A satellite of mass M2 is revolving around earth in a circular orbit at a height of R3 from earth surface. The angular momentum of the satellite is MGMRx. The value of x is _______, where M and R are the mass and radius of earth, respectively. (G is the gravitational constant)

orbital velocity v0=GM4R/3=3GM4RAngular momentum of satellite =M2v04R3=M2⋅3GM4R⋅4R3=MGMR3x = 3

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A satellite of mass $\frac{M}{2}$ is revolving around earth in a circular orbit at a height of $\frac{R}{3}$ from earth surface. The angular momentum of the satellite is $M \sqrt{\frac{GMR}{x}}$ . The value of x is _______, where M and R are the mass and radius of earth, respectively. (G is the gravitational constant)

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answer is 3.

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orbital velocity $v_{0} = \sqrt{\frac{GM}{4 R / 3}} = \sqrt{\frac{3 GM}{4 R}}$
Angular momentum of satellite $= \frac{M}{2} v_{0} \frac{4 R}{3}$

$\begin{array}{l} = \frac{M}{2} \cdot \sqrt{\frac{3 GM}{4 R}} \cdot \frac{4 R}{3} \\ = M \sqrt{\frac{GMR}{3}} \end{array}$
x = 3

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