GRE GR 1777, Problem 017 Solution

17. Classical Mechanics (Stellar Dynamics)

Solution
In Keplar's Thrid law, the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

T^2 \propto a^3
where T is the orbital period, and a is the semi-major axis of the planet.

For system 1 the radius of the orbit is a, and for system 2 the radius of the orbit is 4a.

The ratio of the period of system 1 and 2 can be described as 
\Big( \frac{T_1}{T_2} \Big)^2 = \Big( \frac{a_1}{a_2} \Big)^3 = \Big( \frac{a}{4a} \Big)^3 = \Big( \frac{1}{64} \Big)

Therefore, 
\frac{T_1}{T_2} = \frac{1}{8}
Answer
(D)
Reference
https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_Of_Kepler's

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