# 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.

## 1 Answer

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 $$G$$.