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Question

The Jupiter's period of revolution around the Sun is 12 times that of the Earth. Assuming the planetary orbits to be circular, find the acceleration of Jupiter in the heliocentric reference frame.

A
2×104m/s2
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B
4.2×104m/s2
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C
2.2×104m/s2
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D
4×104m/s2
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Solution

The correct option is C 2.2×104m/s2
if T is time period of the planet and r the distance of the planet from the sun,
then from kepler's third law
T2=kr3 ,

Let, Time period of jupiter be T1 and for earth be T2 , distance of jupiter from sun be r1 and of earth from sun be r2 ,

r1 = r2(T1/T2)3/2

T1/T2=12

G=6.67×1011 and r2 = 1.4×1011m , M =1.97×1030kg

So acceleration of jupiter is a = GM/r1 = 2.2×104m/s2

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