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. If the mass is removed further away such that OP becomes 2h, by what factor the force of gravitation will decrease, if h = r ?

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Alok Kumar Singh | Contributor-Level 10

8 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Gravitational force at h height F = $\frac{G M m h}{{{(r}^{2} + h^{2})}^{3 / 2}}$

When mass is displaced upto distance 2h then

F'= $\frac{G M m 2 h}{{{[r}^{2} + {(2 h)]}^{2}]}^{\frac{3}{2}}}$

= $\frac{2 G M m h}{{(r^{2} + 4 h^{2})}^{\frac{3}{2}}}$

When h = r then F= $\frac{G M m h}{[{r^{2} + {(r)}^{2}]}^{3 / 2}}$

So F = $\frac{G M m}{2 \sqrt[]{2 r^{2}}}$

F'= $\frac{2 G M m r}{{(r^{2} + 4 r^{2})}^{\frac{3}{2}}} = \frac{2 G M m}{5 \sqrt[]{5 r^{2}}}$

$\frac{F^{'}}{F} = \frac{4 \sqrt[]{2}}{5 \sqrt[]{5}}$

F' = $\frac{4 \sqrt[]{2}}{5 \sqrt[]{5}} F$

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