
orbital velocity v0=4R/3GM=4R3GM Angular momentum of satellite =2Mv034R=2M⋅4R3GM⋅34R=M3GMRx=3 (ii) Since mass of satellite is comparable to the mass of earth.
(34R)2 G.M. 2M=2Mω2⋅98Rω=128R381GM Angular momentum of satellite about common centre of mass, $\begin{aligned}
& L=\frac{M}{2} \cdot\left(\frac{8 R}{9}\right)^2 \cdot \omega \
& L=M \sqrt{\operatorname{GMR}\left(\frac{8}{81}\right)} \
& x=\frac{81}{8} \simeq 10
\end{aligned}$