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JEE Main 2019 · Physics · Gravitation · medium

A satellite of mass m moves in a circular orbit at a height R above the Earth's surface, where R is the radius of the Earth. It is moved to a…

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A satellite of mass mm moves in a circular orbit at a height RR above the Earth's surface, where RR is the radius of the Earth. It is moved to a circular orbit at a height 3R3R above the surface. If gg is the acceleration due to gravity at the surface, the increase in its total mechanical energy is
  1. A. mgR4\dfrac{mgR}{4}
  2. B. 3mgR8\dfrac{3mgR}{8}
  3. C. mgR16\dfrac{mgR}{16}
  4. D. mgR8\dfrac{mgR}{8}
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Answer: (D) mgR8\dfrac{mgR}{8}

The total energy of a satellite in a circular orbit of radius rr is E=−GMm2rE=-\dfrac{GMm}{2r}. The orbit radii are r1=2Rr_1 = 2R and r2=4Rr_2=4R (height plus Earth's radius). Therefore ΔE=GMm2(12R−14R)=GMm8R\Delta E = \dfrac{GMm}{2}\left(\dfrac1{2R}-\dfrac1{4R}\right)=\dfrac{GMm}{8R}. Using GM=gR2GM = gR^2 this becomes ΔE=mgR8\Delta E = \dfrac{mgR}{8}. Computing only the change in potential energy, GMm(12R−14R)=mgR/4GMm(\tfrac1{2R}-\tfrac1{4R})=mgR/4, ignores the decrease in kinetic energy.
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