Question

A point P is on the axis of a fixed ring of mass M and radius R, at a distance $2$R from the centre O. A small particles starts from P and reaches O under the gravitational attraction only. Its speed at O will be.

• A. Zero
• B. $\sqrt{\frac{2GM}{R}}$
• C. $\sqrt{\frac{2GM}{R}(\sqrt{5}-1)}$
• D. $\sqrt{\frac{2GM}{R}(1-\frac{1}{\sqrt{5}})}$

Answer 1

Answer: (D)

$\sqrt{\frac{2GM}{R}(1-\frac{1}{\sqrt{5}})}$

We know gravitational P.E at distance $x$ on axis of ring of radius $R$ is $V=\frac{-GMm}{\sqrt{R^2+x^2}}$

$\therefore V_p=\frac{-GMm}{\sqrt{R^2+4R^2}}=\frac{-GMm}{R\sqrt{5}}{V}_o=\frac{-GMm}{{\sqrt{R}}^2}=\frac{-GMm}{R}$

According to conservation of energy

$\frac{1}{2}m{v}^2-\frac{GMm}{R}=\frac{-GMm}{R\sqrt{5}}=\frac{GMm}{R}(1+\frac{1}{\sqrt{5}})V=\left[\frac{2GM}{R}\right(1-\frac{1}{\sqrt{5}}){]}^{\frac{1}{2}}$