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JEE Main20199 Jan 2019Evening ShiftPhysicsGravitationActual

The energy required to take a satellite to a height h above the Earth surface (radius of Earth = 6.4 × 10 3 km ) is E 1 , and the kinetic energy required for the satellite to be in a circular orbit at this height is E 2 . The value of h for which E 1 and E 2 are equal, is

Options

  1. A1.28 × 10 4 k m
  2. B6.4 × 10 3 k m
  3. C3.2 × 10 3 k m
  4. D1.6 × 10 3 k m

Correct answer

C. 3.2 × 10 3 k m

Step-by-step solution

U surface   + E 1 = U h K E of satellite is zero at earth surface and at height h - G M e m R e + E 1 = - G M e m R e + h E 1 = G M e m 1 R e - 1 R e + h E 1 = G M e m R e + h × h R e Gravitational attraction F G = m a C = m v 2 R e + h ⇒ m v 2 R e + h = G M e m R e + h 2 m v 2 = G M e m R e + h E 2 = m v 2 2 = G M e m 2 R e + h E 1 = E 2 h R e = 1 2 ⇒ h = R e 2 = 3200   km

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