A meteorite approaching a planet of mass M (in the straight line passing through the centre of the planet) collides with an automatic space station orbiting the planet in a circular trajectory of radius R. The mass of the station is ten times as large as the mass of the meteorite. As a result of the collision, the meteorite sticks in the station which goes over to a new orbit with the minimum distance R/2 from the planet. Speed of the meteorite just before it collides with the planet is : .

A s the space station is moving in circular orbit,

$\begin{array}{l}\frac{\mathrm{GM}\left(10\mathrm{m}\right)}{{\mathrm{R}}^2}=\frac{\left(10\mathrm{m}\right){\mathrm{v}}_0^2}{\mathrm{R}}\\ \Rightarrow {\mathrm{v}}_0=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}....\left(\mathrm{i}\right)\end{array}$
Let u be the velocity of meteorite.
Velocity of the space station after collision can be obtained from momentum conservation.
$\begin{array}{l}\mathrm{mu}=(10\mathrm{m}+\mathrm{m}){\mathrm{v}}_1\Rightarrow {\mathrm{v}}_1=\frac{\mathrm{u}}{11}\\ 10\mathrm{m}\cdot {\mathrm{v}}_0=(10\mathrm{m}+\mathrm{m}){\mathrm{v}}_2\Rightarrow {\mathrm{v}}_2=\frac{10}{11}{\mathrm{v}}_0\end{array}$
Let v be the velocity of space station at closest distance from angular momentum conservation
$10{\mathrm{mv}}_0\times \mathrm{R}=11\mathrm{mv}\frac{\mathrm{R}}{2}\Rightarrow \mathrm{v}=\frac{20{\mathrm{v}}_0}{11}$
from energy conservation
$\begin{array}{l}\frac{1}{2}\times \left(11\mathrm{m}\right)\left({\mathrm{v}}_1^2+{\mathrm{v}}_2^2\right)-\frac{\mathrm{GM}\left(11\mathrm{m}\right)}{\mathrm{R}}=\frac{1}{2}\times \left(11\mathrm{m}\right){\mathrm{v}}^2-\frac{\mathrm{GM}.11\mathrm{m}}{\mathrm{R}/2}\\ \Rightarrow {\left(\frac{\mathrm{u}}{11}\right)}^2+{\left(\frac{10{\mathrm{v}}_0}{11}\right)}^2-\frac{2\mathrm{GM}}{\mathrm{R}}={\left(\frac{20{\mathrm{v}}_0}{11}\right)}^2-\frac{4\mathrm{GM}}{\mathrm{R}}\\ \Rightarrow \frac{{\mathrm{u}}^2}{{11}^2}=\frac{400{\mathrm{v}}_0^2}{{11}^2}-\frac{100{\mathrm{v}}_0^2}{{11}^2}-\frac{2\mathrm{GM}}{\mathrm{R}}\\ \Rightarrow {\mathrm{u}}^2=\frac{\mathrm{GM}}{\mathrm{R}}(400-100-242)=58\frac{\mathrm{GM}}{\mathrm{R}}\end{array}$
$\mathrm{u}=\sqrt{\frac{58\mathrm{GM}}{\mathrm{R}}}$