Solve for r in the gravitational force equation $F_g = \dfrac{G m_1 m_2}{r^2}$ given $G = 6.67\times

What you are solving for

You are given Newton’s law of gravitation and asked to find the separation distance $r$ between two masses when the gravitational force $F_g$ is known.

Isolating $r$ in $F_g = \dfrac{Gm_1m_2}{r^2}$

Start with $$F_g = \frac{Gm_1m_2}{r^2}$$ Multiply both sides by $r^2$: $$F_g r^2 = Gm_1m_2$$ Divide by $F_g$: $$r^2 = \frac{Gm_1m_2}{F_g}$$ Take the square root: $$r = \sqrt{\frac{Gm_1m_2}{F_g}}$$

Plugging in the numbers

Compute the numerator: $$Gm_1m_2 = (6.67\times10^{-11})(50)(360) = (6.67\times10^{-11})(18000)=1.2006\times10^{-6}$$ Now divide by $F_g$: $$\frac{1.2006\times10^{-6}}{1.334\times10^{-11}} = 9.0\times10^4$$ So, $$r = \sqrt{9.0\times10^4} = 300\ \text{m}$$

Quick reasonableness check

Because $F_g$ is in newtons and $G$ has units $\text{N·m}^2/\text{kg}^2$, the expression $\dfrac{Gm_1m_2}{F_g}$ has units of $\text{m}^2$, and the square root correctly gives meters.