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Q. A satellite is orbiting the earth in a circular orbit of radius $r$. Its

KEAMKEAM 2018Gravitation

Solution:

Orbital velocity of the satellite is $v=\sqrt{\frac{G_E}{r}}$ where $M_E$ is the mass of the earth
Kinetic energy, $K =\frac{1}{2} mv ^2=\frac{ GM _{ E } m }{2 r }$
where $m$ is the mass of the satellite. $K \propto \frac{1}{ r }$
Hence, option (a) is incorrect. Angular momentum, $L = mvr$
$ =m \sqrt{\frac{ GM _{ E }}{ r } r } $
$ =m \sqrt{G_{ E } r } \therefore L \propto \sqrt{ r } $
Hence, option (b) is incorrect.
Linear momentum, $p=m v=m \sqrt{\frac{ GM _{ E }}{ r }} \therefore P \propto \frac{1}{\sqrt{ r }}$
Hence, option (c) is incorrect.
Frequency of revolution, $v=\frac{1}{ T }=\frac{1}{2 \pi} \sqrt{\frac{ GM _{ E }}{ r ^3}}$ $ \therefore v \propto \frac{1}{ r ^{3 / 2}} $