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

(A) The gravitational force $F$ acting on a mass $m$ placed at a distance $h$ along the axis of a circular ring of mass $M$ and radius $r$ is given by:
$F = \frac{GMmh}{(r^2 + h^2)^{3/2}}$
When the mass is moved to a distance $2h$,the new force $F'$ is:
$F' = \frac{GMm(2h)}{(r^2 + (2h)^2)^{3/2}} = \frac{2GMmh}{(r^2 + 4h^2)^{3/2}}$
Given $h = r$,we substitute this into the expressions:
$F = \frac{GMmr}{(r^2 + r^2)^{3/2}} = \frac{GMmr}{(2r^2)^{3/2}} = \frac{GMm}{2\sqrt{2}r^2}$
$F' = \frac{2GMmr}{(r^2 + 4r^2)^{3/2}} = \frac{2GMmr}{(5r^2)^{3/2}} = \frac{2GMm}{5\sqrt{5}r^2}$
Now,calculating the ratio $F'/F$:
$\frac{F'}{F} = \frac{2GMm}{5\sqrt{5}r^2} \times \frac{2\sqrt{2}r^2}{GMm} = \frac{4\sqrt{2}}{5\sqrt{5}}$
Thus,the force decreases by a factor of $\frac{4\sqrt{2}}{5\sqrt{5}}$.