# Kepler’s Law

## What are Kepler’s Laws of Planetary motion?

To explain the motion of planets, Kepler formulated three laws which are popularly known as Kepler’s law of planetary motion. These are –

1. First law of Elliptical orbits (Law of Ellipses).
2. Second law of areas (Law of Equal Areas).
3. Third law of periods (Law of Harmonies).

### Kepler’s First Law of Elliptical Orbit

Kepler’s law states that, each planet revolves around the sun in an elliptical orbit with the sun situated at one of the two foci.

• This law is also called as Kepler’s law of elliptical orbits.
• Planets move around the sun in elliptical orbits as shown in figure.

An ellipse has two foci S and S' . The sun remains located at one focus S .

The points P and A  on the orbit are called the perihelion and the aphelion which represents the closest and farthest distances from the sun respectively.

### Kepler’s Second Law of Areas

Kepler’s law states that, the radius vector drawn from the sun to a planet sweeps out equal areas in equal intervals of time.

• This means that, the areal velocity ( area covered per unit time ) of a planet around the sun is constant.

Consider about a planet moving in an elliptical orbit as shown in figure.

• Suppose time taken by planet to move from A to B  is same as time taken to move from C  to D .
• From Kepler’s second law, the swept area ASB and CSD  covered in equal time must be equal.
• Clearly, the planet covers a larger distance CD  when it is near the sun than AB  when it is farther away in the same interval of time.

• Hence, the linear velocity of a planet is more when it is closer to the sun.

### Kepler’s Third Law of Period

Kepler’s law states that, the square of the period of revolution of a planet around the sun is proportional to the cube of the semi-major axis of its elliptical orbit.

Consider about the figure as shown above.

• Major axis of the elliptical orbit is ( 2a ) .
• Therefore, semi-major axis will be ( a ) .

If ( T ) is the period of revolution of a planet, then –

T^2 \propto a^3

Or, \quad T^2 = K a^3 where K is a constant of proportionality.

## Satellite

A satellite is a small body which continuously revolves by its own around a much larger body in a stable orbit due to effect of gravitational force of attraction.

Based upon their origin and type of motion, satellites are of following three types –

1. Natural satellite.
2. Artificial satellite.
3. Geostationary satellite.

### Natural Satellite

A satellite created by nature itself is called a natural satellite.

Earth is regarded as a natural satellite of Sun and Moon is a natural satellite of the earth itself.

### Artificial Satellite

A satellite made by human is called an artificial satellite.

The first artificial satellite is SPUTNIK-I made by Russia in 1957 . After that many artificial satellites have been put into orbits around the earth to study various phenomenon in the outer regions of the earth’s atmosphere.

To put a satellite into an orbit around the earth, we need to give it two velocities

1. A minimum vertical velocity called escape velocity, to take the satellite to a suitable height.
2. After the satellite has ascended the required height, it is given a suitable horizontal velocity to make it to move in circular orbit around the earth.

### Geostationary Satellite

If a satellite is made to revolve around earth from West to East with a period of revolution of 24 hours in a circular orbit concentric and co-planar with the equatorial plane of the earth, its relative velocity with respect to the earth will be zero.

Such a satellite is called geostationary satellite because it appears stationary to an observer on the earth.

## Escape Velocity

Escape velocity is the minimum velocity by which a body must be projected vertically upwards so that it may just escape beyond the effect of gravitational field of earth.

Consider about a body of mass ( m ) lies at point P  at a distance ( x )  from its centre as shown in figure.

• The gravitational force of attraction on the body at point P  is –

F = \left ( \frac { G M m }{ x^2 } \right )

• The small work done in moving the body through a small distance ( PQ = dx ) against the gravitational force is given by –

d W = F dx = \left ( \frac { G M m }{ x^2 } \right ) dx

• The total work done in moving the body from surface of the earth to a region beyond the gravitational field of the earth i.e. work done in moving the body from \left ( x = R \right ) to \left ( x = \infty \right )  will be –

W = \int {d W} = \int\limits_{ R }^{ \infty } \left ( \frac { G M m }{ x^2 } \right ) dx = ( G M m ) \int\limits_{ R }^{ \infty } x^{ -2 } dx

Or, \quad W = ( G M m ) \left [ - \frac { 1 }{ x } \right ]_{ R }^{ \infty } = ( G M m ) \left ( - \frac { 1 }{ \infty } + \frac { 1 }{ R } \right ) = \left ( \frac { G M m }{ R } \right )

Or, \quad W = \left ( \frac { G M m }{ R } \right )

If ( v_e ) is the escape velocity of the body, then the kinetic energy imparted to the body must be just sufficient to perform the work ( W ) .

• Therefore, \quad \left ( \frac { 1 }{ 2 } \right ) m v_e^2 = \left ( \frac { G M m }{ R } \right )

Or, \quad v_e^2 = \left ( \frac { 2 G M }{ R } \right )

• Hence, escape velocity \quad v_e = \left ( \sqrt \frac { 2 G M }{ R } \right ) …….. (1)

But, \quad g = \left ( \frac { G M }{ R^2 } \right ) \quad Putting this value in equation (1)

• Therefore, \quad v_e = \sqrt { \left ( 2 g R \right ) } ……… (2)

## Orbital Velocity

Orbital velocity is defined as the required velocity to put a satellite into its orbit around the earth.

Let –

• ( M ) is the mass of the earth.
• ( R ) is the radius of the earth.
• ( m ) is the mass of satellite.
• ( v_0 ) is the orbital velocity of the satellite.
• ( h ) is the height of satellite above the earth’s surface.

According to laws of gravitation, the force of gravity on the satellite is –

F = \left [ \frac { G M m }{ \left ( R + h \right )^2 } \right ] .

This gravitational force provides necessary centripetal force to the satellite to keep it in its orbit.

• Centripetal force on a body is given by –

F = \left [ \frac { m v_0^2 }{ \left ( R + h \right ) } \right ]

Therefore, \left [ \frac { m v_0^2 }{ \left ( R + h \right ) } \right ] = \left [ \frac { G M m }{ \left ( R + h \right )^2 } \right ]

Or, \quad v_0^2 = \left [ \frac { G M }{ \left ( R + h \right ) } \right ]

• Therefore, orbital velocity \quad v_0 = \sqrt { \frac { G M }{ \left ( R + h \right ) }} ……….. (1)

If ( g )  is the acceleration due to gravity on the earth’s surface, then –

g = \left ( \frac { G M }{ R^2 } \right )

Or, \quad G M = g R^2

Putting this in equation (1), we have –

v_0 = \sqrt { \frac { g R^2 }{ ( R + h )}} = \sqrt { \frac { g R }{ \left ( 1 + \frac { h }{ R } \right ) }}

When, satellite is close to the earth’s surface, \left ( h << R \right ) .

• Therefore, \quad \left ( \frac { h }{ R } \right ) = 0

Or, \quad v_0 = \sqrt { g R }

From equation (1), it is clear that –

1. Orbital velocity is independent of mass of satellite.
2. It decreases with increases in the radius \left ( R + h \right ) .
3. It depends on the mass and radius of the planet about which the satellite revolves.