Question:

Two spheres each of mass $M$ and radius $R$ are separated by a distance of $r$. The gravitational potential at the midpoint of the line joining the centres of the spheres is

Updated On: Jul 7, 2022
  • $ - \frac{GM}{r}$
  • $ - \frac{2GM}{r}$
  • $ - \frac{GM}{2r}$
  • $ - \frac{4GM}{r}$
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The Correct Option is D

Solution and Explanation

Let $P$ be the midpoint of the line joining the centres of the spheres, The gravitational potential at point $P$ is $V_{p} = \frac{GM}{r/2} - \frac{GM}{r/2}$ $= - \frac{2GM}{r} - \frac{2GM}{r}=- \frac{4GM}{r}$
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Concepts Used:

Gravitational Potential Energy

The work which a body needs to do, against the force of gravity, in order to bring that body into a particular space is called Gravitational potential energy. The stored is the result of the gravitational attraction of the Earth for the object. The GPE of the massive ball of a demolition machine depends on two variables - the mass of the ball and the height to which it is raised. There is a direct relation between GPE and the mass of an object. More massive objects have greater GPE. Also, there is a direct relation between GPE and the height of an object. The higher that an object is elevated, the greater the GPE. The relationship is expressed in the following manner:

PEgrav = mass x g x height

PEgrav = m x g x h

Where,

m is the mass of the object,

h is the height of the object

g is the gravitational field strength (9.8 N/kg on Earth) - sometimes referred to as the acceleration of gravity.