \(6 \times 10^9\) J
- \(-3 \times 10^9\) J
- \(-6 \times 10^{+9}\) J
- \(3 \times 10^{+9}\) J