Answer:
Gravitational force
on an object of mass m at point P on the axis of die circular ring of radius r
and mass M is given by
\[F=\frac{GMmh}{{{({{r}^{2}}+{{h}^{2}})}^{3/2}}}\]
Where
h = 2h, then
\[F'\,=\,\frac{GMm\,\times
\,2h}{{{({{r}^{2}}+4{{h}^{2}})}^{3/2}}}\]
If \[h=r,\]
then
\[F=\frac{GMmr}{2\sqrt{2}\,{{r}^{3}}}\,=\frac{GMm}{2\sqrt{2}\,{{r}^{2}}}\]
and
\[F'=\frac{2GMmr}{{{(5{{r}^{2}})}^{3/2}}}\,=\frac{2\,GMm}{5\sqrt{5}\,{{r}^{2}}\,}\]
\[\therefore
\]\[\frac{F'}{F}=\,\,\frac{4\sqrt{2}}{5\sqrt{5}}\] or \[F'\,=\frac{4\sqrt{2}}{5\sqrt{5}}\,F\]
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