Question

What is the gravitational potential energy of a particle of mass $m$ kept at a distance $x$ from the centre of a disc of mass $M$ on its axis? The radius of the disc is $R$.

• A. $-\frac{2GMm}{R^2}(\sqrt{R^2+x^2}-x)$.
• B. $-\frac{4GMm}{R^2}(\sqrt{R^2+x^2}-x)$.
• C. $-\frac{6GMm}{R^2}(\sqrt{R^2+x^2}-x)$.
• D. $-\frac{9GMm}{R^2}(\sqrt{R^2+x^2}-x)$.

Answer 1

The correct option is A $-\frac{2GMm}{R^2}(\sqrt{R^2+x^2}-x)$.

Let's break the disc into small rings

$dm-$ mass of the ring, SO

$dm=2\pi rdr\times \frac{M}{{\pi R}^2}$

So, potential

$dP=\frac{-Gdm}{\sqrt{x^2+r^2}}$

So,

$dP=\frac{G.2\pi xtan\varphi x{sec}^2\varphi d\varphi }{\left(x\sqrt{1+{tan}^2\varphi }\right)\pi R^2}$

$\Rightarrow P=\frac{-2GMx}{R^2}{\int }_0^{{\tan }^{-1}\left(R/x\right)}\tan \varphi \sec \varphi d\varphi$

$\Rightarrow P=\frac{-2GMx}{R^2}(1-\frac{1}{\cos \left({\tan }^2\left(R/x\right)\right)})$

$\Rightarrow P=\frac{-2GM}{R^2}(\sqrt{R^2+x^2}-x)$