The potential energy of a satellite of mass m | vedclass

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(B) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{G M_e m}{r}$.H

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$-\frac{m g R_e}{2}$

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(B) The gravitational potential energy $U$ of a body of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{G M_e m}{r}$.Here,the satellite is at a height $h = R_e$ above the surface of the Earth.Therefore,the distance from the center of the Earth is $r = R_e + h = R_e + R_e = 2 R_e$.Substituting this into the potential energy formula,we get $U = -\frac{G M_e m}{2 R_e}$.We know that the acceleration due to gravity at the surface of the Earth is $g = \frac{G M_e}{R_e^2}$,which implies $G M_e = g R_e^2$.Substituting $G M_e = g R_e^2$ into the expression for $U$,we get $U = -\frac{(g R_e^2) m}{2 R_e} = -\frac{m g R_e}{2}$.

The potential energy of a satellite of mass $m$ revolving at a height $R_e$ above the surface of the Earth,where $R_e$ is the radius of the Earth,is:

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