Question

A particle of mass $m$ is kept on the axis of a fixed circular ring of mass $M$ and radius $R$ at a distance $x$ from the centre of the ring. Find the maximum gravitational force between the ring and the particle.

• A. $(F_x{)}_{max}=\frac{2GMm}{3\sqrt{3}{R}^2}$.
  
• B. $(F_x{)}_{max}=\frac{3GMm}{2\sqrt{3}{R}^2}$.
• C. $(F_x{)}_{max}=\frac{4GMm}{5\sqrt{3}{R}^2}$.
• D. $(F_x{)}_{max}=\frac{6GMm}{7\sqrt{3}{R}^2}$.

Answer 1

The correct option is A $(F_x{)}_{max}=\frac{2GMm}{3\sqrt{3}{R}^2}$.

Mass $M$ is placed at point $P$ consider a mass element of mass $dx$ of mass $dm$ at a point $T$.

Let the distance $\frac{b}{w}$ $P$ and $T$ ber

Mass per unit length is $=\frac{M}{2\pi R}$

Mass of element $dm=\frac{M}{2\pi R}\times dx$

Gravitational force on mass $m$ due to mass element

$dF=GM\frac{\left(M/2\pi R\right)dx}{r^2}$

Now due to symmetry force perpendicular to the axis due to all elements will cancel each other

$F_y=\phi d{F}_y=0$

While component along $x-$axis will add to give$F=\phi F\cos \theta$

$F=\frac{{GM}_m}{2\pi R}\phi \frac{dx}{r^2}\left(\frac{x}{r}\right)$

$r=\sqrt{r^2+R^2}$

On integrating with proper limit $(x=0to2\pi R)$ and further solving $x<<R$

we get $x=\frac{2{GM}_m}{3\sqrt{3}{R}^2}$