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BITSAT 2020 Paper

Section: Physics

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Question : 40 of 150

Marks: +1, -0

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

$-m g R_c$
$\frac{-mg R_c}{2}$
$\frac{-mg R_c}{3}$
$\frac{-mg R_c}{4}$

Solution:

At a height h above the surface of earth the gravitational potential energy of the particle of mass m is

U_h=-\frac{GM_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_c}=-\frac{G M_c m}{2 R_c}=\frac{-m g R_c}{2}

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