AP Physics C: Mechanics · Unit 6: Energy and Momentum of Rotating Systems · Lesson 6.6

Deep Dive: Motion of Orbiting Satellites

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
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Conservation Laws in Orbit

A satellite orbiting a much more massive central body — a moon around a planet, a planet around a star — is really a two-object system interacting only through gravity. When the satellite's mass is negligible compared to the central object's mass, the central object's own motion is negligible too: for all practical purposes, it just sits there while the satellite moves around it.

What governs that motion are the same conservation laws you've used all unit — energy and angular momentum — now applied to a system bound by gravity instead of a spring, a collision, or a rigid connection.

Toggle between a circular orbit and an elliptical orbit to see which quantities stay constant throughout the motion.

Mvcircular orbit — r, v, U, K, L all constant
QuantityConstant?
Total mechanical energy (E)✓ constant
Angular momentum (L)✓ constant
Gravitational PE (U)✓ constant
Kinetic energy (K)✓ constant
🔑In a circular orbit, the satellite's distance from the central body never changes, so its gravitational PE, kinetic energy, angular momentum, and total mechanical energy are all individually constant. In an elliptical orbit, the satellite speeds up near perigee (closest approach) and slows down near apogee (farthest point) — so U and K trade off against each other — but the total mechanical energy E and the angular momentum L stay constant throughout.

Recall from Unit 2 that a satellite in a circular orbit has its centripetal acceleration provided entirely by gravity, which links the orbital period and radius to the central body's mass: T² = 4π²r³/GM. That relationship still holds here — it's the circular-orbit special case of the same gravitational interaction driving everything in this lesson.

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Gravitational PE and Total Energy

Before we can talk about a satellite's energy, we need a zero point. By convention, the gravitational potential energy of a satellite–central-object system is defined to be zero when the satellite is an infinite distance from the central object:

Ug = −Gm1m2 / r
U = 0rU → 0 as r → ∞U → −∞ as r → 0U = −Gm₁m₂/r
💡Because Ug is defined this way, it's always negative for any finite separation — it climbs toward zero as r grows and plunges toward negative infinity as r shrinks. A more negative Ug means a more tightly bound system.

For a satellite in a circular orbit specifically, there's a fixed relationship between its kinetic energy and the system's potential energy — you can derive it by setting the gravitational force equal to the centripetal force requirement. The result:

K = −½U

That lets you write the system's total energy — kinetic plus potential — in terms of just one of them:

Etotal = U + K = ½U = −GMm / 2r

A 1000 kg satellite circles an Earth-mass central body (M ≈ 5.97 × 10²⁴ kg). Slide the orbital radius from low Earth orbit toward geostationary altitude and watch U, K, and Etotal all move together.

r (×10⁶ m)7
U = −GMm/r = -56.92 GJK = −U/2 = 28.46 GJEtotal = U/2 = -28.46 GJ
⚠️Etotal is negative for any bound circular orbit — that negative sign is exactly what makes it a bound orbit. A satellite with negative total energy can't escape to infinity; it needs help from an external force to change that.
ExampleGuided Example — Total Energy of a Geostationary Satellite

A 1000 kg satellite orbits Earth (M ≈ 5.97 × 10²⁴ kg) in a circular geostationary orbit at r = 4.22 × 10⁷ m. Find the system's total mechanical energy.

Step 1Start from the circular-orbit relationship
For a circular orbit, K = −½U, so Etotal = U + K = U − ½U = ½U.
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Escape Velocity

What if a satellite has just enough speed to break free of the central body entirely? The escape velocity is defined as the speed at which the mechanical energy of the satellite–central-object system is exactly zero:

Etotal = ½mv²esc − GMm/r = 0
Mv = vesc — reaches r → ∞ with speed → 0v < vesc — falls back into a bound orbit

Solving that equation for vesc gives the escape velocity from a distance r away from a central body of mass M:

vesc = √(2GM / r)
🔑When a satellite reaches exactly escape velocity, it moves away from the central body forever, its speed approaching zero only in the limit as its distance approaches infinity — it never actually stops, and it never comes back. Any speed less than vesc means the system's total energy is negative, so the satellite stays gravitationally bound.

Adjust the central body's mass and the launch distance from its center to see how escape velocity compares to the speed needed for a circular orbit at that same distance.

M (×10²⁴ kg)5.97
r (×10⁶ m)6.37
vorbit = √(GM/r) = 7.91 km/svesc = √(2GM/r) = 11.18 km/s

vesc = √2 · vorbit — escaping takes about 41% more speed than staying in a circular orbit at that same distance.

ExampleGuided Example — Escaping Earth's Surface

Find the escape velocity for an object launched from Earth's surface (M ≈ 5.97 × 10²⁴ kg, R ≈ 6.37 × 10⁶ m).

Step 1Set total energy to zero
½mv²esc − GMm/r = 0, so the m's cancel and we're solving for vesc alone.
← Back to Lesson 6.6That's Unit 6 complete. Up next: Unit 7 — Oscillations.