A launching pad with a spaceship is moving along a circular orbit of the moon, whose radius $R$ is triple that of the moon $R_{m}$ . The ship leaves the launching pad with a relative velocity equal to the launching pad's initial orbital velocity $V_{0}$ and the launching pad then falls to the moon. If the angle $θ = c o s^{- 1} \frac{n}{\sqrt{10}}$ with the horizontal is the angle at which the launching pad crashes into the surface and its mass is twice that of the spaceship, then the value of $n$ is

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Solution

Let $V_{r}$ be the relative velocity of the spaceship w.r.t the launching pad. Thus,
$V_{r} = V_{0}$
$\Rightarrow V_{s} - V_{l} = V_{0}$ -----(i) (where $V_{s}$ and $V_{l}$ are the velocities of the spaceship and the launching pad after splitting)

Using conservation of momentum along the tagential direction we can say

$(m + 2 m) V_{0} = m V_{s} + 2 m V_{l}$ -----(ii)

Using (i) and (ii) we get
$V_{l} = \frac{2}{3} V_{o}$ and
$V_{s} = \frac{5}{3} V_{o}$

Let $v$ and $θ$ be the velocity and the angle at which the launching pad strikes the surface of the moon.

On applying conservation of total energy on the launching pad of mass $2 m$ ,

$K_{i} + U_{i} = K_{f} + U_{f}$
$\frac{1}{2} (2 m) {(\frac{2}{3} v_{0})}_{0}^{2} + [\frac{- G M (2 m)}{3 R_{m}}] = \frac{1}{2} (2 m) v^{2} + [\frac{- G M (2 m)}{R_{m}}]$ --- (iii)

We know orbital velocity $v_{0} = \sqrt{\frac{G M}{R_{0}}} = \sqrt{\frac{G M}{3 R_{m}}}$

Substituting in equation (iii), we get

$v = \sqrt{\frac{40 G M}{27 R_{m}}}$

By applying conservation of angular momentum, we can say

$(2 m) (\frac{2}{3} V_{0}) (3 R_{m}) = (2 m) (v) (R_{m}) (s i n (90 - θ))$
$⟹ θ = c o s^{- 1} (\frac{3}{\sqrt{10}}$ )
$∴$ value of $n$ is 3

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