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JEE Main 2016 · Physics · Gravitation · medium

A satellite of mass m revolves in a circular orbit of radius 2R around the earth, where R is the radius of the earth and g the acceleration due to…

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A satellite of mass mm revolves in a circular orbit of radius 2R2R around the earth, where RR is the radius of the earth and gg the acceleration due to gravity at its surface. The minimum extra energy required to shift it to a circular orbit of radius 3R3R is
  1. A. mgR6\dfrac{mgR}{6}
  2. B. mgR12\dfrac{mgR}{12}
  3. C. mgR4\dfrac{mgR}{4}
  4. D. mgR2\dfrac{mgR}{2}
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Answer: (B) mgR12\dfrac{mgR}{12}

The total energy of a satellite in a circular orbit of radius rr is E=−GMm2rE=-\dfrac{GMm}{2r}. Therefore E2R=−GMm4RE_{2R}=-\dfrac{GMm}{4R} and E3R=−GMm6RE_{3R}=-\dfrac{GMm}{6R}. The extra energy needed is ΔE=GMmR(14−16)=GMm12R\Delta E=\dfrac{GMm}{R}\left(\dfrac14-\dfrac16\right)=\dfrac{GMm}{12R}. Using GM=gR2GM=gR^2 gives ΔE=mgR12\Delta E=\dfrac{mgR}{12}. Taking only the change of potential energy, GMmR(12−13)=mgR6\dfrac{GMm}{R}\left(\dfrac12-\dfrac13\right)=\dfrac{mgR}{6}, ignores the change in kinetic energy and is the trap here.
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