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A body is projected from the surface of earth with the velocity three times the escape velocity. Find the velocity beyond the gravitational field of the earth. |
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Answer» According to the conservation law of energy, the energy at the surface of the earth is = -GMem/Re + 1/2m(3ve)2 The energy beyond the gravitational field of the earth, = 1/2mv2 (where v is the velocity beyond the earth's gravity). Hence, -GMem/Re + 9/2mve2 = 1/2mv2 or, -1/2mve2 + 9/2mve2 = 1/2mv2 or, 8ve2 = v2 v = √2 ve = 2√2 x (11.2) kms-1 = 22.4√2 kms-1. |
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