Question

The potential energy of a satellite of mass $m$ and revolving at a height $R_{c}$ above the surface of earth where $R_{c}=$ radius of earth, is

• A. $-mgR_c$
• B. $\frac{-m g R_{c}}{2}$
• C. $\frac{-m g R_{c}}{3}$
• D. $\frac{-m g R_{c}}{4}$

Answer 1

Answer: (B)

At a height $h$ above the surface of earth the gravitational potential energy of the particle of mass $m$ is
$U_{h}=-\frac{G M_{e} m}{R_{c}+h}$
Where $M_{c} R_{c}$ are the mass $\&$ radius of earth respectively.
In this question, since $h=R_{c}$
So $U_{h=R_{t}}=-\frac{G M_{c} m}{2 R_{c}}=\frac{-m g R_{c}}{2}$