Answer

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**Hint:**use the gravitational potential energy formula, which is given by

$ U=-\dfrac{GMm}{r} $ Here, m is mass of object ,

M is mass of earth

At the surface of earth r = R

Then gravitational potential energy, $ $ $ U=-\dfrac{GMm}{R} $

Change in potential energy,

$ \Delta u={{u}_{f}}-{{u}_{i}} $

Use the above formula to find the required result.

**Complete step by step solution**

We have given the body of mass m,

And mass of earth by M

At the surface of earth, Gravitational potential energy is given by,

$ {{U}_{i}}=-\dfrac{GMm}{R} $

If h is the height from the surface then, r = h+R , here, r is distance taken from the surface of earth.

Also, we have given height is equal to twice the radius of earth i.e., h = 2R

Then, r = $ \text{2R}+\text{R} $

= 3R

The final gravitational potential energy is given by,

$ {{U}_{f}}=-\dfrac{GMm}{3R} $

Change in potential energy is given by,

$ \Delta u={{u}_{f}}-{{u}_{i}} $

$ =-\dfrac{GMm}{3R}-\text{(}\dfrac{-GMm}{R}) $

$ =-\dfrac{GMm}{3R}+\dfrac{GMm}{R} $

$ \Delta u=\dfrac{2}{3}\dfrac{GMm}{R} $ -------- (1)

Use the relation between g(gravity) and G (Gravitational constant).

$ g=\dfrac{GM}{{{R}^{2}}} $

Hence, The eq. (1) becomes

$ \Delta u=\dfrac{2}{3}\times \dfrac{g{{R}^{2}}m}{R} $

$ \Delta u=\dfrac{2}{3}mgR $ this is the required result.

**So option (D) is the correct answer.**

**Note**

Negative sign show that the potential energy is due to attractive gravitational force exerted by earth on the body

We know $ u=-\dfrac{GMm}{r} $ , if the value of r increases, the gravitational potential energy increases as it becomes less negative. When r is infinity, gravitational potential energy becomes zero.

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