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If $R$ is the radius of orbit of a satellite, then the kinetic energy of the satellite is
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The correct answer is:
$\propto \frac{1}{R}$
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.
$\mathrm{KE}=\frac{G M m}{2 R}$
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.
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