# A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g0, the value of acceleration due to gravity at the earth's surface, is

Gravitational potential energy at height 'h' -

$U_{h}=-\frac{GMm}{R+h}$

$U_{h}=-\frac{gR^{2}m}{R+h}$

$U_{h}\rightarrow$ Potential energy at height $h$

$R\rightarrow$ Radius of earth

- wherein

$U_{h}=-\frac{mgR}{1+\frac{h}{R}}$

total energy of satelliteis equal to half of its potential energy.

$\therefore E= \frac{-G m_{s}.Me}{2(R+h)}$

$\because g.=\frac{GMe}{R_{2}}$  or  $GMe = g_{0}R^{2}$

$E =\frac{ g_{0} m_{s}.R^{2}}{2(R+h)}$

### Preparation Products

##### JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
##### Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 4999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
##### Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-