Gravitation

JEE Main level, single correct, medium

199 of 223

Question 199 of 223

A planet (P1)(P_1) is moving around the star of mass 2M2M in the orbit of radius RR. Another planet (P2)(P_2) is moving around another star of mass 4M4M in a orbit of radius 2R2R. Ratio of time periods of revolution of P2P_2 and P1P_1 is _______.

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Answer: (B) 22

Using Kepler's third law, the time period of revolution of a planet is given by T=2πr3GMsT = 2\pi \sqrt{\dfrac{r^3}{GM_s}}.

For planet P1P_1, the orbital radius is r=Rr = R and the mass of the star is Ms=2MM_s = 2M:
T1=2πR3G(2M)T_1 = 2\pi \sqrt{\dfrac{R^3}{G(2M)}}

For planet P2P_2, the orbital radius is r=2Rr = 2R and the mass of the star is Ms=4MM_s = 4M:
T2=2π(2R)3G(4M)=2π8R34GM=2π2R3GMT_2 = 2\pi \sqrt{\dfrac{(2R)^3}{G(4M)}} = 2\pi \sqrt{\dfrac{8R^3}{4GM}} = 2\pi \sqrt{\dfrac{2R^3}{GM}}

Taking the ratio of T2T_2 and T1T_1:
T2T1=2R3GMR32GM=2×2=2\dfrac{T_2}{T_1} = \dfrac{\sqrt{\dfrac{2R^3}{GM}}}{\sqrt{\dfrac{R^3}{2GM}}} = \sqrt{2 \times 2} = 2

Answer: 22

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