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JEE Main 2017 · Physics · Gravitation · medium

An artificial satellite of mass 500 kg moves in a circular orbit at a height equal to the radius of the Earth above its surface. Taking R = 6.4×10^6…

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An artificial satellite of mass 500 kg500\ \mathrm{kg} moves in a circular orbit at a height equal to the radius of the Earth above its surface. Taking R=6.4×106 mR = 6.4\times10^{6}\ \mathrm{m} and g=10 m s−2g = 10\ \mathrm{m\,s^{-2}} at the surface, the minimum extra energy that must be supplied to the satellite so that it escapes from the Earth's gravitational field is
  1. A. 8×109 J8\times10^{9}\ \mathrm{J}
  2. B. 1.6×1010 J1.6\times10^{10}\ \mathrm{J}
  3. C. 3.2×1010 J3.2\times10^{10}\ \mathrm{J}
  4. D. 4×109 J4\times10^{9}\ \mathrm{J}
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Answer: (A) 8×109 J8\times10^{9}\ \mathrm{J}

The orbit radius is r=2Rr = 2R. The total energy in a circular orbit is E=−GMm2r=−GMm4RE = -\dfrac{GMm}{2r} = -\dfrac{GMm}{4R}. To escape, the total energy must be raised to zero, so the extra energy is GMm4R=mgR4\dfrac{GMm}{4R} = \dfrac{mgR}{4} using GM=gR2GM = gR^2. Hence ΔE=500×10×6.4×1064=8×109 J\Delta E = \dfrac{500\times10\times6.4\times10^6}{4} = 8\times10^{9}\ \mathrm{J}. The value 3.2×1010 J3.2\times10^{10}\ \mathrm{J} is mgRmgR, the energy needed to escape from the surface, not from the orbit.
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