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 the object at point P is
`F = (GMmh)/((r^(2) + h^(2))^((3)/(2)))`
when h = r
`F = (GMmr)/((r^(2) + r^(2))^((3)/(2))) = (GMm)/(2sqrt(2) r^(2))`
When h becomes 2h
`F. = (GMm 2h)/(r^(2) + 4h^(2))^((3)/(2))`
When h = r
And `F. = (2GMm r)/(r^(2) + 4r^(2))^((3)/(2)) = (2GMm)/(5sqrt(5) r^(2))`
`:. (F.)/(F) = (4sqrt(2))/(5sqrt(5))`
or `F. = (4sqrt(2))/(5sqrt(5)) F`