Sphere of mass `M` and radius `R` is surrounded by a spherical shell of mass `2M` and radius `2R` as shown. A small particle of mass `m` is released from rest from a height `h(ltltR)` above the shell. There is hole in the shell.
What time will it take to move from `A` to `B`?

Options

A. `=(R^(2))/(sqrt(GMh))`
B. `gt(R^(2))/(sqrt(GMh))`
C. `lt(R^(2))/(sqrt(GMh))`
D. none of these

Answer

C

Explanation

`r_(1)=a-ae=a(1-e)`
`r_(2)=a+ae=a(1+e)....(i)`
`mV_(1)r_(1)sin90^(@)=mV_(2)r_(2) sin 90^(@)`
`V_(1)r_(1)=V_(2)r_(2)......(ii)`
According to conservation of energy at `P` and `A`
`1/2mV_(1)^(2)-(GMm)/(r_(1))=1/2mV_(2)^(2)-(GMm)/(r_(2))....(iii)`
From `(i),(ii)` and `(iii) V_(1)=sqrt((GM)/a((1+e)/(1-e)))`