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Question

(A) The gravitational potential energy $U$ of a satellite of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.

Vedclass · Accepted Answer

$-0.5 \, mgR_e$

Vedclass · Answer

(A) The gravitational potential energy $U$ of a satellite of mass $m$ at a distance $r$ from the center of the Earth is given by $U = -\frac{GMm}{r}$.Given height $h = 6.4 	imes 10^6 \, m$. Since the radius of the Earth $R_e \approx 6.4 	imes 10^6 \, m$,we have $h = R_e$.The distance from the center of the Earth is $r = R_e + h = R_e + R_e = 2R_e$.Substituting this into the formula: $U = -\frac{GMm}{2R_e}$.Using the relation $GM = gR_e^2$,where $g$ is the acceleration due to gravity at the Earth's surface:$U = -\frac{(gR_e^2)m}{2R_e} = -\frac{1}{2} mgR_e = -0.5 \, mgR_e$.

What is the potential energy of a satellite at a height of $6.4 \times 10^6 \, m$ from the surface of the Earth? ($R_e$ = radius of the Earth)

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