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.

Options

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))))`

Answer

A

Explanation

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))))`.