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7.Gravitation
normal
The potential energy of a satellite of mass $m$ and revolving at a height $R_e$ above the surface of earth where $R_e =$ radius of earth, is
A
$-m\, g\, R_e$
B
$\frac{{ - m\,g\,{R_e}}}{2}$
C
$\frac{{ - m\,g\,{R_e}}}{3}$
D
$\frac{{ - m\,g\,{R_e}}}{4}$
Solution
At a height $\mathrm{h}$ above the surface of earth the gravitational potential energy of the particle of mass $\mathrm{m}$ is
$\mathrm{U}_{\mathrm{h}}=-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}}$
Where $\mathrm{M}_{\mathrm{e}} $ and $ \mathrm{R}_{\mathrm{e}}$ are the mass and radius of earth respectively.
In this question, since $\mathrm{h}=\mathrm{R}_{\mathrm{e}}$
So $U_{h-R_{e}}=-\frac{G M_{e} m}{2 R_{e}}=\frac{-m g R_{e}}{2}$
Standard 11
Physics
