Question

A uniform ring of mass $2\,m$ and radius $a$ is placed directly above the uniform sphere of mass $M$ and of same radius. The centre of ring is at distance $\sqrt{3}a$ from the centre of sphere. The gravitational force exerted by the sphere on the ring is $N\frac{GMm}{a^{2}}$ units. Write the value of $\left[100\, N\right],$ where $\left[\right]$ is the greatest integer function.

Answer 1

$dF=G\frac{Mdm}{4 a^{2}}$
$F=\Sigma dFcos\theta $
$=\displaystyle \sum \frac{G M d m}{4 a^{2}}cos\theta $
$=\frac{G M}{4 a^{2}}\cdot \frac{\sqrt{3 a}}{2 a}\displaystyle \sum dm$
$=\frac{\sqrt{3} G M}{8 a^{2}}\cdot 2m$
$=\frac{\sqrt{3}}{4}\cdot \frac{GMm}{a^{2}}$
$\therefore F=0.43\frac{GMm}{a^{2}}$
$\therefore N=0.43$
Alternate method

Gravitational field $\left(\right.\bar{E}\left.\right)=\frac{G 2 m d}{\left(a^{2} + d^{2}\right)^{3 / 2}}$
$=\frac{\sqrt{3} \times 2 GM}{8 a^{2}}$
$\therefore \bar{E}=\frac{\sqrt{3}}{4}\frac{Gm}{a^{2}}$
Force $=M\bar{E}=\frac{\sqrt{3}}{4}\frac{G M m}{a^{2}}$
$\therefore F=0.433\frac{G M m}{a^{2}}$
$N=0.43$
$\left[100\, N\right]=43$