__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 –

- First law of Elliptical orbits (Law of Ellipses).
- Second law of areas (Law of Equal Areas).
- 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

**which represents the closest and farthest distances from the sun respectively.**

*aphelion*__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
- Clearly, the planet covers a larger distance CD
- 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 –

- Natural satellite.
- Artificial satellite.
- 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* –

- A minimum vertical velocity called escape velocity, to take the satellite to a suitable height.
- 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

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 )

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 –

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

See numerical problems based on this article.