Question

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 the minimum distance $R/2$ from the planet. Speed of the meteorite just before it collides with the planet is:

• A. $\sqrt{\frac{58GM}{R}}$
• B. $\sqrt{\frac{38GM}{R}}$
• C. $\sqrt{\frac{28GM}{R}}$
• D. $\sqrt{\frac{18GM}{R}}$

Answer 1

The correct option is A $\sqrt{\frac{58GM}{R}}$


$\begin{array}{c}\frac{GM\left(10m\right)}{R^2}=\frac{\left(10m\right)v_0^2}{R}\\ \Rightarrow v_0=\sqrt{\frac{GM}{R}}\end{array}$
Let u be the velocity of meteorite. Velocity of the space station after collsion can be obtained from momentum conservation.
$mu=(10m+m)v_1⟹v_1=\frac{u}{11}10m\cdot v_0=(10m+m)v_2⟹v_2=\frac{10}{11}{v}_0$

Let $v$ be the velocity of space station at closest distance.

From angular momentum conservation $10m{v}_0\times R=11mv\frac{R}{2}⟹v=\frac{20v_0}{11}$

from energy conservation $\frac{1}{2}\times \left(11m\right)(v_1^2+v_2^2)-\frac{GM\left(11m\right)}{R}=\frac{1}{2}\times \left(11m\right)v^2-\frac{GM\cdot 11m}{R/2}$

$\Rightarrow {\left(\frac{u}{11}\right)}^2+{\left(\frac{10v_0}{11}\right)}^2-\frac{2GM}{R}={\left(\frac{20v_0}{11}\right)}^2-\frac{4GM}{R}$

$\Rightarrow \frac{u^2}{{11}^2}=\frac{400v_0^2}{{11}^2}-\frac{100v_0^2}{{11}^2}-\frac{2GM}{R}$

$\Rightarrow u^2=\frac{GM}{R}(400-100-242)=58\frac{GM}{R}$

So, $u=\sqrt{\frac{58GM}{R}}$