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A planet has period $T$ at distance $R$ from a star. Using Kepler's third law, its period at distance $2R$ would be:
A$\sqrt{8}\,T$
B$\sqrt{4}\,T$
C$\sqrt{2}\,T$
D$\sqrt{6}\,T$
Answer & Solution
Correct answer: A. $\sqrt{8}\,T$
1. Kepler's third law: $\dfrac{T^2}{r^3} = K$ (constant) for the same star.
2. So $\dfrac{T_1^2}{R^3} = \dfrac{T_2^2}{(2R)^3}$.
3. $(2R)^3 = 8R^3$, giving $T_2^2 = T_1^2 \times \dfrac{8R^3}{R^3} = 8T^2$.
4. Taking the square root, $T_2 = \sqrt{8}\,T$.
5. $\sqrt{4}\,T = 2T$ wrongly scales the period linearly with distance and is the trap.
_Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 1 "Gravitation", p.24_
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