A planet of mass `m` moves along an ellipse around the sun so that its maximum and minimum distance from the sun are equal to `r_(1)` and `r_(2)` respectively. Find the angular momentum of this planet relative to the centre of the sun. mass of the sun is `M`.

Explanation

If `v_(1)` and `v_(2)` are the velocity of planet at its apogee and perigee respectively then according to conservation of angular momentum, we have
`mv_(1) r_(1) = m v_(2) r_(2)`
or `v_(1) r_(1) = v_(2) r_(2)`

As the total energy of the planet is also constant, we have
`-(GMm)/(r_(1))+(1)/(2)mv_(1)^(2)=-(GMm)/(r_(2))+(1)/(2)mv_(2)^(2)`
Where M is the mass of the sun.
or `G M[(1)/(r_(1))-(1)/(r_(2 ))]=(v_(2)^(2))/(2)-(v_(1)^(2))/(2)`
or `G M((r_(1)-r_(2))/(r_(1)r_(2)))=(v_(1)^(2)r_(1))/(2r_(2)^(2))-(v_(1)^(2))/(2)`
or `G M((r_(1)-r_(2))/(r_(1)r_(2)))=(v_(1)^(2))/(2)((r_(1)^(2))/(r_(2)^(2))-1)`
`= (v_(1)^(2))/(2)((v_(1)^(2)-v_(2)^(2))/(v_(2)^(2)))`
or `= v_(1)^(2)=(2GM(r_(1)-r_(2))r_(2)^(2))/(r_(1)r_(2)(r_(1)^(2)-r_(2)^(2)))=(2Gr_(2))/(r_(1)(r_(1)+r_(2)))`
or `v_(1) = sqrt([(2GMr_(2))/(r_(1)(r_(1)+r_(2)))])`
Now Angular momentum of planet is given as
`L = mv_(1) r_(1)`
`= m sqrt([(2GMr_(1)r_(2))/((r_(1)+r_(2)))])`