Question:

The change in potential energy when a body of mass m m is raised to a height nR nR from earths surface is (R = radius of the earth)

Updated On: Jun 6, 2022
  • mgR=n(n1) mgR=\frac{n}{(n-1)}
  • mgR mgR
  • mgR=n(n+1) mgR=\frac{n}{(n+1)}
  • mgR=n2(n2+1) mgR=\frac{{{n}^{2}}}{({{n}^{2}}+1)}
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The Correct Option is C

Solution and Explanation

Change in potential energy
ΔU=U2U1 \Delta U={{U}_{2}}-{{U}_{1}}
\therefore ΔU=GMm(R+nR)+GMmR \Delta U=-\frac{GMm}{(R+nR)}+\frac{GMm}{R}
Or ΔU=GMmR(1+n)+GMmR \Delta U=-\frac{GMm}{R(1+n)}+\frac{GMm}{R}
Or ΔU=GMmR[11+n+1] \Delta U=\frac{GMm}{R}\left[ -\frac{1}{1+n}+1 \right]
Or ΔU=(R2g)mR×n(1+n) \Delta U=\frac{({{R}^{2}}g)m}{R}\times \frac{n}{(1+n)}
[g=GMR2] \left[ \because g=\frac{GM}{{{R}^{2}}} \right]
Or ΔU=mgR(nn+1) \Delta U=mgR\left( \frac{n}{n+1} \right)
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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.