The satellite of mass m revolving in a circular orbit of radius r around the earth has kinetic energy E. Then its angular momentum will be

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$\frac{E}{2 m^{2}}$

$\sqrt{\frac{E}{{mr}^{2}}}$

$\sqrt{2 Emr}$

$\sqrt{2 {Emr}^{2}}$

answer is C.

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Detailed Solution

$\begin{array}{l} L = m V_{0} r; V_{0} = \sqrt{\frac{GM}{r}}, \\ E = \frac{GMm}{2 r} \Rightarrow GM = \frac{2 rE}{m} \\ ∴ V_{0} = \sqrt{\frac{2 rE}{mr}} = \sqrt{\frac{2 E}{m}} \\ L = m (\sqrt{\frac{2 E}{m}}) \cdot r = \sqrt{2 {mEr}^{2}} \end{array}$

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