Gravitation

JEE Main level, single correct, medium

179 of 223

Question 179 of 223

If the distance of the earth from Sun is 1.5×106km1.5\times {10}^{6}km, then the distance of an imaginary planet from Sun, if its period of revolution is 2.832.83 years is:

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Answer: (C) 3×106km3\times {10}^{6}km

From Kepler's law of periods, we know that
T2∝R3⇒(T1T2)2=(R1R2)3{T}^{2}\propto {R}^{3} \Rightarrow {\left(\dfrac{{T}_{1}}{{T}_{2}}\right)}^{2}={\left(\dfrac{{R}_{1}}{{R}_{2}}\right)}^{3}
⇒(12.83)2=(1.5×106R2)3\Rightarrow {\left(\dfrac{1}{2.83}\right)}^{2}={\left(\dfrac{1.5\times {10}^{6}}{{R}_{2}}\right)}^{3}
⇒R2=[(2.83)2×(1.5×106)3]13\Rightarrow {R}_{2}={\left[{\left(2.83\right)}^{2}\times {\left(1.5\times {10}^{6}\right)}^{3}\right]}^{\dfrac{1}{3}}
=813×1.5×106=3×106km={8}^{\dfrac{1}{3}}\times 1.5\times {10}^{6}=3\times {10}^{6}km

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