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(lt ltR)` above the shell. There is hole in the shell.
In what time will it enter the hole at `A`?

Options

A. `2sqrt((hR^(2))/(GM))`
B. `sqrt((2hR^(2))/(GM))`
C. `sqrt((hR^(2))/(GM))`
D. none of these

Answer

A

Explanation

Just before collision, the total energy of two satellites, is
`E=-(GMm)/(2r)-(GMm)/(2r)-(GMm)/r`
Let orbital velocity is `v_(1)` then from momentum conservation,
`mv-mv=2mxxv_(1)rArr v_(1)=0`
As velocity of combined mass just after collision is zero, the combined mass will fall towards earth. at this instant, the total energy of the system only consist of the gravitational potential given by `U=(GMxx2m)/(2r)`