Physics · Gravitation

NEET (UG) 2023 — Question 32

Two bodies of mass $m$ and $9m$ are placed at a distance $R$ . The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( $G$ = gravitational constant):

Option A:
$-\frac{12 G m}{R}$
Option B:
$-\frac{16 G m}{R}$
Correct
Option C:
$-\frac{20 G m}{R}$
Option D:
$-\frac{8 G m}{R}$

Answer: B

Step-by-step solution

Let the point be at distance $x$ from mass $m$ . Field balance: $\dfrac{Gm}{x^2} = \dfrac{G(9m)}{(R-x)^2}$ . Simplify: $\dfrac{1}{x^2} = \dfrac{9}{(R-x)^2}$ $implies $(R-x)/x = 3$$ so $x = R/4$ . Potential at that point: $V = -\dfrac{Gm}{x} - \dfrac{G(9m)}{R-x}$ . Substitute $x = R/4$ : $V = -\dfrac{Gm}{R/4} - \dfrac{9Gm}{3R/4} = -\dfrac{4Gm}{R} - \dfrac{12Gm}{R} = -\dfrac{16Gm}{R}$ . Thus, the gravitational potential is $-\dfrac{16Gm}{R}$ , which corresponds to option B.

Answer key and solution verified before publishing.

Practise Gravitation

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: NEET (UG) 2023
Paper: 2023 paper
Subject: Physics
Chapter: Gravitation
Topic: Gravitational Potential Energy and Potential

A satellite is orbiting just above the surface of the earth with period T . If d is the density of the earth and G is the universal…

Answer: B

Step-by-step solution

Practise Gravitation

More Gravitation questions from this paper