The potential energy of a satellite of mass $m$ revolving at height $R$ above the surface of the earth where $R =$ radius of earth is
$- m g R$
$- \frac{m g R}{2}$
$- \frac{m g R}{4}$
$- \frac{m g R}{3}$

Question

The potential energy of a satellite of mass m revolving height R above the surface of the earth where R - radius of earth is-mgR-dfrac -mgR-2-dfrac -mgR-4-dfrac -mgR-3-

Toppr Admin · Accepted Answer

PE-x2212-GMm2R-x2212-x2212-GMmR-x2212-GMm2RAlso-xA0- -xA0- -xA0- -xA0-g-GMR2Putting the value of-xA0-GMR in equation -1- we get-PE-x2212-gmR2

The potential energy of a satellite of mass m revolving at height R above the surface of the earth where R= radius of earth is−mgR−mgR2−mgR4−mgR3

PE=−GMm2R−(−GMmR)=−GMm2RAlso, g=GMR2Putting the value of GMR in equation (1) we get,PE=−gmR2

The potential energy of a satellite of mass $m$ revolving at height $R$ above the surface of the earth where $R =$ radius of earth is
$- m g R$
$- \frac{m g R}{2}$
$- \frac{m g R}{4}$
$- \frac{m g R}{3}$

$P E = \frac{- G M m}{2 R} - (\frac{- G M m}{R}) = \frac{- G M m}{2 R}$
Also,
$g = \frac{G M}{R^{2}}$
Putting the value of $\frac{G M}{R}$ in equation $(1)$ we get,
$P E = \frac{- g m R}{2}$