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 $M \sqrt{\frac{G M R}{x}}$. The value of $x$ is ______ , where $M$ and $R$ are the mass and radius of earth, respectively. ( G is the gravitational constant)

Solution

$\begin{aligned} & \text { orbital velocity } v_0=\sqrt{\frac{\mathrm{GM}}{4 \mathrm{R} / 3}}=\sqrt{\frac{3 \mathrm{GM}}{4 \mathrm{R}}} \\ & \text { Angular momentum of satellite }=\frac{\mathrm{M}}{2} \mathrm{v}_0 \frac{4 R}{3} \\ & =\frac{M}{2} \cdot \sqrt{\frac{3 \mathrm{GM}}{4 \mathrm{R}}} \cdot \frac{4 \mathrm{R}}{3} \\ & =M \sqrt{\frac{\mathrm{GMR}}{3}} \\ & x=3\end{aligned}$ (ii) Since mass of satellite is comparable to the mass of earth.
$\begin{aligned} & \frac{\text { G.M. } \frac{\mathrm{M}}{2}}{\left(\frac{4 \mathrm{R}}{3}\right)^2}=\frac{\mathrm{M}}{2} \omega^2 \cdot \frac{8 \mathrm{R}}{9} \\ & \omega=\sqrt{\frac{81 \mathrm{GM}}{128 \mathrm{R}^3}}\end{aligned}$ Angular momentum of satellite about common centre of mass, $\begin{aligned} & L=\frac{M}{2} \cdot\left(\frac{8 R}{9}\right)^2 \cdot \omega \\ & L=M \sqrt{\operatorname{GMR}\left(\frac{8}{81}\right)} \\ & x=\frac{81}{8} \simeq 10 \end{aligned}$

Hence, the correct answer is 3.