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# If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass m from the earth’s surface to a height equal to the radius of earth is:A) \$\dfrac{{mgR}}{4}\$B) \$\dfrac{{mgR}}{2}\$C) \$mgR\$D) \$2mgR\$

Last updated date: 20th Jun 2024
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Hint: The potential energy of a body is the energy possessed by the virtue of its position in the gravitational field. The gravitational potential energy of an object is given by the following equation:
\$U = - \dfrac{{GMm}}{r}\$
Where G=gravitational constant, M=mass of the earth, m=mass of the body, and r=distance of body from the centre of the earth. Calculate the two potential energies one at surface of earth and other at height equal to R, then gain in potential energy is given by difference in two potential energies.

Complete step by step solution:
Step1: Calculate the gravitational potential energy at the surface of the earth.
The gravitational potential energy of an object is given by the following equation-
\$U = - \dfrac{{GMm}}{r}\$
Where G=gravitational constant, M=mass of the earth, m=mass of the body, and r=distance of body from the centre of the earth.
At earth’s surface r=R
Therefore,
\${U_1} = - \dfrac{{GMm}}{R}\$
Step2: calculate the gravitational potential energy at a height of R.
Therefore, total distance from centre of earth (r)= R+R=2R
Gravitational potential energy at this height is given by-
\${U_2} = - \dfrac{{GMm}}{{2R}}\$

Step3: Calculate the gain in potential energy.
Therefore,
\$\Delta U = {U_2} - {U_1}\$
\$\Delta U = - \dfrac{{GMm}}{{2R}} + \dfrac{{GMm}}{R}\$
\$ \Rightarrow \Delta U = \dfrac{{GMm}}{{2R}}\$
Also, acceleration due to gravity is given by-
\$g = \dfrac{{GM}}{{{R^2}}}\$
\$ \Rightarrow GM = g{R^2}\$
Substituting above, we get –
\$\Delta U = \dfrac{{(g{R^2})m}}{{2R}}\$
\$\Delta U = \dfrac{{mgR}}{2}\$
Which is gain in the potential energy, hence option (B) is the correct answer.