InterviewSolution
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InterviewSolution
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A planet of mass m moves along an ellipse around the sum of mass M so that its maximum and minimum distances from sum are a and b respectively. Prove that the angular momentum L of this planet relative to the centre of the sun is `L=msqrt((2GGMab)/((a+b)))` |
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Answer» Angular momentum at maximum distance from sum=angular momentum at minimum distance from sun `mv_(1)a=mv_(2)bimpliesv_(1)=(v_(2)b)/(a)` by applying conservation of energy `(1)/(2)mv_(1)^(2)-(GMm)/(a)=(1)/(2)mv_(2)^(2)-(GMm)/(b)` from above equations `v_(1)=sqrt((2GMb)/(a(a+b)))` angular momentum of planet `L=mv_(1)a=masqrt((2GMb)/((a+b)a))` |
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