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 lauching pad after leaving the ship is v' from conservation linear momentum
`m(v_(0)+v')+2mv' =3mv_(0) rArr v'=(3)/(2)v_(0)` …(i)
applying conservation of mechanical enmergy.
`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)` ...(ii)
orbitle velocity `v_(0) = sqrt((GM)/(3R))` ...(iii)
from (2) & (3) `-v_(0)^(2)+(1)/(2)v'^(2)=-3v_(0)^(2)+(1)/(2)v^(2) rArr v=(2sqrt(10))/(3)v_(0)` ...(iv)
from conservation of angular momentum
`2mvR cos theta = 2mv'xx3R rArr 2mxx(2sqrt(10))/(3)v_(0)xxR cos theta = 2mxx(2)/(3)v_(o)xx3R`
by solving , we get `cos theta = (3)/(sqrt(10))`