What Is Two-Body Gravitational Potential Energy? (HL)

HL (AHL), D.1 Gravitational fields: For two point masses, gravitational potential energy is the work an external agent must do to assemble the system slowly from infinite separation. It is

$$E_{\text{p}}=-\frac{Gm_1m_2}{r},$$

where $G$ is the gravitational constant and $r$ is the centre-to-centre separation. Because the reference value is defined as $E_{\text{p}}=0$ at infinity, every finite separation produces a negative value.

Reasoning and Mechanism

Gravity is attractive and does positive work as the masses move together. To assemble the system slowly, an external agent applies an opposing force and does negative work, so the system's potential energy decreases:

$$E_{\text{p}}(r)-E_{\text{p}}(\infty) =-\int_{\infty}^{r}\vec{F}\cdot d\vec{r} =-\frac{Gm_1m_2}{r}.$$

Do not confuse gravitational potential $V$ with potential energy. Potential is energy per unit test mass; the two-body potential energy depends on both masses, with $E_{\text{p}}=mV$ when $V$ is produced by the other mass.

For example, masses of $5.0\ \text{kg}$ and $10\ \text{kg}$, separated by $2.0\ \text{m}$, have

$$E_{\text{p}}=-\frac{(6.67\times10^{-11})(5.0)(10)}{2.0} =-1.7\times10^{-9}\ \text{J}.$$

When only gravity acts, the decrease in potential energy appears as an equal increase in kinetic energy, because mechanical energy is conserved. Conversely, separating the masses while keeping their speeds unchanged requires an external energy transfer equal to the positive change in potential energy. This connects the equation directly to orbital and escape-energy calculations in the HL syllabus.

The effect of changing separation is therefore:

If the same masses move from $2.0\ \text{m}$ to $4.0\ \text{m}$, then $E_{\text{p,final}}=-8.3\times10^{-10}\ \text{J}$. Therefore, , which must be supplied if kinetic energy is unchanged.

The common misconception is that negative potential energy means energy is negative in an absolute sense. Instead, the sign indicates a bound system relative to the chosen zero at infinity. The expression $mg\Delta h$ is only a near-surface approximation for a nearly uniform field.

Exam Technique

In a Calculate response, use centre-to-centre separation, retain the negative sign, and use the data-booklet value $G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$. For a change, calculate .