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A body of mass ‘m’ taken from the earth surface to the height equal to twice the radius (R) of the earth. The change in potential energy of the body will be,
(A) 2mgR
(B) $ \dfrac{1}{3}mgR $
(C) 3mgR
(D) $ \dfrac{2}{3}mgR $

Last updated date: 29th Feb 2024
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IVSAT 2024
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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.

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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