AP Physics C: Mechanics · Unit 6 · Topic 5 of 5 · 6 min lesson

Orbits

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.

Quiz this topic · 5 questions

The big idea

Gravity provides the centripetal force for a circular orbit, giving v = √(GM/r). In any orbit, gravity points at the planet’s center, so it makes no torque about it: angular momentum is conserved. Total energy E = K + U is conserved too.

1. Circular orbits

v = √(GM/r)from GMm/r² = mv²/r

T² = (4π²/GM) r³Kepler’s third law

U = −GMm/rzero at infinity

K = GMm/(2r)

E = −GMm/(2r)negative: bound

A higher orbit is slower (smaller K) but has more total energy (E is less negative). Moving up takes energy from the rockets.

Escape speed: E = 0, so ½mv² = GMm/R and v_esc = √(2GM/R), √2 times the circular orbital speed at that radius.

2. Elliptical orbits

At the closest and farthest points, the velocity is perpendicular to r, so L = mvr. Closer means faster.

At the two ends of the ellipse, v ⊥ r, so conserving L gives r₁v₁ = r₂v₂.

Energy is also conserved: ½mv₁² − GMm/r₁ = ½mv₂² − GMm/r₂. Together these two equations solve most orbit problems.

3. Example, then try it

Example

A satellite orbits Earth (GM = 4.0 × 10¹⁴ m³/s²) at r = 6.8 × 10⁶ m. Find its speed and period.

v = √(4.0 × 10¹⁴ ÷ 6.8 × 10⁶) ≈ 7.7 × 10³ m/s.

T = 2πr/v = 2π(6.8 × 10⁶) ÷ 7.7 × 10³ ≈ 5.6 × 10³ s, about 93 minutes.

Try it · work it on paper first

A 500 kg satellite moves from a circular orbit of radius 7.0 × 10⁶ m to one twice as large. Using GM = 4.0 × 10¹⁴ m³/s², find its kinetic and total energy in each orbit, the energy its rockets must supply, and how its period changes.

Lower orbit: K = GMm/(2r) = (4.0 × 10¹⁴)(500) ÷ (1.4 × 10⁷) ≈ 1.43 × 10¹⁰ J; E = −K ≈ −1.43 × 10¹⁰ J.

Higher orbit (2r): K ≈ 7.1 × 10⁹ J; E ≈ −7.1 × 10⁹ J. Slower, but more total energy.

Energy needed: ΔE ≈ −7.1 × 10⁹ − (−1.43 × 10¹⁰) ≈ +7.1 × 10⁹ J.

Period: T ∝ r^3/2, so it grows by 2^3/2 ≈ 2.8 times.

Answer: K halves, E rises by 7.1 × 10⁹ J, and the period grows 2.8×.

4. Common mistakes

“Higher orbit, more kinetic energy.”Higher orbits are slower; only the total energy is greater.

Using mgh for orbits.Use U = −GMm/r.

Using r₁v₁ = r₂v₂ at any two points.It holds where v ⊥ r (the closest and farthest points); elsewhere use L = mvr sin θ.

Check yourself

At which point of an elliptical orbit is a satellite moving fastest?

Angular momentum is conserved, and at the ends of the ellipse L = mvr. The smallest r gives the largest v.

Ready to practice?

5 questions on Orbits, each with a hint and an explanation for every wrong answer.

Quiz this topic All of Unit 6