Kepler's third law relates the period of a planet

to its orbital radius r, the constant G in Newton's law of gravi-

tation (F = Gm1m2/r^2), and the mass of the sun M ... What com-

bination of these factors gives the correct dimensions for the

period of a planet?

√ (r³ / (MG))

Explanation:

The dimensions of each variable are:

r = [m]

G = [m³/kg/s²]

M = [kg]

Multiplying M and G eliminates kilograms:

MG = [m³/s²]

The radius cubed divided by MG eliminates meters:

r³ / (MG) = [s²]

The square root gives us seconds:

√ (r³ / (MG)) = [s]