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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Physics
Chapter
Gravitation

BITSAT 2020 · Physics · Gravitation

The potential energy of a satellite of mass m and revolving at a...

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

Options

  1. A−mgRe-\mathrm{m} \mathrm{g} \mathrm{R}_{\mathrm{e}}
  2. B−mgRe2\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{2}
  3. C−mgRe3\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{3}
  4. D−mgRe4\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{4}

Correct answer

B. −mgRe2\frac{-\mathrm{mg} \mathrm{R}_{\mathrm{e}}}{2}

Step-by-step solution

At a height h above the surface of earth the gravitational potential energy of the particle of mass m is Uh=−GMemRe+h\mathrm{U}_{\mathrm{h}}=-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}} Where Me  and ReM_{e} \ \text{ and } R_{e} are the mass \ radius of earth respectively. In this question, since h=Re\mathrm{h}=\mathrm{R}_{\mathrm{e}} So Uh−Re=−GMem2Re=−mgRe2U_{h-R_{e}}=-\frac{G M_{e} m}{2 R_{e}}=\frac{-m g R_{e}}{2}
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