A launching pad with a spaceship is moving along a circular orbit of the moon, whose radius $R$ is triple that of moon $R m .$ The ship leaves the launching pad with a relative velocity equal to the launching pad's initial orbital velocity ${\to v}_{0}$ and the launching pad then falls to the moon. The angle $θ$ with the horizontal at which the launching pad crashes into the surface if its mass is twice that of the spaceship $m .$

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Solution

Here, the momentum of launcher spaceship is conserved so initial momentum $=$ final momentum

Mass of spaceship is $m \to$ Mass of launcher is $2 m$

Initial momentum $= (m + 2 m) v_{o}$

Final momentum $= m (V + v_{o}) + 2 m V$

so $, 3 m v_{o} = m v_{o} + 3 m V \Rightarrow V = \frac{2 v_{o}}{3}$

Now, final horizontal velocity $= V_{h} = \frac{2 v_{o}}{3} ($ since gravity acts downwards $)$

orbital velocity $v_{o} = (\frac{G M}{3 R})^{0.5} \to$

Change in P.E. $=$ change in vertical K.E.

therefore, $\frac{- G M (2 m)}{3 R} + \frac{- G M (2 m)}{R} = \frac{(2 m) (V_{v})^{2}}{2}$

solving this we get $V_{v} = (\frac{4 G M}{3 R})^{0.5} = 2 v_{o}$

$cos θ = \frac{V_{h}}{\sqrt{V_{h}^{2} + V_{h}^{2}}} = \frac{\frac{2 v_{o}}{3}}{\sqrt{{(\frac{2 v_{o}}{3})}^{2} + {(2 v_{o})}_{o}^{2}}} = \frac{3}{\sqrt{10}}$

So $, x = 3$

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