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The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and Radius R in a circular orbit at an altitude of 2R is . |
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Answer» Correct option is: \(\frac{5\,GMm}{6R}\) Given mass of satellite = M mass of the surface = M Radius = R Altitude h = 2R Gravitational potential energy = \(-\frac{GMm}{r}\) Gravitational potential energy at Altitude = \(-\frac{GMm}{r+h}\) = \(-\frac{GMm}{R+2R}\) = \(-\frac{GMm}{3R}\) Orbital velocity Vo2 = \(\frac{GM}{R+h}\) Vo2 = \(\frac{GM}{3R}\) total potential energy E = kinetic energy + potential energy Ef = \(\frac{1}{2}\)mvo2 + \((-\frac{GMm}{3R})\) ∵ Vo2 = \(\frac{GM}{3R}\) Ef = \(\frac{1}{2}\) \(\frac{GMm}{3R}-\frac{GMm}{3R}\) Ef = \(\frac{GMm-2\,GMm}{6R}\) Ef = \(-\frac{GMm}{6R}\) Ei = Ef the minium energy required = \(\frac{GMm}{R}-\frac{GMm}{6R}\) = \(\frac{6\,GMm-GMm}{6R}\) Minimum energy required = \(\frac{5\,GMm}{6R}\) |
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