# What are the units of the gravitational constant of the universe $G$?

This is a homework question from a precalculus class that I'm a TA for.

The force $$F$$ between two bodies of mass $$m_1$$ kilograms and $$m_2$$ kilograms whose centers of mass are a distance $$r$$ meters apart is given by the formula $$\begin{equation*} F = \frac{G m_1 m_2}{r^2} \,, \end{equation*}$$ where $$G$$ is the gravitational constant of the universe. What are the units of the gravitational constant $$G$$?

I wanted to write up a thorough solution to this exercise for my class, and figured I'd post it online to help anyone else who may wander across it.

Since we want to figure out what the units of $$G$$ are, first we should rewrite the given formula with $$G$$ isolated: $$\begin{equation*} F = \frac{G m_1 m_2}{r^2} \quad\implies\quad G = \frac{F r^2}{m_1 m_2} \end{equation*}$$ Since this is an equality, the units on the left-hand side (so the units of $$G$$) will be the same as the units on the right-hand side. We recall that since $$F=ma$$ the units of force are $$\mathrm{kg}(\mathrm{m}/\mathrm{s}^2)$$. So the units of the right-hand side must be $$\begin{equation*} \frac{\mathrm{kg}\frac{\mathrm{m}}{\mathrm{s}^2}\,\mathrm{m}^2} {\mathrm{kg}\,\mathrm{kg}} \,, \end{equation*}$$ which we can write more cleanly as $$\begin{equation*} \frac{\mathrm{m}^3}{\mathrm{kg}\,\mathrm{s}^2} \,. \end{equation*}$$ So these are the units of $$G$$.