If the distance of the earth from Sun is 1.5 × 10 6 km. Then the… | JEE Main 2023 (24 Jan, Shift 2)

Q10JEE Main 2023 · 24 JAN · SHIFT 2Medium

Kepler's Laws & Planetary Motion

If the distance of the earth from Sun is 1.5 $\times$ 10 $6$ km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is:

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Answer: (d)

Kepler's third law: $T^2 \propto r^3$ , so $r_2 = r_1 \left(\frac{T_2}{T_1}\right)^{2/3}$ .
Earth: $T_1 = 1$ year, $r_1 = 1.5 \times 10^{6}$ km (value as given).
$\left(2.83\right)^{2} \approx 8$ , so $(2.83)^{2/3} = 8^{1/3} = 2$ .
$r_2 = 2 \times 1.5 \times 10^{6} = 3 \times 10^{6}$ km
So the answer is (d) $3 \times 10^{6}$ km.

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