Gravity and Orbits

Section 5 of 6

Conservation Laws

Part 5: Conservation Laws in Orbital Mechanics

Energy Conservation: Predicting Orbital Fate

An orbiting object has two forms of mechanical energy: kinetic (from motion) and potential (from position in the gravitational field). A feature that surprises many students is that gravitational potential energy is negative. The total mechanical energy is

$E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r},$

where $K = \tfrac{1}{2}mv^2$ is the kinetic energy (always positive) and $U = -GMm/r$ is the gravitational potential energy (always negative).

Gravitational potential energy

The energy stored in the gravitational field of two bodies, $U = -GMm/r$ with zero defined at infinity. It is negative because a bound object sits below the zero level — in a “well” it would need energy to climb out of.

Deep Dive: Why Is Potential Energy Negative?

We choose zero energy at infinity. Infinitely far from any mass, there’s no gravitational influence: $U = 0$ as $r \to \infty$ .

Gravity pulls you into a “well.” As you fall toward a mass, gravity does work on you, speeding you up — trading potential energy for kinetic. Since you started at zero and gravity is giving you kinetic energy, your potential energy must go negative.

Think of it as debt. Negative $U$ means you “owe” energy to escape. The more negative (closer to the mass), the deeper the hole. At $r = \infty$ , $U = 0$ (free); at finite $r$ , $U < 0$ (bound); closer in, $U$ is more negative. This is why the sign of total energy $E = K + U$ tells you whether you can escape: if $E < 0$ , your kinetic “savings” can’t pay off your potential “debt.”

Both terms have units of erg: $[K] = \mathrm{g}\,(\mathrm{cm/s})^2 = \mathrm{g\,cm^2/s^2} = \mathrm{erg}$ , and $[U] = (\mathrm{cm^3\,g^{-1}\,s^{-2}})(\mathrm{g})(\mathrm{g})/\mathrm{cm} = \mathrm{g\,cm^2/s^2} = \mathrm{erg}$ ✓. Total energy is conserved: as an object moves closer to the central mass it speeds up, converting potential to kinetic, but $E$ stays constant.

Elliptical orbit with energy indicators showing kinetic and potential energy trading off around the orbit while total energy remains constant. — Figure 7Conservation of Energy in Orbits: total energy E = K + U is constant. At perihelion, high K and low U; at aphelion, low K and high U.cococubed.com

Central body with multiple orbital paths showing bound elliptical orbits and unbound hyperbolic trajectories, labeled to distinguish orbit types by total energy. — Figure 8Orbit Classification by Energy: negative total energy gives bound orbits (ellipses); zero energy, parabolic escape; positive energy, hyperbolic escape.cococubed.com

Energy	Meaning	Orbit Type
$E < 0$	Bound	Closed (ellipse or circle)
$E = 0$	Marginally bound	Parabolic (just escapes)
$E > 0$	Unbound	Hyperbolic (escapes with speed to spare)

For any bound orbit (circular or elliptical) with semi-major axis $a$ , the total energy is $E = -GMm/(2a)$ . For a circular orbit, $a = r$ .

Pause & Predict #5

Reason about total mechanical energy.

A comet approaches the Sun with total energy E > 0 (and doesn't hit the Sun). Will it return, or escape to infinity?

If a spacecraft in circular orbit fires its engines to speed up, what happens to its total energy — more bound or less bound?

Check below.

Multiple choice

A comet approaches the Sun with total energy $E>0$ (and doesn’t hit the Sun). Will it return, or escape to infinity?

Return on a bound orbitEscape to infinity

Multiple choice

A spacecraft in a circular orbit fires its engines to speed up. Does it become more bound or less bound?

More boundLess bound (larger semi-major axis)

Angular Momentum Conservation: Why Kepler’s Second Law Works

Angular momentum measures “rotational inertia” — how hard it is to change an object’s orbiting motion. For mass $m$ moving in a circle of radius $r$ , its magnitude is $L = mvr$ , where $v$ is the speed perpendicular to the radius. Equivalently, $L = mr^2\dot{\phi}$ , where $\dot{\phi}$ is the angular velocity (how fast the angle changes, in radians per second). The two forms agree because $v = r\dot{\phi}$ for circular motion.

Angular momentum

A conserved measure of orbiting/spinning motion, $L = mvr$ for circular motion. A central force exerts no torque, so $L$ stays constant — the deep reason behind Kepler’s Second Law.

Unit check: $[L] = [m][v][r] = \mathrm{g}\,(\mathrm{cm/s})\,\mathrm{cm} = \mathrm{g\,cm^2/s}$ ✓. Why is angular momentum conserved? Because gravity is a central force — it acts along the line connecting the masses, so it exerts no torque about the center and cannot change $L$ .

Torque

A twisting influence that changes angular momentum. A central force produces zero torque about the center, so it leaves $L$ unchanged.

Diagram showing angular momentum vectors and the relationship between orbital radius and velocity, demonstrating that smaller radius requires larger velocity to conserve L. — Figure 9Conservation of Angular Momentum: L = mvr remains constant. When r decreases, v must increase, explaining why planets speed up near the Sun.cococubed.com

This immediately explains Kepler’s Second Law:

Gravity is a central force — it points directly toward the Sun.
A central force exerts no torque about the center.
No torque means angular momentum $L$ stays constant.
The area swept in time $dt$ is a thin pie-slice, $dA \approx \tfrac{1}{2}r^2\,d\phi$ .
Since $L = mr^2\dot{\phi}$ is constant, the areal velocity $dA/dt = \tfrac{1}{2}r^2\dot{\phi} = L/(2m)$ is also constant.
Constant $dA/dt$ is exactly Kepler’s Second Law: equal areas in equal times.

Areal velocity

The rate at which the planet-Sun line sweeps out area, $dA/dt = L/(2m)$ . Angular-momentum conservation makes it constant — which is Kepler’s Second Law.

When a planet is closer to the Sun (smaller $r$ ), it must move faster angularly to keep $L = mr^2\dot{\phi}$ constant — like an ice skater spinning faster when pulling their arms in.

Pause & Predict #6

A planet's elliptical orbit has r_p = 1 AU at perihelion and r_a = 4 AU at aphelion.

At which point is the planet moving faster: perihelion or aphelion?

Using L = m r^2 phi-dot = const, by what factor is the angular velocity larger at perihelion than aphelion?

Is the linear speed ratio the same as the angular velocity ratio? Why or why not?

Check below.

Problem

Check the three predictions above for the orbit with $r_p = 1$ AU and $r_a = 4$ AU. (a) Where is the planet faster, perihelion or aphelion? (b) Using $L = mr^2\dot{\phi} = \text{const}$ , by what factor is the angular velocity larger at perihelion? (c) Is the linear-speed ratio the same as the angular-velocity ratio?

The Virial Theorem: A Deep Connection

For gravitationally bound systems there’s a remarkable relationship between kinetic and potential energy — but it applies only to truly static systems (a star in hydrostatic equilibrium) or periodic systems averaged over a complete orbit. Writing $\langle K \rangle$ for the time-averaged kinetic energy, the virial theorem states:

Equivalently, $\langle K \rangle = -\tfrac{1}{2}\langle U \rangle$ . The kinetic energy is exactly half the magnitude of the (negative) potential energy. (We meet the virial theorem in full when we build stellar structure in Module 3.)

Verify it for a circular orbit, where instantaneous equals average. The kinetic energy is $K = \tfrac{1}{2}mv_{orb}^2 = \tfrac{1}{2}m\,(GM/r) = GMm/(2r)$ , and the potential energy is $U = -GMm/r$ . Then $K = GMm/(2r) = -\tfrac{1}{2}(-GMm/r) = -\tfrac{1}{2}U$ ✓. The total energy is $E = K + U = GMm/(2r) - GMm/r = -GMm/(2r)$ , i.e. $E = -K$ .

The same relationship that governs planetary orbits applies across the universe — when systems are in equilibrium. It sets stellar structure (thermal energy balances gravitational binding), reveals galaxy masses including dark matter (from stellar motions), and determines when gas clouds collapse.