Question

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

• A. $\frac{3\sqrt{2}}{4\sqrt{3}}$
• B. $\frac{5\sqrt{2}}{\sqrt{3}}$
• C. $\frac{4\sqrt{3}}{5}$
• D. $\frac{4\sqrt{2}}{5\sqrt{5}}$

Answer 1

The correct option is D $\frac{4\sqrt{2}}{5\sqrt{5}}$

Gravitational force acting on an object of mass $m$, placed at a point $\text{P}$ at a distance $h$ along the line joining centre of ring and mass $m$ is given by,

$F=m{E}_g...\left(1\right)$

Where, $E_g$ is a gravitational field at point $\text{P}$ due to ring can be calculated as,

$E_g=\frac{GMh}{(r^2+h^2{)}^{3/2}}...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ we get,

$F=\frac{GMmh}{(r^2+h^2{)}^{3/2}}$

When, the mass is moved to distance $2h$, the force acting on the mass $m$ will be,

$F^{\prime }=\frac{GMm\left(2h\right)}{{[r^2+(2h{)}^2]}^{3/2}}$

$\Rightarrow F^{\prime }=\frac{2GMmh}{(r^2+4h^2{)}^{3/2}}$

When $h=r$,

$\Rightarrow F=\frac{GMmr}{(r^2+r^2{)}^{3/2}}=\frac{GMm}{2\sqrt{2}{r}^2}$

and

$\Rightarrow F^{\prime }=\frac{2GMmr}{(r^2+4r^2{)}^{3/2}}=\frac{2GMm}{5\sqrt{5}{r}^2}$

So,

$\frac{F^{\prime }}{F}=\frac{\frac{2GMm}{5\sqrt{5}{r}^2}}{\frac{GMm}{2\sqrt{2}{r}^2}}$

$\Rightarrow \frac{F^{\prime }}{F}=\frac{4\sqrt{2}}{5\sqrt{5}}$

$\therefore F^{\prime }=\frac{4\sqrt{2}}{5\sqrt{5}}F$

Hence, option (d) is correct answer.