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A planet of mass m moves along an ellips...

A planet of mass m moves along an ellipse around the Sun so that its maximum and minimum distances from the Sun are equal to `r_1` and `r_2` respectively. Find the angular momentum M of this planet relative to the centre of the Sun.

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The correct Answer is:
A
If `v_1` and `v_2` are the velocities of planet at its aphelion and perihelion respectively then according to conservation of angular momentum, we have `mv_1r_1 =m v_2r_2`
or `v_1r_1=v_2r_2`
As the total energy of the planet is also constant, we have `-(GM m)/r_1 + 1/2m v_1^2 =-(GM m)/r_2 + 1/2m v_2^2`
Where M is the mass of the sun.
or `GM[1/r_2-1/r_1]=v_2^2/2-v_1^2/2`
or `GM((r_1-r_2)/(r_1r_2))=(v_1^2 v_1^2)/(2r_2^2)-v_1^2/2`
or `GM((r_1-r_2)/(r_1r_2))=v_1^2/2 (r_1^2/r_2^2-1)=v_1^2/2 ((r_1^2-r_2^2)/r_2^2)`
or `v_1^2=(2GM(r_1-r_2)r_2^2)/(r_1r_2(r_1^2-r_2^2))=(2G r_2)/(r_1(r_1+r_2))` or `v_1=sqrt([(2GM r_1r_2)/(r_1+r_2)])`
`L=m v_1r_1 " " therefore " " L=msqrt((2GMr_1r_2)/(r_1+r_2)) " " therefore ` P=16
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