If $R$ is the radius of orbit of a satellite, then the kinetic energy of the satellite is

If $R$ is the radius of orbit of a satellite, then the kinetic energy of the satellite is
  1. $\propto \frac{1}{R}$
  2. $\propto \frac{1}{\sqrt{R}}$
  3. $\propto R$
  4. $\propto \frac{1}{R^{3 / 2}}$

Solution

We know that, the kinetic energy of satellite revolving around a planet of radius of orbit $R$ is $\mathrm{KE}=\frac{G M m}{2 R}$ where, $M=$ mass of planet, $m=$ mass of satellite $R=$ radius of orbit $G=$ universal gravitational constant. $\therefore \quad \mathrm{KE} \propto \frac{1}{R}$ Hence, $\mathrm{KE}$ is inversely proportional to radius of orbit.

Asked in: AP EAMCET 2021 (24 Aug Shift 2)

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