A launching pad with a spaceship is moving along a circular orbit of the moon, whose radius `R` is triple that of moon `Rm`. The ship leaves the launching pad with a relative velocity equal to the launching pad's initial orbital velocity `vec(v)_(0)` and the launching pad then falls to the moon. Determine the angle `theta` with the horizontal at which the launching pad crashes into the surface it its mass is twice that of the spaceship `m`.

Answer

`cos theta=3/(sqrt(10))`

Explanation

Let velocity of launching pad after leaving the ship is `v'` from conservation of linear momentum
`m(v_(0)+v')+2mv'=3mv_(0)`
`v'=2/3v_(0)...(1)`
Applying conservation of mechanical energy.
`U_(i)+K_(i)=U_(f)+K_(f)` (here `v` is the velocity by which pad strike,s the surface
`-((Gm)(2m))/(3R)+1/2(2m)v'^(2)=(Gm(2m))/R+1/2(2m)v^(2)...(2)`
orbital velocity `v_(0)=sqrt((GM)/(3R))...(3)`
From `(2)` & `(3)` `-v_(0)^(2)+1/2v'^(2)=-3v_(0)^(2)+1/2v^(2)`
`v=(2sqrt(10))/3 v_(0)....(4)`
from conservation of angular momentum
`2mvR cos theta=2mv'xx3R`
`2mxx(2sqrt(10))/3v_(0)xxR cos theta=2mxx2/3xxv_(0)xx3R`
by solving, we get `cos theta=3/(sqrt(10))`