InterviewSolution
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InterviewSolution
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Derive an expression for the escape speed of a body from the surface of the Earth. Express it in terms of the mean density of the Earth. |
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Answer» Solution :For projection speed EQUAL to the escape speed `(v_(E))`, the kinetic energy of the body equals the binding energy. `therefore (1)/(2) mv_(e)^(2) = (GMM)/(R )` where m is the mass of the body, G is the gravitational CONSTANT, M is the mass of the Earth and R is the radius of the Earth. `therefore v_(e)=sqrt((2GM)/(R ))` The mean density of the Earth, `rho = ("mass")/("volume") = (M)/((4)/(3)piR^(3))` `therefore M = (4)/(3) pi R^(3) rho` `thereforev_(e)=sqrt((2G)/(R)xx(4)/(3)piR^(3)rho)=2Rsqrt((2piGrho)/(3))` |
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