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NEET PHYSICSMedium

A body of mass mm is taken from the Earth’s surface to the height equal to twice the radius (RR) of the Earth. The change in potential energy of the body will be:

A

23mgR\frac{2}{3}mgR

B

3mgR3mgR

C

13mgR\frac{1}{3}mgR

D

2mgR2mgR

Step-by-Step Solution

  1. Gravitational Potential Energy Formula: The gravitational potential energy of a mass mm at a distance rr from the center of the Earth (mass MM) is given by U=−GMmrU = -\frac{GMm}{r} [Eq. 7.23].
  2. Initial State: At the Earth's surface, r1=Rr_1 = R. Therefore, Ui=−GMmRU_i = -\frac{GMm}{R}.
  3. Final State: At a height h=2Rh = 2R above the surface, the distance from the center is r2=R+h=R+2R=3Rr_2 = R + h = R + 2R = 3R. Therefore, Uf=−GMm3RU_f = -\frac{GMm}{3R}.
  4. Change in Potential Energy (ΔU\Delta U): ΔU=Uf−Ui\Delta U = U_f - U_i ΔU=(−GMm3R)−(−GMmR)\Delta U = \left( -\frac{GMm}{3R} \right) - \left( -\frac{GMm}{R} \right) ΔU=GMmR−GMm3R=GMmR(1−13)=2GMm3R\Delta U = \frac{GMm}{R} - \frac{GMm}{3R} = \frac{GMm}{R} \left( 1 - \frac{1}{3} \right) = \frac{2GMm}{3R}
  5. Substitution using Gravity (gg): We know that g=GMR2g = \frac{GM}{R^2}, which implies GM=gR2GM = gR^2. Substituting this into the expression for ΔU\Delta U: ΔU=2(gR2)m3R=23mgR\Delta U = \frac{2(gR^2)m}{3R} = \frac{2}{3}mgR
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