Newton’s law of gravitation in the motion of satellite

STATEMENT:

Consider a satellite of mass “$m$” revolving around Earth of mass “$M$” at height “$h$” from the surface of Earth or distance “$r$” form the Earth’s center.

Hence, From figure:

$R=$ Radius of Earth .

$r=R+h=$ Radius of satellites orbit.

For uniform Circular motion of satellite around earth the action-reaction form must be equal. Hence,

DERIVATION:

$\text{Centripetal force = Gravitational force}$

OR

$\boxed{F_C=F_G}$ ……..(i)

Since,

$F_C=\frac{mV^2}{r}$ & $F_G=G\frac{mM}{r^2}$

Substitute the values in equation (i):

$\frac{mV^2}{r} = G\frac{mM}{r^2}$

By simplifying we get:

${V^2}= G\frac{M}{r}$

Since, $r= R+h$.

Therefore, ${V^2}= G\frac{M}{R+h}$

Take square root on both sides:

$\boxed{{V}= \sqrt {\frac{GM}{R+h}}}$

The above equation gives the velocity of satellites while orbiting around Earth.

TIME PERIOD:

“The time required for a satellite to complete one revolution around the Earth in its orbit is called its ‘Time Period ( $T$ )’”.

The time period of a satellite can be calculated as follows:

DERIVATION:

We know that

$\boxed{T=\frac{2\pi r }{V}}$ ………… (i)

Also, the velocity of satellite is given by the equation:

$\boxed{{V}= \sqrt {\frac{GM}{R+h}}}$ ……….. (ii

Put value of “$V$” from equation (ii) into (i).

$T=\Large \frac{2\pi r }{\sqrt {\frac{GM}{R+h}}}$

$T=2\pi r \Large \sqrt{\frac{R+h}{GM}}$

Since, $r=R+h$

Hence,

$T=2\pi r \Large \sqrt{\frac{r}{GM}}$

$T=2\pi \Large \sqrt{\frac{r.r^2}{GM}}$

$\boxed{T=2\pi \sqrt{\frac{r^3}{GM}}}$

This equation gives the time period for a satellite orbiting around the Earth.