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NEET PHYSICSGRAVITATIONEasy

Question

Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass mm when taken to a height hh from the surface of the earth (of radius RR and mass MM), is given by:

A

GMmR+h-\frac{GMm}{R+h}

B

GMmhR(R+h)\frac{GMmh}{R(R+h)}

C

mghmgh

D

GMmR+h\frac{GMm}{R+h}

Step-by-Step Solution

  1. Formula: The gravitational potential energy of a mass mm at a distance rr from the center of the Earth is given by U(r)=GMmrU(r) = -\frac{GMm}{r} [Eq. 7.23, 7.24].
  2. Initial State: At the surface of the Earth, the distance from the center is RR. Thus, the initial potential energy is Ui=GMmRU_i = -\frac{GMm}{R}.
  3. Final State: At a height hh above the surface, the distance from the center is r=R+hr = R + h. Thus, the final potential energy is Uf=GMmR+hU_f = -\frac{GMm}{R+h}.
  4. Change in Potential Energy (ΔU\Delta U): ΔU=UfUi\Delta U = U_f - U_i ΔU=(GMmR+h)(GMmR)\Delta U = \left( -\frac{GMm}{R+h} \right) - \left( -\frac{GMm}{R} \right) ΔU=GMmRGMmR+h\Delta U = \frac{GMm}{R} - \frac{GMm}{R+h} ΔU=GMm(1R1R+h)\Delta U = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) ΔU=GMm(R+hRR(R+h))\Delta U = GMm \left( \frac{R+h-R}{R(R+h)} \right) ΔU=GMmhR(R+h)\Delta U = \frac{GMmh}{R(R+h)} (Note: This expression reduces to mghmgh only when hRh \ll R).

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.

PHYSICSGRAVITATIONassuminggravitationalpotentialenergyobject

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