Problem 66
Easy Gravitational Field

What is the mutual gravitational potential energy of two…

What is the mutual gravitational potential energy of two point masses $m_1$ and $m_2$ a distance $r_0$ apart?

Explanation of Method

Mutual energy of two point masses.

Step 1: Idea

Potential energy is defined only up to a constant, so we must first choose where it is zero. For gravity the natural choice is when the masses are infinitely far apart, since then they no longer interact. The energy at separation $r_0$ is then minus the work we would have to do to pull the masses apart from $r_0$ to infinity.

Step 2: Work to separate them

At separation $r$ the attractive force is $\gamma m_1m_2/r^2$; we must pull with an equal force outward. Since the force changes with $r$, we integrate: \[A = \int_{r_0}^{\infty}\gamma\frac{m_1m_2}{r^2}\,dr = \gamma m_1m_2\left[-\frac1r\right]_{r_0}^{\infty},\]

\[ A = \gamma\frac{m_1m_2}{r_0}. \]

This work raises the energy from $E_p(r_0)$ to $E_p(\infty) = 0$, so $E_p(r_0) = -A$: $E_p = -\gamma\frac{m_1m_2}{r_0}$.

The energy is negative for every finite separation and becomes more negative as the masses approach. This is the mathematical expression of attraction: the system \"falls\" toward lower energy by bringing the masses together.

Check

$ -dE_p/dr = -\gamma m_1m_2/r^2 $: an attractive force of the correct size $\checkmark$

$E_p$ $r$ $r_0$ $E_p = -\gamma\frac{m_1m_2}{r_0}$
Mutual potential energy against separation: zero at infinite distance, increasingly negative as the masses approach. The work needed to separate them completely is $\gamma m_1m_2/r_0$.

Result

\[\boxed{\,E_p = -\gamma\frac{m_1m_2}{r_0}\,}\]