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- 32 days
- 8 days
- 18 days
- 9 days

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Solution

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Kepler's third low planetary motion the square of the period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.

∴T2∝R3

TsTc=(RsRc)3/2

given Rs=4Rc

TsTc(4RcRc)3/2=8

for Tc=1day

Ts=8 days.

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