Physics - Gravitation Question with Solution | TestHub

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