§ 13.2 — Gravitational Potential Energy

Energy in the Gravitational Field

Gravitational potential energy is the energy stored in the relative positions of masses in a gravitational field. Unlike the simple \( mgh \) formula used near Earth's surface, which assumes a uniform gravitational field, we need a more general expression when dealing with varying gravitational fields over large distances.

The gravitational potential energy of a system of two point masses is a measure of how much work would be required to bring them from infinite separation to their current separation (or equivalently, how much work the gravitational force can do as they move infinitely far apart).

General Expression for Gravitational Potential Energy

For two masses \( m_1 \) and \( m_2 \) separated by distance \( r \), the gravitational potential energy is:

\[ U(r) = -G\frac{m_1 m_2}{r} \]

The negative sign indicates that the gravitational force is attractive—the potential energy is most negative (lowest) when the masses are closest together. At infinite separation (\( r \to \infty \)), the potential energy approaches zero, which we take as our reference point.

This contrasts with the near-surface approximation where \( U = mgh \) with \( h \) measured upward from an arbitrary reference. The exact form \( U(r) = -G\frac{m_1 m_2}{r} \) is valid regardless of scale.

Mechanical Energy Conservation

For conservative forces (including gravity), mechanical energy is conserved:

\[ E = K + U = \frac{1}{2}mv^2 - G\frac{Mm}{r} = \text{constant} \]

where \( M \) is the central body (e.g., Earth or Sun) and \( m \) is the test mass. This relationship, combined with orbital dynamics, determines whether an object remains bound to a massive body or escapes to infinity.

Potential Energy in a Gravitational Field

A related concept is the gravitational potential \( \Phi(r) \), defined as the potential energy per unit mass:

\[ \Phi(r) = -G\frac{M}{r} \]

This quantity is independent of the test mass and represents an intrinsic property of the gravitational field. It proves useful when analyzing how multiple masses contribute to the field, since potentials superpose (unlike vector forces which require component-wise addition).

A Parallel Problem: Satellite Release

A satellite of mass \( m = 2500 \, \text{kg} \) orbits Earth at an altitude of \( h = 400 \, \text{km} \). It is given an impulse that increases its speed by \( \Delta v = 50 \, \text{m/s} \) in the direction of orbital motion. Determine whether this impulse allows escape from Earth's gravitational field, and calculate the final velocity at infinity (if escape occurs).

Given:

• Mass of Earth: \( M_E = 5.972 \times 10^{24} \, \text{kg} \)
• Radius of Earth: \( R_E = 6.371 \times 10^6 \, \text{m} \)
• Gravitational constant: \( G = 6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \)

Solution:

First, find the orbital velocity at altitude \( h = 400 \, \text{km} \):

\[ r = R_E + h = 6.371 \times 10^6 + 0.4 \times 10^6 = 6.771 \times 10^6 \, \text{m} \]

\[ v_{\text{orbit}} = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{6.771 \times 10^6}} \]

\[ v_{\text{orbit}} = \sqrt{5.89 \times 10^7} = 7670 \, \text{m/s} \]

After the impulse, the satellite's speed is:

\[ v = v_{\text{orbit}} + \Delta v = 7670 + 50 = 7720 \, \text{m/s} \]

The escape velocity at this altitude is:

\[ v_{\text{esc}} = \sqrt{\frac{2GM_E}{r}} = \sqrt{2} \times v_{\text{orbit}} = 1.414 \times 7670 = 10,850 \, \text{m/s} \]

Since \( v = 7720 \, \text{m/s} < v_{\text{esc}} = 10,850 \, \text{m/s} \), the satellite does not achieve escape velocity and remains bound to Earth.

To find the apogee (highest point) of the new elliptical orbit, apply energy conservation:

\[ \frac{1}{2}mv^2 - G\frac{M_E m}{r} = \frac{1}{2}mv_a^2 - G\frac{M_E m}{r_a} \]

where \( v_a \) is the velocity at apogee and \( r_a \) is the apogee distance. Additionally, angular momentum is conserved at the point of impulse (closest approach):

\[ mvr = mv_a r_a \]

\[ v_a = \frac{vr}{r_a} \]

Substituting into the energy equation and solving:

\[ \frac{1}{2}v^2 - G\frac{M_E}{r} = \frac{1}{2}\frac{v^2r^2}{r_a^2} - G\frac{M_E}{r_a} \]

\[ v^2\left(1 - \frac{r^2}{r_a^2}\right) = 2GM_E\left(\frac{1}{r} - \frac{1}{r_a}\right) \]

Solving numerically yields \( r_a \approx 7.6 \times 10^6 \, \text{m} \) (altitude \( \approx 1230 \, \text{km} \)), representing a higher elliptical orbit.

Real-World Implications

Understanding gravitational potential energy is crucial for mission planning in spaceflight. It determines the energy cost of orbital transfers (Hohmann transfers), escape trajectories, and the stability of satellites. In astrophysics, the concept explains why neutron stars and black holes have such extreme properties—their enormous masses compressed into small volumes create potential energy wells of tremendous depth.

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