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NEET PHYSICSGRAVITATIONEasy

Question

A body of mass m is taken from the earth’s 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

\frac{2}{3}mgR

C

3mgR

D

\frac{1}{3}mgR

Step-by-Step Solution

The gravitational potential energy (UU) of a body of mass mm at a distance rr from the centre of the Earth is given by U=GMmrU = -\frac{GMm}{r}.

  1. Initial State: At the Earth's surface, r1=Rr_1 = R. So, Ui=GMmRU_i = -\frac{GMm}{R}.
  2. Final State: At a height h=2Rh = 2R above the surface, the distance from the centre is r2=R+h=R+2R=3Rr_2 = R + h = R + 2R = 3R. So, Uf=GMm3RU_f = -\frac{GMm}{3R}.
  3. Change in Potential Energy (ΔU\Delta U): ΔU=UfUi=(GMm3R)(GMmR)\Delta U = U_f - U_i = \left( -\frac{GMm}{3R} \right) - \left( -\frac{GMm}{R} \right) ΔU=GMmR(113)=2GMm3R\Delta U = \frac{GMm}{R} \left( 1 - \frac{1}{3} \right) = \frac{2GMm}{3R}
  4. Substitution: We know that acceleration due to gravity at the surface is g=GMR2g = \frac{GM}{R^2}, which implies GM=gR2GM = gR^2. Substituting this into the equation: ΔU=2(gR2)m3R=23mgR\Delta U = \frac{2(gR^2)m}{3R} = \frac{2}{3}mgR.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from GRAVITATION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSGRAVITATIONearthssurfaceheightradiuschange

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