NotesShala Class 11 · Physics

Chapter07

Gravitation

Planetary Motion and the Law of Gravitation · Gravity, Field and Potential · Escape Speed and Satellites

Reading key Definition Note Remember Derivation Strategy Formulas sit on graph paper · coloured words are the key terms

Part 1

Planetary Motion and the Law of Gravitation

All objects in the universe attract one another. This pull is called the force of gravitation. Galileo was the first to see that all bodies fall towards the Earth with a constant acceleration. Their masses make no difference. Kepler believed that the heavenly bodies move in a mathematical order. He thought this order was not like anything on the Earth.

1.1Theory of Planetary Motion

From early times, people watched the heavenly bodies, such as the Sun, the Moon, the stars and the planets. They believed that all of these bodies go around the Earth. They thought the Earth was fixed at the centre of the universe.

iGeocentric Model

In this model, the Earth is at the centre. All the heavenly bodies move in orbits around it.

It is the earliest recorded model. Ptolemy gave it about 2000 years ago.

iiHeliocentric Model

In this model, the Sun is at the centre. The planets revolve around the Sun.

It is a more elegant model. Galileo supported it after his experiments and studies of the Moon and the planets.

Tycho Brahe (1546–1601) observed the planets with his naked eyes for his whole life. He kept a careful record of what he saw. His assistant, Johannes Kepler, collected and studied this data. From it, Kepler found three neat laws. Today we call them Kepler's laws.

Tycho Brahe's data→Kepler's study→Kepler's three laws

1.2Kepler's First Law (Law of Orbits)

Kepler's laws are three laws of science. They describe how the planets move around the Sun.

Law

Law of Orbits

Kepler's first law: Every planet moves around the Sun in an elliptical orbit. The Sun is at one focus of the ellipse.

This law tells us that the Sun–Earth distance does not stay the same. It keeps changing while the Earth moves along its orbit.

In the figure, a planet traces an ellipse around the Sun $S$ . The planet comes closest to the Sun at point $P$ . It goes farthest from the Sun at point $A$ .

Ellipse traced by a planet around the Sun

S

: major axis

AP = 2a

, minor axis

BC = 2b

PPerihelion

The point of the orbit where the planet is nearest to the Sun.

AAphelion

The point of the orbit where the planet is farthest from the Sun.

Length of the major axis = Sun–planet distance at perihelion + Sun–planet distance at aphelion

AP = SP + SA

Length of the major axis

AP = 2a

Semi-major axis, which is half of the distance

AP

a = \dfrac{AP}{2}

1.3Kepler's Second Law (Law of Areas)

Law

Law of Areas

Kepler's second law: The line from the Sun to a planet, called the radius vector, sweeps out equal areas in equal intervals of time. In other words, the planet's areal velocity stays constant. For this, the speed of the planet changes along its orbit.

In the figure, a planet moves on an elliptical orbit around the Sun. The radius vector sweeps out the areas $A_1$ , $A_2$ and $A_3$ in equal intervals of time.

So, by Kepler's second law:

A_1 = A_2 = A_3

Law of areas: equal areas

A_1

A_2

A_3

are swept in equal times

Note

Areal velocity is the area that the radius vector sweeps out per unit time.

Kepler's second law is the same as the law of conservation of angular momentum. To see this, take a small area $\Delta A$ swept in a small time interval $\Delta t$ .

Let $\vec r$ be the position of the planet from the Sun. In time $\Delta t$ , the radius vector turns through a small angle $\Delta\theta$ . The force $\vec F$ on the planet points towards the Sun.

Small area

\Delta A

swept by the radius vector in time

\Delta t

Derivation

Areal Velocity of a Planet Is Constant

The thin sector has radius $r$ and arc length $r\,\Delta\theta$ . Its area is:
$\Delta A = \tfrac{1}{2}\,r\,(r\,\Delta\theta)$
Divide both sides by the time interval $\Delta t$ :
$\dfrac{\Delta A}{\Delta t} = \tfrac{1}{2}\,r^{2}\,\dfrac{\Delta\theta}{\Delta t}$
Now let $\Delta t \to 0$ . The ratio $\dfrac{\Delta\theta}{\Delta t}$ becomes the angular velocity $\omega$ :
$\dfrac{dA}{dt} = \tfrac{1}{2}\,r^{2}\omega \qquad \left[\because\ \omega = \displaystyle\lim_{\Delta t \to 0}\dfrac{\Delta\theta}{\Delta t}\right] \qquad \ldots(i)$
The angular momentum of the planet (mass $m$ ) about the Sun at this instant is:
$L = m\,r^{2}\omega \qquad \ldots(ii)$
Combine Eqs. (i) and (ii):
$\dfrac{dA}{dt} = \dfrac{L}{2m} \qquad \text{or} \qquad L = 2m \times \dfrac{dA}{dt} \qquad \ldots(iii)$
The gravitational force on the planet acts along the line through the Sun. So the angle between $\vec r$ and $\vec F$ is $180°$ , and the torque is zero:
$\vec\tau_{\text{ext}} = \vec r \times \vec F = rF\sin 180°\ \hat n = 0 \qquad \therefore\ \tau_{\text{ext}} = 0$
There is no external torque. So the angular momentum does not change with time:
$L = \text{constant} \qquad \left[\because\ \dfrac{dL}{dt} = \tau_{\text{ext}} = 0\right] \qquad \ldots(iv)$
From Eqs. (iii) and (iv), the areal velocity stays constant. This proves Kepler's second law.
$\text{Areal velocity},\ \dfrac{dA}{dt} = \dfrac{L}{2m} = \text{constant}$

Areal Velocity of a Planet

\dfrac{dA}{dt} = \dfrac{L}{2m} = \text{constant}

where, $L$ = angular momentum of the planet about the Sun, $m$ = mass of the planet

1.4Kepler's Third Law (Law of Periods)

Law

Law of Periods

Kepler's third law: For every planet, the square of its time period of revolution around the Sun is directly proportional to the cube of the semi-major axis of its orbit.

Let $T$ be the time period of revolution of a planet. In the figure, the Sun is at the focus $F$ of the orbit. Here, $r_1$ is the shortest distance of the planet from the Sun. $r_2$ is its longest distance from the Sun.

T^2 \propto a^3 \qquad \ldots(i)

Sun at the focus

F

AF = r_1

FB = r_2

FC = c

and

CB = a

Derivation

Law of Periods in Terms of

r_1

and

r_2

Kepler's third law for a planet with time period $T$ and semi-major axis $a$ is:
$T^{2} \propto a^{3} \qquad \ldots(i)$
From the figure, the major axis $AB$ is made of the two parts $AF$ and $FB$ :
$AB = AF + FB \qquad \Rightarrow \qquad 2a = r_1 + r_2$
So the semi-major axis of the orbit is:
$a = \dfrac{r_1 + r_2}{2}$
Put this value of $a$ in Eq. (i):
$T^{2} \propto \left(\dfrac{r_1 + r_2}{2}\right)^{3}$

Kepler's Third Law

T^2 \propto a^3 \qquad\qquad T^{2} \propto \left(\dfrac{r_1 + r_2}{2}\right)^{3}

where, $a$ = semi-major axis, $r_1$ = shortest distance of the planet from the Sun (perihelion), $r_2$ = longest distance of the planet from the Sun (aphelion)

Note

Kepler's laws are true not only for the solar system. They also hold for artificial satellites and for moons that go around the planets.

Kepler's Law of Periods for the Solar System

In the table, the last column $T^2/r^3$ is nearly the same for every planet. This confirms the law.

Planet	Semi-major axis $r$ ( $10^{10}$ m)	Time period $T$ (years)	$T^2/r^3$ ( $10^{-34}\ \text{y}^2\,\text{m}^{-3}$ )
Mercury	5.79	0.241	2.99
Venus	10.8	0.615	3.00
Earth	15.0	1.00	2.96
Mars	22.8	1.88	2.98
Jupiter	77.8	11.9	3.01
Saturn	143	29.5	2.98
Uranus	287	84.0	2.98
Neptune	450	165.0	2.99
Pluto*	590	248.0	2.99

*Pluto is now classed as a dwarf planet. Its data still fits the law.

1.5Universal Law of Gravitation

Law

Newton's Law of Gravitation

Newton's law of gravitation: Every body in the universe attracts every other body with a force. This force is directly proportional to the product of their masses ( $m_1 m_2$ ). It is inversely proportional to the square of the distance $r$ between their centres. The force acts along the line joining the centres of the two bodies.

Take two bodies of masses $m_1$ and $m_2$ . The distance between their centres is $r$ . Let $F$ be the magnitude of the gravitational force between them. Newton's law says:

F \propto m_1 m_2 \;\;\ldots(i) \qquad F \propto \dfrac{1}{r^2} \;\;\ldots(ii)

Combining (i) and (ii), $\;\therefore\ F \propto \dfrac{m_1 m_2}{r^2}$

Two bodies pull each other with equal and opposite forces

Newton's Law of Gravitation

F = G\,\dfrac{m_1 m_2}{r^2} \qquad \ldots(iii)

where, $G$ = constant of proportionality. It is called the gravitational constant or the universal gravitational constant.

Value of

G

in the SI system

G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}

Value of

G

in the CGS system

G = 6.67 \times 10^{-8}\ \text{dyne cm}^2\,\text{g}^{-2}

Derivation

Dimensional Formula of

G

Rearrange Eq. (iii) to get $G$ :
$G = \dfrac{F\,r^2}{m_1 m_2}$
Write the dimensions of force, distance and mass:
$G = \dfrac{[\text{MLT}^{-2}][\text{L}^2]}{[\text{M}^2]} = [\text{MLT}^{-2}][\text{L}^2][\text{M}^{-2}]$
So the dimensional formula of the universal gravitational constant is:
$G = [\text{M}^{-1}\text{L}^{3}\text{T}^{-2}]$

Concept

Physical Meaning of

G

Let $m_1 = m_2 = 1$ unit and $r = 1$ unit. Then Eq. (iii) gives $F = \dfrac{G \times 1 \times 1}{(1)^2}$ , so $F = G$ .

So $G$ is numerically equal to the force of attraction between two bodies of unit mass each, kept a unit distance apart.

Important Points about Gravitational Force

1Always Attractive

Gravitational force always attracts. Electric and magnetic forces can attract or repel.

2Medium Has No Effect

It does not depend on the medium between the bodies. Electric and magnetic forces do depend on the medium.

3Conservative Force

Gravitational force is a conservative force.

G

Never Changes

The value of $G$ does not depend on the nature or the size of the masses.

5Point Mass outside a Shell

A uniform hollow spherical shell pulls a point mass outside it as if all its mass were at its centre. So gravity has spherical symmetry.

6Point Mass inside a Shell

A uniform hollow spherical shell puts zero force on a point mass kept inside it.

1.6Vector Form of Newton's Law of Gravitation

Take two particles $A$ and $B$ with masses $m_1$ and $m_2$ . Their displacement vectors and the forces between them are named like this:

$\vec r_{12}$ = displacement vector from $A$ to $B$
$\vec r_{21}$ = displacement vector from $B$ to $A$ , so $\vec r_{21} = -\vec r_{12}$
$\vec F_{12}$ = gravitational force on $A$ due to $B$
$\vec F_{21}$ = gravitational force on $B$ due to $A$

Displacement vectors and forces between particles

A

and

B

Derivation

Vector Form and Central Force

Let $\hat r_{21}$ be the unit vector that points towards $A$ . In vector form, the force on $A$ due to $B$ is:
$\vec F_{12} = -\,G\,\dfrac{m_1 m_2}{r_{21}^{2}}\,\hat r_{21} \qquad \ldots(i)$
The minus sign means that $\vec F_{12}$ is opposite to $\hat r_{21}$ . So gravity pulls the particles together. It is an attractive force.
In the same way, let $\hat r_{12}$ be the unit vector that points towards $B$ . The force on $B$ due to $A$ is:
$\vec F_{21} = -\,G\,\dfrac{m_1 m_2}{r_{12}^{2}}\,\hat r_{12}$
But $\hat r_{21} = -\hat r_{12}$ and $r_{21}^{2} = r_{12}^{2}$ . Put these in:
$\vec F_{21} = G\,\dfrac{m_1 m_2}{r_{21}^{2}}\,\hat r_{21} \qquad \ldots(ii)$
Compare Eqs. (i) and (ii). Both forces point towards the centres of the particles. So gravitational force is a central force.
$\vec F_{12} = -\,\vec F_{21}$

Vector Form of the Law

\vec F_{12} = -\,G\,\dfrac{m_1 m_2}{r_{21}^{2}}\,\hat r_{21}

where, $\hat r_{21}$ = unit vector pointing towards $A$ . The negative sign means the force is attractive.

Key $\vec F_{12} = -\vec F_{21}$ . Gravity acts along the line of centres, so it is a central force

1.7Principle of Superposition

Principle

Let $\vec F_1, \vec F_2, \ldots, \vec F_n$ be the forces due to the masses $m_1, m_2, m_3, \ldots, m_n$ . Each force follows the universal law of gravitation. The principle of superposition says that each of these forces acts independently. The other bodies have no effect on it.

The force on $m_1$ due to $m_2$ acts along the vector $\vec r$ . Here $\vec r_1$ and $\vec r_2$ are the positions of the two particles, measured from the origin $O$ .

\vec r = \vec r_2 - \vec r_1

Force on

m_1

due to

m_2

acts along

{\vec r = \vec r_2 - \vec r_1}

Now let many masses $m_2, m_3, m_4, \ldots, m_n$ pull on $m_1$ . Each force $\vec F_{12}, \vec F_{13}, \ldots, \vec F_{1n}$ points from $m_1$ towards the other mass.

The resultant force on $m_1$ is the vector sum of all these forces.

Superposition of gravitational forces on

m_1

Derivation

Resultant Gravitational Force on

m_1

The resultant force $\vec F$ on $m_1$ is the vector sum of all the separate forces:
$\vec F = \vec F_{12} + \vec F_{13} + \vec F_{14} + \ldots + \vec F_{1n}$
Write each force using the vector form of Newton's law of gravitation:
$\vec F = -\,G\,\dfrac{m_1 m_2}{r_{12}^{2}}\,\hat r_{21} \;-\; G\,\dfrac{m_1 m_3}{r_{13}^{2}}\,\hat r_{31} \;-\; \ldots \;-\; G\,\dfrac{m_1 m_n}{r_{1n}^{2}}\,\hat r_{n1}$
Take $-\,Gm_1$ common. The resultant force is:
$\vec F = -\,G m_1\left[\dfrac{m_2}{r_{12}^{2}}\,\hat r_{21} + \dfrac{m_3}{r_{13}^{2}}\,\hat r_{31} + \ldots + \dfrac{m_n}{r_{1n}^{2}}\,\hat r_{n1}\right]$

Resultant Force by Superposition

\vec F = -\,G m_1\left[\dfrac{m_2}{r_{12}^{2}}\,\hat r_{21} + \dfrac{m_3}{r_{13}^{2}}\,\hat r_{31} + \ldots + \dfrac{m_n}{r_{1n}^{2}}\,\hat r_{n1}\right]

It gives the total pull on $m_1$ from all the other masses. Each force acts on its own, free of the others.

1.8Gravitational Constant (Cavendish's Experiment)

The English scientist Henry Cavendish was the first to measure the value of $G$ by experiment. He did this in 1798.

Two small lead spheres (mass $m$ each) sit at the ends of a rod of length $L$ . A fine quartz fibre holds the rod at its middle. This set-up is a torsion balance.

Two large lead spheres (mass $M$ each) are brought near the small ones, on opposite sides of the rod.

Cavendish's torsion balance

Derivation

Measuring

G

with the Torsion Balance

Each small sphere is pulled towards the big sphere next to it. Let $R$ be the distance between their centres. The force on each small sphere is:
$F = \dfrac{GMm}{R^{2}}$
The two forces are equal and opposite. So the net force on the rod is zero. But they produce a gravitational torque on it:
$\text{Gravitational torque} = \text{gravitational force} \times \text{length of the bar} = \dfrac{GMm}{R^{2}} \times L$
This torque turns the rod through an angle $\theta$ , with the fibre as the axis. So the fibre gets twisted through $\theta$ . Let $\tau$ be the restoring torque per unit twist of the fibre:
$\text{Restoring torque} = \tau\,\theta$
In equilibrium, the restoring torque equals the gravitational torque:
$\tau\,\theta = \dfrac{GMm}{R^{2}} \times L$
Every quantity on the right side is known from the experiment. So $G$ can be found:
$G = \dfrac{R^{2}\,\tau\,\theta}{M\,m\,L}$

Gravitational Constant (Cavendish)

G = \dfrac{R^{2}\tau\theta}{MmL}

Later experiments have measured $G$ more accurately. Its accepted value is $G = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$ .

Note

$G$ is a universal constant. Its value does not change with the medium between the bodies. It also does not change with temperature, pressure and so on.

Good to know

Newton's Principia

Kepler had stated his third law by 1619. Newton announced the universal law of gravitation behind it about seventy years later, in 1687. It came out in his great book Philosophiae Naturalis Principia Mathematica, usually called the Principia.

Part 2

Gravity, Field and Potential

2.1Acceleration due to Gravity of the Earth

Definition

The acceleration of a body that falls freely under gravity is called the acceleration due to gravity. It is equal to the force of gravity on a unit mass kept on or near the Earth's surface. Its symbol is $g$ . On the Earth's surface, $g = 9.8\ \text{m/s}^2$ .

Value of

g

for all bodies at one place, whatever their massesSame ✓

Value of

g

at different places on the Earth's surfaceDiffers

Value of

g

with altitude and depthVaries

For a point outside the Earth, the Earth pulls as if its whole mass were at its centre.

Imagine the Earth as a set of concentric shells. A point mass $m$ is at $P$ , a distance $r$ from the centre $O$ . Point $P$ lies outside the sphere of radius $r$ . It lies inside every shell of radius larger than $r$ .

Finding

g

: the Earth as concentric shells

Derivation

g

on the Surface of the Earth

The inner sphere has radius $r$ and mass $m_r$ . It pulls $m$ as if the mass $m_r$ were at the centre. The shells of radius larger than $r$ put no force on $m$ . So the force on $m$ at $P$ is:
$F = \dfrac{G\,m\,(m_r)}{r^{2}} \qquad \ldots(i)$
Write the masses using the density $\rho$ of the Earth. Here $M$ is the mass and $R$ the radius of the Earth:
$M = \dfrac{4\pi}{3}R^{3}\rho \qquad\quad m_r = \dfrac{4\pi}{3}\rho\, r^{3}$
Put the value of $m_r$ in Eq. (i):
$F = \dfrac{Gm}{r^{2}}\cdot\dfrac{4\pi}{3}\rho\,r^{3} = Gm\,\dfrac{M}{R^{3}}\,r$
On the surface of the Earth, $r = R$ :
$F = G\,\dfrac{Mm}{R^{2}}$
By Newton's second law, $F = mg$ . So:
$g = \dfrac{F}{m} = \dfrac{GM}{R^{2}}$

Acceleration due to Gravity

g = \dfrac{GM}{R^2}

where, $M$ = mass of the Earth, $R$ = radius of the Earth. With $M \approx 6 \times 10^{24}$ kg and $R = 6.4 \times 10^{6}$ m, we get $g = 9.8\ \text{m/s}^2$ .

Note

On the surface of the Moon, $g$ is $\dfrac{1}{6}$ of its value on the surface of the Earth.
$g$ does not depend on the mass, shape or size of the body. It depends on the mass and radius of the Earth.

2.2Variation of g with Altitude

A body of mass $m$ lies on the surface of the Earth. The Earth has mass $M$ and radius $R$ .

Now the body is taken to a height $h$ above the surface. Let the acceleration due to gravity there be $g_h$ . The centre of the body is now at a distance $(R+h)$ from the centre of the Earth.

Body at a height

h

above the Earth's surface

Derivation

g

at a Height

h

At the surface of the Earth:
$g = \dfrac{GM}{R^{2}} \qquad \ldots(i)$
At height $h$ , the body is at a distance $(R+h)$ from the centre of the Earth:
$g_h = \dfrac{GM}{(R+h)^{2}} \qquad \ldots(ii)$
Divide Eq. (ii) by Eq. (i):
$\dfrac{g_h}{g} = \dfrac{GM}{(R+h)^{2}} \times \dfrac{R^{2}}{GM} = \dfrac{R^{2}}{(R+h)^{2}} \qquad \ldots(iii)$
Write $(R+h)$ as $R\left(1+\dfrac{h}{R}\right)$ :
$\dfrac{g_h}{g} = \dfrac{R^{2}}{R^{2}\left(1+\frac{h}{R}\right)^{2}} = \left(1+\dfrac{h}{R}\right)^{-2} \qquad\Rightarrow\qquad g_h = \dfrac{g}{\left(1+\frac{h}{R}\right)^{2}}$
If $h \ll R$ , then $h/R$ is very small. Expand by the binomial theorem. Drop the squares and higher powers of $h/R$ :
$\dfrac{g_h}{g} = 1 - \dfrac{2h}{R}$
So the acceleration due to gravity at height $h$ is:
$g_h = g\left(1 - \dfrac{2h}{R}\right) \qquad \ldots(iv) \qquad\quad \text{or} \qquad g - g_h = \dfrac{2h}{R}\,g \qquad \ldots(v)$

g at Height h

g_h = \dfrac{g}{\left(1+\dfrac{h}{R}\right)^{2}} \qquad\qquad g_h = g\left(1-\dfrac{2h}{R}\right) \quad (h \ll R)

where, $g$ = acceleration due to gravity at the surface, $R$ = radius of the Earth, $h$ = height above the surface

Decrease in g

g - g_h = \dfrac{2hg}{R}

Fractional Decrease

\dfrac{g - g_h}{g} = \dfrac{2h}{R}

Percentage Decrease

\left(\dfrac{g-g_h}{g}\right)\times 100 = \dfrac{2h}{R}\times 100\%

KeyFrom Eq. (v), $g - g_h \propto h$ . So $g$ decreases as the height increases

2.3Variation of g with Depth

Take the Earth as a homogeneous sphere, that is, a sphere of uniform density. Let $\rho$ be its mean density. A body on the surface has acceleration due to gravity $g$ .

Now take the body to a depth $d$ below the surface. The gravity on it now comes only from the inner solid sphere of radius $(R-d)$ .

Body at a depth

d

below the Earth's surface

Derivation

g

at a Depth

d

At the surface, $g = \dfrac{GM}{R^2}$ . The mass of the Earth is $M = \frac{4}{3}\pi R^{3}\rho$ . So:
$g = \dfrac{G \times \frac{4}{3}\pi R^{3}\rho}{R^{2}} = \dfrac{4}{3}\,\pi\, G R \rho \qquad \ldots(i)$
At depth $d$ , only the inner solid sphere pulls the body. It has radius $(R-d)$ and mass $M'$ :
$g_d = \dfrac{G M'}{(R-d)^{2}}$
The mass of the inner solid sphere is $M' = \frac{4}{3}\pi (R-d)^{3}\rho$ . Put it in:
$g_d = \dfrac{G}{(R-d)^{2}} \times \dfrac{4}{3}\pi (R-d)^{3}\rho = \dfrac{4}{3}\,\pi\, G\,(R-d)\,\rho \qquad \ldots(ii)$
Divide Eq. (ii) by Eq. (i):
$\dfrac{g_d}{g} = \dfrac{\frac{4}{3}\pi G (R-d)\rho}{\frac{4}{3}\pi G R \rho} = \dfrac{R-d}{R} = 1 - \dfrac{d}{R}$
So the acceleration due to gravity at depth $d$ is given below. Here $(g - g_d)$ is the decrease in the value of $g$ .
$g_d = g\left(1 - \dfrac{d}{R}\right) \qquad\quad \text{or} \qquad g - g_d = \dfrac{d}{R}\,g$

g at Depth d

g_d = g\left(1 - \dfrac{d}{R}\right)

where, $g$ = acceleration due to gravity at the surface, $R$ = radius of the Earth, $d$ = depth below the surface

Decrease in g

g - g_d = \dfrac{g\,d}{R}

Fractional Decrease

\dfrac{g - g_d}{g} = \dfrac{d}{R}

Percentage Decrease

\left(\dfrac{g-g_d}{g}\right)\times 100 = \dfrac{d}{R}\times 100\%

Key $(g - g_d) \propto d$ . So $g$ decreases as we go down below the Earth's surface

Remember

At the Centre of the Earth

At the centre $O$ of the Earth, $d = R$ . So $g_d = g\left(1-\dfrac{R}{R}\right) = 0$ . So $g$ becomes zero at the Earth's centre.

So the weight of a body is zero at the centre of the Earth. But its mass stays the same.

Note

Effect of the Earth's Rotation

The rotation of the Earth also changes $g$ . Let a body of mass $m$ be at a place of latitude $\lambda$ . Let $\omega$ be the angular speed of the Earth. Then the body has an apparent acceleration due to gravity given by:

g' = g - \omega^2 R \cos^2\lambda

2.4Intensity of Gravitational Field

Definition

The intensity of the gravitational field at a point is the force on a unit mass placed at that point.

A body of mass $M$ has a gravitational field around it. To find the field intensity at a point $P$ , place a test mass $m$ at $P$ .

The test mass must be very small. Then it does not change the field at all. Let $\vec F$ be the force on $m$ at $P$ . Then the intensity is:

I = \dfrac{F}{m} = \dfrac{\;\dfrac{GMm}{r^2}\;}{m}

Test mass

m

in the field of a body of mass

M

Intensity of Gravitational Field

I = \dfrac{GM}{r^2}

where, $r$ = distance of the point $P$ from the centre of the body that produces the field

r = \infty

, then

I = 0

. The field intensity is zero only at an infinite distance from the body.

I = 0

For a unit mass on the surface of the Earth, the force on it is its weight

w

. So

I = \dfrac{w}{m} = \dfrac{mg}{m}

I = g

2.5Gravitational Potential

Definition

The gravitational potential at a point in a gravitational field is the work done to move a unit mass from infinity up to that point, with no acceleration.

V = -\dfrac{W}{m} = -\displaystyle\int \dfrac{\vec F \cdot d\vec r}{m} = -\displaystyle\int \vec I \cdot d\vec r \qquad \left[\text{as}\ \dfrac{F}{m} = I\right]

Here, $W$ is the work done by the gravitational force as a body of mass $m$ comes from infinity to the point. $I$ is the intensity of the field. From this relation:

Field from Potential

I = -\dfrac{dV}{dr}

The negative gradient of the potential gives the field intensity. Potential is a scalar function of position. Its rate of change with distance gives the intensity.

The negative sign shows that the field intensity points in the direction in which the potential decreases.

Nature

Scalar quantity.

Units

SI: $\text{J kg}^{-1}$ (joule/kg). CGS: $\text{erg g}^{-1}$ (erg/g).

Dimensions

$V = \dfrac{W}{m} = \dfrac{[\text{ML}^2\text{T}^{-2}]}{[\text{M}]}$

$= [\text{M}^0\text{L}^2\text{T}^{-2}]$

Expression for Gravitational Potential

Take the Earth as a perfect sphere of radius $R$ and mass $M$ . Its mass acts as if it were all at its centre $O$ .

We find the potential at a point $Q$ , where $OQ = r$ and $r > R$ . Take two points $A$ and $B$ on the same line, with $OA = x$ and $AB = dx$ .

Finding the gravitational potential at

Q

Derivation

Gravitational Potential at a Point

At point $A$ , the gravitational pull on a body of unit mass is:
$F = \dfrac{GM \times 1}{x^{2}} = \dfrac{GM}{x^{2}}$
Move the unit mass without acceleration through the small distance $AB = dx$ . The small work done is:
$dW = F\,dx = \dfrac{GM}{x^{2}}\,dx$
The total work done in bringing the body from infinity to the point $Q$ is:
$W = \displaystyle\int_{\infty}^{r} \dfrac{GM}{x^{2}}\,dx = -\left[\dfrac{GM}{x}\right]_{\infty}^{r}$
Put in the limits:
$W = -\,GM\left[\dfrac{1}{r} - \dfrac{1}{\infty}\right] = \dfrac{-\,GM}{r}$
This is the work done on a unit mass, so $V = W$ . The gravitational potential at $Q$ is:
$V = \dfrac{-\,GM}{r} \qquad \ldots(i)$

Gravitational Potential

V = -\,\dfrac{GM}{r}

where, $V$ = gravitational potential at the point $Q$ , $r$ = distance of $Q$ from the Earth's centre

iAt Infinity

r = \infty \;\Rightarrow\; V_Q = 0

The gravitational potential is maximum (zero) at infinity.

iiAt the Surface

r = R \;\Rightarrow\; V_Q = -\dfrac{GM}{R}

This is the potential on the surface of the Earth.

Note

The potential due to many particles is the scalar sum of their separate potentials. So the potentials simply add, like ordinary numbers.

The formula below gives the potential at a point $P$ due to masses $M_1, M_2, M_3, \ldots$ kept at distances $r_1, r_2, r_3, \ldots$ from it.

Superposition of potentials at

P

Potential due to Many Masses

V = V_1 + V_2 + V_3 + \ldots = \dfrac{-GM_1}{r_1} - \dfrac{GM_2}{r_2} - \ldots = -\,G \sum_{i=1}^{i=n} \dfrac{M_i}{r_i}

2.6Gravitational Potential Energy

Definition

The work done against the gravitational force in bringing a body from infinity to a point is called the gravitational potential energy of the body at that point.

A body of mass $m$ is at a point $P$ in the gravitational field of a body of mass $M$ . Let $r$ be the distance of $P$ from the centre $O$ . The total work done against gravity in moving the body from infinity to the distance $r$ is:

W = -\,\dfrac{GmM}{r}

This work is equal to the gravitational potential energy $U$ of the mass $m$ .

Mass

m

at a distance

r

from the centre

O

Gravitational Potential Energy

U = -\,\dfrac{GMm}{r} \qquad\qquad U = \left(-\dfrac{GM}{r}\right) \times m

where, $r$ = distance of the body of mass $m$ from the centre of the body of mass $M$

Potential energy=Mass×Gravitational potential

Special Cases

iSystem of Three Particles

Take three particles of masses $m_1$ , $m_2$ and $m_3$ in an isolated system. By the superposition principle, the total gravitational potential energy of the system is:

U = -\left(\dfrac{Gm_1m_2}{r_{12}} + \dfrac{Gm_2m_3}{r_{23}} + \dfrac{Gm_1m_3}{r_{13}}\right)

Derivation

(ii) Body Raised to a Height

h

A body of mass $m$ is lifted from the Earth's surface to a height $h$ above it.

The change in potential energy, or the work done against gravity, in moving from $r_1$ to $r_2$ is:
$W = \Delta U = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$
Take $r_1 = R$ (the surface) and $r_2 = R + h$ (at height $h$ ):
$\Delta U = GMm\left(\dfrac{1}{R} - \dfrac{1}{R+h}\right)$
Simplify, and use $\dfrac{GM}{R^2} = g$ :
$\Delta U = \dfrac{GMmh}{R^{2}\left(1+\frac{h}{R}\right)} = \dfrac{mgh}{1+\frac{h}{R}}$
Expand $(1+h/R)^{-1}$ by the binomial theorem:
$\Delta U = mgh\left(1+\dfrac{h}{R}\right)^{-1} = mgh\left(1 - \dfrac{h}{R}\right)$
Since $h \ll R$ , the term $h/R$ is very small. We can neglect it:
$\Delta U = mgh$

iiiHeight Equal to the Earth's Radius

Now lift the body to a height $h = R$ , that is, one Earth radius above the surface. The work done is:

W = \dfrac{mgR}{1 + R/R} = \dfrac{1}{2}\,mgR

Part 3

Escape Speed and Satellites

3.1Escape Speed

Definition

The escape speed is the minimum speed with which a body must be thrown vertically upwards from the Earth's surface so that it just escapes the Earth's gravitational field. The body then never returns on its own. The same idea holds for any other planet.

iUsual Projectile (

v < v_e

)

A body thrown upwards normally slows down. It stops for a moment and then falls back to the Earth.

iiMinimum Initial Speed (

v = v_e

)

The body keeps moving upwards forever. It comes to rest only at infinity.

A body of mass $m$ is at a distance $x$ from the centre of the Earth. Let $M$ be the mass and $R$ the radius of the Earth.

By Newton's law of gravitation, the Earth pulls the body with a gravitational force $F = \dfrac{GMm}{x^2}$ .

Finding the escape speed

Derivation

Escape Speed

At a distance $x$ , the gravitational pull on the body is:
$F = G\,\dfrac{Mm}{x^{2}}$
The work done to move the body up through a small distance $dx$ is:
$dW = F\,dx = \dfrac{GMm}{x^{2}}\,dx$
Find the total work done to take the body from the surface ( $x = R$ ) to $x = \infty$ . Here $GMm$ is a constant:
$W = \displaystyle\int dW = GMm\int_{R}^{\infty} x^{-2}\,dx = -\,GMm\left[\dfrac{1}{x}\right]_{R}^{\infty}$
Put in the limits, using $\dfrac{1}{\infty} = 0$ :
$W = -\,GMm\left[\dfrac{1}{\infty} - \dfrac{1}{R}\right] = \dfrac{GMm}{R}$
Let $v_e$ be the escape speed. The kinetic energy given to the body must be equal to this work:
$\dfrac{1}{2}\,m v_e^{2} = \dfrac{GMm}{R}$
Solve for $v_e$ :
$v_e^{2} = \dfrac{2GM}{R} \qquad\Rightarrow\qquad v_e = \sqrt{\dfrac{2GM}{R}} \qquad \ldots(i)$

KeyEscape speed depends on the mass and radius of the planet. It does not depend on the mass of the body

In Terms of

g

Since $g = \dfrac{GM}{R^2}$ , we have $GM = gR^2$ . So ${v_e = \sqrt{\dfrac{2gR^2}{R}}}$ :

v_e = \sqrt{2gR} \qquad \ldots(ii)

In Terms of Mean Density

\rho

Since $M = \dfrac{4}{3}\pi R^3 \rho$ , we get ${v_e = \sqrt{\dfrac{2G}{R}\times\dfrac{4}{3}\pi R^3\rho}}$ :

v_e = R\sqrt{\dfrac{8}{3}\pi G\rho} \qquad \ldots(iii)

Escape Speed (Three Forms)

v_e = \sqrt{\dfrac{2GM}{R}} \;=\; \sqrt{2gR} \;=\; R\sqrt{\dfrac{8}{3}\pi G \rho}

where, $R$ = radius of the Earth (planet), $g$ = acceleration due to gravity at its surface, $\rho$ = mean density of the Earth (planet)

3.2Escape Speed from Conservation of Energy

Suppose the object reaches infinity with a speed $v_f$ . The total energy of an object is its potential energy plus its kinetic energy.

Derivation

Escape Speed Using Energy Conservation

Let $W_1$ be the gravitational potential energy of the object at infinity. Its total energy at infinity is:
$E(\infty) = W_1 + \dfrac{m v_f^{2}}{2} \qquad \ldots(i)$
Suppose the object was thrown with speed $v_i$ from a point at a distance $(h+R)$ from the centre of the Earth. Its energy at the start is:
$E(h+R) = \dfrac{1}{2}\,m v_i^{2} - \dfrac{GmM}{h+R} + W_1 \qquad \ldots(ii)$
By the principle of conservation of energy, Eqs. (i) and (ii) are equal:
$\dfrac{m v_i^{2}}{2} - \dfrac{GmM}{(h+R)} = \dfrac{m v_f^{2}}{2}$
The minimum speed is the one with which the object just reaches infinity. Then $\dfrac{mv_f^2}{2} = 0$ :
$\dfrac{1}{2}\,m\,(v_i^{2})_{\min} = \dfrac{GmM}{h+R}$
For an object thrown from the surface of the Earth, $h = 0$ :
$(v_i)_{\min} = \sqrt{\dfrac{2GM}{R}}$
Use the relation $g = GM/R^{2}$ :
$\text{Minimum speed},\ (v_i)_{\min} = \sqrt{2gR}$

Escape Speed of the Earth

(v_i)_{\min} = \sqrt{2gR}

With $g = 9.8\ \text{m/s}^2$ and $R = 6.4 \times 10^{6}$ m, $\;(v_i)_{\min} = 11.2\ \text{km/s}$

Note

Orbital Velocity

The orbital velocity $(v_o)$ of a satellite is the velocity a satellite must be given to place it in its orbit around the Earth. For an orbit close to the Earth's surface, $v_o = \sqrt{gR}$ . So:

v_e = \sqrt{2}\,v_o

3.3Earth Satellites

Definition

A satellite is a body that keeps revolving in an orbit around a much larger body.

iNatural Satellite

e.g. The Moon

iiArtificial Satellite

e.g. INSAT-1B

Note

Condition for a Satellite

The centre of the orbit of the satellite must be at the centre of the Earth. In other words, the satellite must move along a great circle of the Earth.

Derivation

Orbital Speed and Time Period of a Satellite

A satellite of mass $m$ moves in a circular orbit at a height $h$ . It needs a centripetal force towards the centre:
$F_c = \dfrac{mv^{2}}{R+h} \qquad \ldots(i)$
The gravitational force of the Earth provides this centripetal force:
$F_g = \dfrac{GmM}{(R+h)^{2}} \qquad \ldots(ii)$
Equate Eqs. (i) and (ii). This shows that $v$ decreases as $h$ increases:
$v^{2} = \dfrac{GM}{R+h}$
For an orbit very close to the surface, $h = 0$ . Using $g = \dfrac{GM}{R^2}$ :
$v^{2} = \dfrac{GM}{R} = gR$
In one revolution, the satellite covers a distance $2\pi(R+h)$ at speed $v$ . Using $v^2 = \dfrac{GM}{R+h}$ :
$T = \dfrac{2\pi(R+h)}{v} = \dfrac{2\pi(R+h)^{3/2}}{\sqrt{GM}} \qquad \ldots(iii)$
Square both sides of Eq. (iii). The result is Kepler's law of periods:
$T^{2} = k\,(R+h)^{3} \qquad \left[\because\ k = \dfrac{4\pi^{2}}{GM}\right]$

Orbital Speed and Period

v = \sqrt{\dfrac{GM}{R+h}} \qquad\qquad T^{2} = k\,(R+h)^{3},\ \ k = \dfrac{4\pi^{2}}{GM}

where, $R$ = radius of the Earth, $h$ = height of the orbit above the surface, $M$ = mass of the Earth

Orbit Just above the Earth's Surface

Here $h$ is tiny compared with $R$ , so we can neglect it. Put $h = 0$ and call the period $T_0$ :

T_0 = 2\pi\sqrt{R/g} = 2\pi\sqrt{\dfrac{6.4\times10^{6}}{9.8}} \cong 85\ \text{min}

3.4Energy of an Orbiting Satellite

A satellite going around a planet in its orbit has two kinds of energy. It has potential energy because of its position against the Earth's pull. It has kinetic energy because of its orbital motion.

Derivation

Energy of an Orbiting Satellite

Let $m$ be the mass of the satellite and $v$ its orbital velocity. Using $v^2 = \dfrac{GM}{r}$ with $r = R+h$ , its kinetic energy is:
$K = \dfrac{1}{2}mv^{2} = \dfrac{1}{2}m\,\dfrac{GM}{r} \qquad\Rightarrow\qquad K = \dfrac{GMm}{2(R+h)} \qquad \ldots(i)$
Its potential energy is mass × gravitational potential, that is, $U = mV$ :
$U = -\,\dfrac{GMm}{R+h} \qquad \ldots(ii)$
The total mechanical energy is $E = K + U$ :
$E = \dfrac{GMm}{2(R+h)} - \dfrac{GMm}{(R+h)}$
So the total energy of the orbiting satellite is:
$E = -\,\dfrac{GMm}{2(R+h)} \qquad \ldots(iii)$

Energy of an Orbiting Satellite

K = \dfrac{GMm}{2(R+h)} \qquad U = -\dfrac{GMm}{R+h} \qquad E = -\dfrac{GMm}{2(R+h)}

where, $m$ = mass of the satellite, $M$ = mass of the Earth, $(R+h)$ = radius $r$ of the orbit

Remember

Why Is

E

Always Negative?

A satellite always stays at a finite distance from the Earth. So its total energy can never be positive or zero.

The graph shows the kinetic, potential and total energies against the orbit radius $r$ . It follows these relations:

(i) Kinetic energy

K = -\,(\text{total energy})

(ii) Potential energy

U = 2\,(\text{total energy})

(iii) Potential energy

U = -2\,(\text{kinetic energy})

Energies of a satellite against the orbit radius

r

A satellite can also move in an elliptical orbit with semi-major axis $a$ . Its total energy is then also constant:

E = \dfrac{-\,GMm}{2a} = \text{constant}

Perigee (

r_{\min}

)KE = max · PE = min

Apogee (

r_{\max}

)KE = min · PE = max

Energy at different points of an elliptical orbit

3.5Weightlessness

Definition

Static or True Weight

The force with which the Earth pulls a body towards its centre is called the weight of the body. When the body is at rest with respect to the Earth, its weight equals the force of gravity on it. This weight is called its static weight or true weight.

A body in space can feel two gravitational pulls in opposite directions. At one particular point, these two pulls become equal and opposite. The net pull on the body is then zero. This point is called the null point or the zero gravity region. A body there is weightless.

The state of weightlessness is seen in these cases:

1Free Fall

Bodies in free fall under gravity. e.g. a freely falling lift.

2Orbiting Satellite

A satellite revolving in its orbit around the Earth.

3Null Points

Bodies at the null points in outer space.

Weightlessness in a Satellite

Take a satellite that has no gravity of its own. Gravity alone keeps it moving in a circular orbit around the Earth.

The acceleration of the satellite is $\dfrac{GM}{r^2}$ . It is directed towards the Earth's centre. Inside the satellite, a body of mass $m$ rests on a surface.

A satellite in orbit at a height

h

Derivation

Apparent Weight in a Satellite (

N = 0

)

Two forces act on the body: the pull of the Earth, $\dfrac{GMm}{r^2}$ , and the reaction of the surface, $N$ .
Net force = mass × acceleration. The acceleration of the satellite is $a = \dfrac{GM}{r^2}$ :
$\dfrac{GMm}{r^{2}} - N = ma \qquad\Rightarrow\qquad \dfrac{GMm}{r^{2}} - N = m\left(\dfrac{GM}{r^{2}}\right)$
So the surface puts no force on the body. Its apparent weight is zero:
$N = 0$

Remember

Condition for Weightlessness

A body is weightless in a satellite only if the mass of the satellite is negligible. Then the satellite has no gravity of its own.

e.g. The Moon is a satellite of the Earth. But the Moon has its own large mass. It pulls a body kept on its surface. So the weight of the body on the Moon is not zero.

A simple pendulum experiment cannot be done in a weightless state, because

T = 2\pi\sqrt{\dfrac{L}{g}} = \infty

g = 0

End of Chapter 7 · notesshala.in