Let us consider system consisting of a ring and mass $m$ is placed at a distance $h$ at a point $(P)$ mass is shown in figure.


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
$$
\begin{gathered}
F=\frac{G M m \cos \theta}{A P^2}, \quad\left[\because \cos \theta=\frac{h}{\left(r^2+h^2\right)^{1 / 2}}\right] \\
\left.F=\frac{G M m h}{\left(r^2+h^2\right)^{3 / 2}} \quad \text { (along } P O\right) \quad \ldots \text { (i) }
\end{gathered}
$$
When mass is displaced upto distance $2 h$, then
$$
\begin{array}{r}
F^{\prime}=\frac{G M m \times 2 h}{\left[r^2+(2 h)^2\right]^{3 / 2}} \quad\left[\because A P^2=\left(r^2+h^2\right)\right] \\
{[\because h=2 r]}
\end{array}
$$
$$
=\frac{2 G M m h}{\left(r^2+4 h^2\right)^{3 / 2}}
$$
When $h=r$, then from Eq. (i)
$$
\begin{aligned}
&F=\frac{G M m \times r}{\left(r^2+r^2\right)^{3 / 2}} \Rightarrow F=\frac{G M m}{2 \sqrt{2 r^2}} \\
&\text { and } F^{\prime}=\frac{2 G M m r}{\left(r^2+4 r^2\right)^{3 / 2}}=\frac{2 G M m}{5 \sqrt{5 r^2}} \\
&{[\text { [From Eq. (ii) substituting } h=r]} \\
&\therefore \frac{F^{\prime}}{F}=\frac{4 \sqrt{2}}{5 \sqrt{5}} \Rightarrow F^{\prime}=\frac{4 \sqrt{2}}{5 \sqrt{5}} F
\end{aligned}
$$
Gravitational force on $m$ at distance $2 r$ from $O$ is the $\frac{4}{5} \sqrt{\frac{2}{5}}$ time the gravitational force when $m$ is placed at $r$ distance from $O$.