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` ?

Options

A. `(3sqrt(2))/(4sqrt(3))`
B. `(5sqrt(2))/(sqrt(3))`
C. `(4sqrt(3))/(5)`
D. `(4sqrt(2))/(5sqrt(5))`

Answer

D

Explanation

Gravitational force acting on an object of mass m placed at point P at a distance h along the normal through the centre of a circular ring of mass M and radius r is given by
`F = (GMnh)/((r^(2)+h^(2))^(3//2))`
When mass is displaced upto distance 2h, then
`F' = (GMm xx 2h)/((r^(2)+(2h)^(2))^(3//2)) = (2GMmh)/((r^(2)+4h^(2))^(3//2))`
When h = r, then
`F = (GMmxxr)/((r^(2)xxr^(2))^(3//2)) = (GMm)/(2sqrt(2)r^(2))`
and `F' = (2GMmr)/((r^(2)+4^(2))^(3//2)) =(2GMm)/(5sqrt(5)r^(2))`
`:. (F')/(F) = (4sqrt(2))/(5sqrt(5))` or `F' = (4sqrt(2))/(5sqrt(5))F`