Gravitation

JEE Main level, single correct, medium

206 of 223

Question 206 of 223

A planet takes 200200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution?

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Answer: (A) 2525

The Law of Periods: The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit.
T2∝r3{T}^{2}\propto {r}^{3}
Therefore, we can write
T12r13=T22r23\dfrac{{T}_{1}^{2}}{{r}_{1}^{3}}=\dfrac{{T}_{2}^{2}}{{r}_{2}^{3}}
⇒(200)2r3=T22(r4)3\Rightarrow \dfrac{(200{)}^{2}}{{r}^{3}}=\dfrac{{T}_{2}^{2}}{{\left(\dfrac{r}{4}\right)}^{3}}
⇒200×2004×4×4=T22\Rightarrow \dfrac{200\times 200}{4\times 4\times 4}={T}_{2}^{2}
⇒T2=25\Rightarrow {T}_{2}=25 days

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