Question

# A planet of mass m revolves in elliptical orbit around the sun so that its maximum and minimum distances from the sun are equal to ra and rp respectively. Find the angular momentum of this planet relative to the sun.

A
L=mGMrpra(rp+ra)
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B
L=m2GMrpra(rp+ra)
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C
L=M2GMrpra(rp+ra)
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D
L=M(rp+ra)Gmrpra
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Solution

## The correct option is B L=m√2GMrpra(rp+ra)For a planet moving around the sun in an orbit, angular momentum (L) is constant.L=mvara=mvbrb⇒va=vbrbraBy conserving energy between the two points, farthest and nearest.−GMmrb+12mv2b=−GMmra+12mv2a⇒GM(1ra−1rb)=12(v2a−v2b)Substituting va from momentum equation.⇒GM(1ra−1rb)=12[v2b(r2br2a)−v2b]⇒GM[rb−rararb]=v2b2[(rb−ra)(rb+ra)r2a]⇒v2b=2GMra(ra+rb)rb∴L=mvbrb=m√2GMra(ra+rb)rbrb=m√2GMrarbra+rb

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