# 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$$T^2 \propto a^3$
where $T$$T$ is the orbital period, and $a$$a$ is the semi-major axis of the planet.
For system 1 the radius of the orbit is $a$$a$, and for system 2 the radius of the orbit is $4a$$4a$.
The ratio of the period of system 1 and 2 can be described as
$\Big( \frac{T_1}{T_2} \Big)^2 = \Big( \frac{a_1}{a_2} \Big)^3 = \Big( \frac{a}{4a} \Big)^3 = \Big( \frac{1}{64} \Big)$$\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}$$\frac{T_1}{T_2} = \frac{1}{8}$
(D)
https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_Of_Kepler's