Topic 6.6 Notes – Motion of Orbiting Satellites

1. Gravitationally Bound Satellite Systems

When two objects interact only through gravity, the force between them is

$$F = \frac{GMm}{r^{2}}$$

• Always attractive
• Directed along the line connecting their centers
• Depends only on separation $r$

The gravitational potential energy of this two‑object system is defined to be zero at infinity:

$$U = -\frac{GMm}{r}$$

As $r$ decreases, $U$ becomes more negative. That negative sign matters a lot later.

Massive central object approximation

In most AP problems, $M \gg m$ (planet-satellite). Technically both objects orbit their center of mass, but:

• The center of mass lies inside the massive object.
• The massive object’s motion is negligible.
• You treat the central mass as fixed.

That simplifies everything to “satellite orbiting a planet.”

Bound vs unbound

The system is isolated, so:

• Total mechanical energy $E = K + U$ is conserved.
• Angular momentum $L$ of the satellite about the central object is conserved.

Energy tells you the type of motion:

• $E < 0$ → bound orbit
• $E = 0$ → just escapes
• $E > 0$ → unbound flyby

That energy sign shows up constantly in MCQs.

2. Conservation Laws That Constrain Orbits

Gravity is a central force, so two big conservation laws control everything.

Conservation of Mechanical Energy

$$E = K + U = \text{constant}$$

As the satellite moves:

• If it gets closer → $U$ more negative → $K$ increases.
• If it moves farther away → $U$ less negative → $K$ decreases.

The tradeoff between $K$ and $U$ drives speed changes.

Conservation of Angular Momentum

$$L = mvr_\perp = \text{constant}$$

For orbital motion, velocity is perpendicular to radius at closest and farthest points, so $L = mvr$.

If $r$ decreases, $v$ must increase. That’s why satellites speed up near periapsis. This is the physics behind Kepler’s Second Law.

What changes in circular vs elliptical orbits

Students often forget that K and U are constant in circular motion.

3. Energy in Circular Orbits

In a circular orbit, gravity provides the centripetal force:

$$\frac{mv^{2}}{r} = \frac{GMm}{r^{2}}$$

Solving gives orbital speed:

$$v = \sqrt{\frac{GM}{r}}$$

Now plug that into energy expressions.

Kinetic energy

$$K = \frac{1}{2}mv^{2} = \frac{GMm}{2r}$$

Potential energy

$$U = -\frac{GMm}{r}$$

Total energy

$$E = K + U = -\frac{GMm}{2r}$$

Important relationships:

• $K = -\frac{1}{2}U$
• $E = \frac{1}{2}U$
• Total energy is negative

Bigger $r$ means:

• Smaller speed
• Energy closer to zero
• Less tightly bound

A classic AP move is to give you a new orbital radius and ask how energy changes. Remember $E \propto -1/r$.

4. Energy in Elliptical Orbits

In an ellipse, distance and speed vary.

Here’s the geometry you should picture:

Periapsis (closest point)

• Smallest $r$
• Maximum speed
• Maximum $K$
• Most negative $U$

Apoapsis (farthest point)

• Largest $r$
• Minimum speed
• Minimum $K$
• Least negative $U$

Even though $K$ and $U$ change as the planet moves between perihelion and aphelion, total energy stays constant.

For an ellipse with semi‑major axis $a$:

$$E = -\frac{GMm}{2a}$$

This is huge. Total energy depends only on $a$, not where the satellite is in the orbit.

As $a \to \infty$, $E \to 0$. That connects directly to escape.

5. Escape Velocity

Escape velocity is the speed that makes total mechanical energy zero.

Set

$$\frac{1}{2}mv^{2} - \frac{GMm}{r} = 0$$

Solving gives

$$v_{\text{esc}} = \sqrt{\frac{2GM}{r}}$$

Key facts:

• Independent of satellite mass.
• $v_{\text{esc}} = \sqrt{2}\, v_{\text{circular}}$ at the same radius.
• If launched exactly at $v_{\text{esc}}$, the object moves outward forever and its speed approaches zero as $r \to \infty$.

Energy picture:

• $E < 0$ → ellipse or circle
• $E = 0$ → parabolic escape
• $E > 0$ → hyperbolic trajectory

On FRQs, the fastest path is almost always an energy argument, not forces.