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JIPMER 2005
Exam
JIPMER
Year
2005
Subject
Physics
Chapter
Gravitation

JIPMER 2005 · Physics · Gravitation

Radius of orbit of satellite of earth is R . Its kinetic energy is...

Radius of orbit of satellite of earth is RR. Its kinetic energy is proportional to

Options

  1. A1R\frac{1}{R}
  2. B1R\frac{1}{\sqrt{R}}
  3. CRR
  4. D1R3/2\frac{1}{R^{3 / 2}}

Correct answer

A. 1R\frac{1}{R}

Step-by-step solution

Kinetic energy of the satellite KE=12mv02\mathrm{KE}=\frac{1}{2} m v_0^2 ...(1) where v0=GMRv_0=\sqrt{\frac{G M}{R}} Now putting the value of v0v_0 is eq. (1), we get KE=12m(GMR)2=12mGMR\mathrm{KE}=\frac{1}{2} m\left(\sqrt{\frac{G M}{R}}\right)^2=\frac{1}{2} \frac{m G M}{R} Hence, KE∝1R\mathrm{KE} \propto \frac{1}{R}
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