Q125 / 125 · BITSAT 2020 · Physics

The potential energy of a satellite of mass $\mathrm{m}$ and revolving at a height $\mathrm{R}_{\mathrm{e}}$ above the surface of earth where $\mathrm{R}_{\mathrm{e}}=$ radius of earth, is

  • A
    $-\mathrm{m} \mathrm{g} \mathrm{R}_{\mathrm{e}}$
  • Correct answer
    B
    $\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{2}$
  • C
    $\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{3}$
  • D
    $\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{4}$

Solution

At a height h above the surface of earth the gravitational potential energy of the particle of mass m is $\mathrm{U}_{\mathrm{h}}=-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}}$ Where $M_{e} \& R_{e}$ are the mass $\&$ radius of earth respectively. In this question, since $\mathrm{h}=\mathrm{R}_{\mathrm{e}}$ So $U_{h-R_{e}}=-\frac{G M_{e} m}{2 R_{e}}=\frac{-m g R_{e}}{2}$

Hence, the correct answer is option: 'B'.

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