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InterviewSolution
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A planet of mass `m` revolves in elliptical orbit around the sun of mass `M` so that its maximum and minimum distance from the sun equal to `r_(a)` and `r_(p)` respectively. Find the angular momentum of this planet relative to the sun.A. `msqrt((2GMr_(p)r_(a))/((r_(p)+r_(a))))`B. `msqrt((4GMr_(p)r_(a))/((r_(p)+r_(a))))`C. `m sqrt((GMr_(p)r_(a))/((r_(p)+r_(a))))`D. `m sqrt((GMr_(p)r_(a))/(2(r_(p)+r_(a))))` |
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Answer» Correct Answer - A Using conservation of angular momentum, `mv_(p)r_(p)=mv_(a)r_(a)` As velocities are perpendicular to radius vectors at apogee and perigee `rArr v_(p)r_(p)=v_(a)r_(a)` Using conservation of energy, `-(GMm)/(r_(p))+(1)/(2)mv_(p)^(2)=(-GMm)/(r_(a))+(1)/(2)mv_(a)^(2)` By solving, the above equations, `v_(P)=sqrt((2GMr_(a))/(r_(p)(r_(p)+r_(a))))rArr L=mv_(p)r_(p)=msqrt((2GMr_(p)r_(a))/((r_(p)+r_(a))))`. |
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