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Question

A planet of mass m moves along an ellipse around the sun of mass M so that its maximum and minimum distances from sun are a and b respectively. The angular momentum L of this planet about the centre of sun will be:
(sun is stationary.)

A
M2Gmab(a+b)
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B
mb2GMa+b
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C
m2GMab(a+b)
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D
M2Gmab
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Solution

The correct option is C m2GMab(a+b)


The planet is revolving around the sun due to gravitational pull.

Since, we know that τext about the centre of sun is zero on the system of sun and planet. So, we can apply angular momentum conservation on the sun- planet system when planet at maximum and minimum distance from the sun.

Li at max. distance=Lf at min. distance

mv1a=mv2b

v1=v2ba ..........(1)

Since, gravitation field is a conservative field, so we can apply energy conservation.

U1+K1=U2+K2

GMma+12mv21=GMmb+12mv22

Substituting from eq. (1), we get

12m[v21v21a2b2]=GMm[1a1b]

v21(b2a2)=2b2GM(ba)ab

v21=2GMba(a+b)

v1=2GMba(a+b)

Therefore, angular momentum of planet will be

L=mv1a=mv2b

L=ma2GMba(a+b)

Hence, option (c) is correct.
Why this question ?
Tip: The gravitational force exerted by sun will be directed along line joining centre of planet and sun. Thus, τext=0 on system about centre of sun, and we can apply angular momentum conservation

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