Unit 6 · Topic 6.6

6.6 Motion of Orbiting Satellites

A satellite orbiting a much more massive object obeys conservation of energy and of angular momentum. The gravitational potential energy is $U = -\dfrac{GMm}{r}$ , which is zero at infinite distance. Circular orbits keep every energy constant, elliptical orbits trade kinetic for potential energy, and escape velocity makes the total energy zero.

Key terms

gravitational potential energy
circular orbit
elliptical orbit
escape velocity
total mechanical energy
angular momentum

What changed in the 2024 update

Before the Fall 2024 course update, orbits were part of their own Gravitation unit (old Unit 7). That unit is gone: Newton's law of gravitation is now Topic 2.6, circular orbits as a force problem are Topic 2.10, and orbits solved with conservation of energy and angular momentum are here in Topic 6.6. Older videos may still label this "Gravitation."

The setup

Take a satellite of mass m orbiting a central body of mass M, with M much bigger than m. The central body barely moves, so you treat it as fixed. Gravity is the only force, and it always points toward the center.

The gravitational potential energy of the pair is $U = -\frac{GMm}{r}$ where r is the distance between centers. It's zero when they're infinitely far apart and negative everywhere else. Closer means more negative, which means lower energy.

Circular orbits

In a circular orbit, gravity provides the centripetal force (topic 2.10): $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , so $v = \sqrt{\dfrac{GM}{r}}$ . The orbital speed doesn't depend on the satellite's mass.

This makes the energies simple: $K = \frac{GMm}{2r} \qquad U = -\frac{GMm}{r} \qquad E = K + U = -\frac{GMm}{2r}$ So in a circular orbit, $K = -\tfrac{1}{2}U$ and $E = \tfrac{1}{2}U = -K$ .

Since r doesn't change, the speed, K, U, E and the angular momentum mvr all stay constant.

A higher orbit has a slower speed and less kinetic energy, but a less negative total energy. To move a satellite to a higher orbit, you have to add energy.

Elliptical orbits

In an elliptical orbit, r changes. Gravity still points at the center, so it exerts no torque about the center, and the satellite's angular momentum is constant. With no other forces, the total mechanical energy is constant too. But U and K each change.

As the satellite moves closer, U gets more negative and K grows, so it speeds up. It's fastest at the closest point and slowest at the farthest. At those two points, the velocity is perpendicular to r, so angular momentum gives $r_{\text{near}}\,v_{\text{near}} = r_{\text{far}}\,v_{\text{far}}$

Combine that with energy conservation to find unknown speeds or distances. You don't need Kepler's laws by name.

Escape velocity

A launched object escapes for good if it can just reach infinite distance with zero speed. There, U = 0 and K = 0, so the total energy must be zero: $\tfrac{1}{2}mv^2 - \dfrac{GMm}{R} = 0$ . That gives $v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$

If the total energy is negative, the object is bound: it orbits or falls back. If it's zero or positive, it escapes. Escape velocity doesn't depend on the launched object's mass or on the direction of launch, as long as nothing blocks its path.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
A satellite in a low circular orbit
A 1000 kg satellite orbits 400 km above Earth's surface. Earth's mass is 5.97 × 10²⁴ kg and its radius is 6.37 × 10⁶ m. Find the orbital speed, the period and the satellite's kinetic, potential and total energy.
Show the solutionHide the solution
1. Step 1: Use the distance from Earth's center: $r = 6.37\times10^6 + 4.0\times10^5 = 6.77\times10^6$ m.
2. Step 2: Speed: $v = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{6.77\times10^{6}}} \approx 7.67\times10^3$ m/s.
3. Step 3: Period: $T = \dfrac{2\pi r}{v} \approx 5.55\times10^3$ s, about 92 minutes.
4. Step 4: $K = \dfrac{GMm}{2r} \approx 2.94\times10^{10}$ J, $U = -2K \approx -5.88\times10^{10}$ J, and $E = -K \approx -2.94\times10^{10}$ J.
Answer: v ≈ 7.7 km/s, T ≈ 92 min, K ≈ 2.9 × 10¹⁰ J, U ≈ −5.9 × 10¹⁰ J, E ≈ −2.9 × 10¹⁰ J
Example 2Calculator allowed
Escape speed from Earth
Find the escape speed from Earth's surface. Then compare it with the speed of the satellite in the previous example.
Show the solutionHide the solution
1. Step 1: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{2(6.67\times10^{-11})(5.97\times10^{24})}{6.37\times10^{6}}} \approx 1.12\times10^4$ m/s.
2. Step 2: At the same r, escape speed is $\sqrt{2}$ times circular orbit speed, since $\sqrt{2GM/r} = \sqrt{2}\sqrt{GM/r}$ . From the 400 km orbit, escape takes about $\sqrt{2}(7.67) \approx 10.8$ km/s.
3. Step 3: So the satellite would need about 3.2 km/s more speed to escape from its orbit.
Answer: About 11.2 km/s from the surface
Example 3Calculator allowed
Speeds in an elliptical orbit (exam-level)
A satellite's elliptical orbit around Earth (GM = 3.98 × 10¹⁴ m³/s²) has its closest point 7.0 × 10⁶ m from Earth's center and its farthest point 1.4 × 10⁷ m away. Find the satellite's speed at each point.
Show the solutionHide the solution
1. Step 1: Angular momentum (velocity ⊥ radius at both points): $r_n v_n = r_f v_f$ , so $v_f = \dfrac{r_n}{r_f}v_n = \tfrac{1}{2}v_n$ .
2. Step 2: Energy per unit mass: $\tfrac{1}{2}v_n^2 - \dfrac{GM}{r_n} = \tfrac{1}{2}v_f^2 - \dfrac{GM}{r_f}$ .
3. Step 3: Substitute $v_f = \tfrac{1}{2}v_n$ : $\tfrac{1}{2}v_n^2\left(1 - \tfrac{1}{4}\right) = GM\left(\dfrac{1}{r_n} - \dfrac{1}{r_f}\right)$ .
4. Step 4: The right side is $3.98\times10^{14}\left(\dfrac{1}{7.0\times10^6} - \dfrac{1}{1.4\times10^7}\right) \approx 2.84\times10^7$ m²/s². So $v_n^2 \approx 7.58\times10^7$ and $v_n \approx 8.7\times10^3$ m/s.
5. Step 5: Then v_f ≈ 4.4 × 10³ m/s. Check: the speed at the closest point is more than the circular speed at that distance (about 7.5 km/s), as it must be for an elliptical orbit that swings outward.
Answer: About 8.7 km/s at the closest point and 4.4 km/s at the farthest

Common mistakes

Using the altitude above the surface as r. In these formulas, r is measured from the center of the central body.
Using U = mgh for orbits. That only works near a planet's surface; use −GMm/r.
Thinking a satellite in a higher circular orbit moves faster. It moves slower, even though its total energy is higher.
Saying the satellite's kinetic energy is constant in an elliptical orbit. Only the total energy and the angular momentum are.

On the exam

Expect derivations: v for a circular orbit, E = −GMm/2r, or escape velocity. Start from Newton's law of gravitation or energy conservation and show each step.
For elliptical orbits, name both conservation laws you use and why each applies: gravity is the only force (energy) and it points at the center, so it makes no torque (angular momentum).

Connected topics

Videos

Topic 6.6 - Motion of Orbiting Satellites
Lessons With LondotWatch on YouTube (opens in a new tab)
Energy of satellite systems | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Mechanical Energy of a Satellite in Circular Orbit
Flipping PhysicsWatch on YouTube (opens in a new tab)
Angular momentum of satellites | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
AP Physics C - Orbits
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
Deriving Escape Velocity of Planet Earth
Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 6.6 Motion of Orbiting Satellites. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

Satellite X orbits a planet in a circle of radius r. Satellite Y orbits the same planet in a circle of radius 4r. How does Y's speed compare with X's?

A satellite moves in a circular orbit around a planet whose mass is much larger than the satellite's. The satellite's kinetic energy is 5.0 × 10⁹ J. Take the gravitational potential energy to be zero when the satellite is infinitely far away.

Described situation

Question 2 of 4Calculator allowed

What is the gravitational potential energy of the satellite–planet system?

Question 3 of 4Calculator allowed

What is the total mechanical energy of the system?

Question 4 of 4Calculator allowed

Rockets move the satellite to a new circular orbit of larger radius. Compared with the first orbit, what are its kinetic energy and the system's total energy?

0 of 4 answered