A mass `m` is placed at `P` a distance `h` along the normal through the centre `O` of a thin circular ring of mass `M` and radius `r` Fig.
If the mass is removed futher away such that `OP` becomes `2h`, by what factor the force of gravitational will decrease, if `h = r` ?

Explanation

Gravitational force on an object of mass `m` at point `P` distance `h` from centre `O` lying on the axis of the circular ring of radius `r` and mass `M` is given by
`F = (GM mh)/((r^(2) + h^(2))^(3//2))` [See Sol. Q. 15 (HOTS)]
When `h = 2 h`, then `F' = (GM m xx 2h)/((r^(2) +4 h^(2))^(3//2))`
If `h = r`, then `F = (GM mr)/((r^(2) + r^(2))^(3//2)) = (GM m)/(2 sqrt(2) r^(3))` and `F' = (GM mxx 2r)/((r^(2) + 4r^(2))^(3//2)) = (2GM m)/(5sqrt(5) r^(3))`
`:. (F')/(F) = (4sqrt(2))/(5sqrt(5))` or `F' = (4sqrt(2))/(5sqrt(5))`