AP® Physics 1: Algebra-Based review sheet from Aim for Five (aimforfive.com/physics/units/6/6-6)
Unit 6 · Topic 6.6
6.6 Motion of Orbiting Satellites
A satellite orbiting a much more massive object, with only gravity acting, keeps a constant total mechanical energy and a constant angular momentum. Gravitational potential energy is U = −GMm/r, so as an orbit brings a satellite closer it speeds up. Escape velocity, √(2GM/r), is the speed that makes the total energy exactly zero.
Key terms
- gravitational potential energy
- circular orbit
- elliptical orbit
- escape velocity
- conservation of angular momentum
The model
Think of a satellite circling Earth, or a planet circling the Sun. The central object is so much more massive that its own motion is negligible, so you treat it as fixed and track only the satellite.
Only gravity acts, and it always points toward the central object. That gives two conservation laws. No outside force does work on the satellite–planet system, so its total mechanical energy is constant. And gravity points straight at the center, so it exerts no torque about it, which means the satellite's angular momentum about the center is constant.
Gravitational potential energy far from Earth
Near the ground, ΔU = mgΔy works because g barely changes. For orbits it does change, so you use U = −GMm/r, where r is the distance between the centers and G = 6.67 × 10⁻¹¹ N·m²/kg².
This formula sets U = 0 when the objects are infinitely far apart. Since gravity pulls them together, U is negative at every finite distance and rises toward zero as r grows. A more negative U means a more tightly bound satellite.
Circular orbits
In a circular orbit gravity supplies the centripetal force (2.9): GMm/r² = mv²/r, so v = √(GM/r). The satellite's mass cancels. Then K = ½mv² = GMm/(2r), U = −GMm/r and the total is E = −GMm/(2r).
In a circular orbit r doesn't change, so K, U, E and L are all constant. A higher orbit is slower and takes longer, but has more total energy (less negative), so moving a satellite up costs energy.
| Quantity | Circular orbit of radius r |
|---|---|
| speed | v = √(GM/r) |
| kinetic energy | K = GMm/(2r) |
| potential energy | U = −GMm/r |
| total energy | E = −GMm/(2r) |
Elliptical orbits
In an elliptical orbit the distance r changes. Total energy and angular momentum are still constant, but U and K trade: as the satellite falls closer, U drops and K rises, so it moves fastest at its closest point and slowest at its farthest.
At the closest and farthest points, the velocity is perpendicular to the line to the center, so L = mvr there. Angular momentum conservation then gives r₁v₁ = r₂v₂: half the distance means twice the speed. You don't need Kepler's laws by name; these two conservation laws do the work.
Escape velocity
If the total energy is negative, the satellite is bound: it can't get infinitely far away. Escape velocity is the speed that makes the total energy exactly zero: ½mv² − GMm/r = 0, so v_esc = √(2GM/r).
Launched at exactly escape velocity, with only gravity acting, an object keeps moving away and slows down, reaching zero speed only at an infinite distance. Escape velocity doesn't depend on the object's mass, and it's √2 times the circular orbit speed at the same distance.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
A satellite in low orbit
A 1000 kg satellite moves in a circular orbit 400 km above Earth's surface. Earth's mass is 5.97 × 10²⁴ kg and its radius is 6.37 × 10⁶ m; G = 6.67 × 10⁻¹¹ N·m²/kg². Find its speed, its period, and the system's kinetic, potential and total energy.
Show the solutionHide the solution
- Step 1: Use the distance from Earth's center: r = 6.37 × 10⁶ + 4.0 × 10⁵ = 6.77 × 10⁶ m.
- Step 2: v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴ ÷ 6.77 × 10⁶) ≈ 7.67 × 10³ m/s.
- Step 3: Period: T = 2πr/v ≈ 5.55 × 10³ s, about 92 minutes.
- Step 4: U = −GMm/r ≈ −5.88 × 10¹⁰ J. K = ½mv² ≈ 2.94 × 10¹⁰ J. E = K + U ≈ −2.94 × 10¹⁰ J, which is exactly half of U, as expected.
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 velocity from Earth (classic trap)
Using the values above, find the escape velocity from Earth's surface (ignore air resistance) and from the 400 km orbit. Does a heavier rocket need a larger escape velocity?
Show the solutionHide the solution
- Step 1: Set total energy to zero: ½mv² − GMm/r = 0, so v = √(2GM/r).
- Step 2: At the surface: √(2 × 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ ÷ 6.37 × 10⁶) ≈ 1.12 × 10⁴ m/s.
- Step 3: At r = 6.77 × 10⁶ m: ≈ 1.08 × 10⁴ m/s, which is √2 × 7.67 km/s.
- Step 4: The mass m canceled, so no: the escape velocity is the same for any mass. A heavier rocket needs more energy to reach it, but not a higher speed. Using the altitude instead of the distance from Earth's center is another trap.
Answer: About 11.2 km/s from the surface, 10.8 km/s from the orbit; it doesn't depend on the rocket's mass
- Example 3Calculator allowed
Speed at the far end of an elliptical orbit
A satellite in an elliptical orbit around Earth is 7.0 × 10⁶ m from Earth's center at its closest point, moving at 8.71 × 10³ m/s. Its farthest point is 1.4 × 10⁷ m from the center. Find its speed there, and check that energy is conserved. (GM for Earth = 3.98 × 10¹⁴ m³/s².)
Show the solutionHide the solution
- Step 1: At both points the velocity is perpendicular to r, so angular momentum conservation gives r₁v₁ = r₂v₂.
- Step 2: v₂ = (7.0 × 10⁶)(8.71 × 10³) ÷ (1.4 × 10⁷) ≈ 4.36 × 10³ m/s. Twice the distance, half the speed.
- Step 3: Energy per kilogram at the close point: ½v² − GM/r = 3.793 × 10⁷ − 5.686 × 10⁷ ≈ −1.89 × 10⁷ J/kg.
- Step 4: At the far point: 0.948 × 10⁷ − 2.843 × 10⁷ ≈ −1.89 × 10⁷ J/kg. The same to within rounding, so energy is conserved too.
Answer: About 4.4 × 10³ m/s; total energy per kilogram is about −1.89 × 10⁷ J/kg at both points
Common mistakes
- Measuring r from Earth's surface instead of its center in U = −GMm/r and v = √(GM/r).
- Using mgh for orbits. It only works close to the surface, where g is nearly constant.
- Thinking a satellite in an elliptical orbit has constant speed or kinetic energy. Only total energy and angular momentum stay constant.
- Saying escape velocity depends on the object's mass, or that the object stops at some finite height.
On the exam
- Expect questions comparing two orbits: which has greater speed, kinetic energy or total energy? Use v = √(GM/r) and E = −GMm/(2r).
- For elliptical orbits, justify speed changes with both conservation laws: angular momentum (closer means faster) and energy (U falls, so K rises).
- Derivations of v_esc or orbital speed are common; start from conservation of energy or Newton's second law with gravity as the centripetal force.
Connected topics
Videos
Check yourself
4 questions on 6.6 Motion of Orbiting Satellites. Pick an answer to see if you got it, and why.
A satellite moves in an elliptical orbit around a planet with much greater mass. Only gravity acts. Which quantities stay constant during the orbit?
| Point in orbit | Distance from planet's center (m) | Satellite speed (m/s) |
|---|---|---|
| Closest approach | 8.0 × 10⁶ | 6.0 × 10³ |
| Farthest point | 1.6 × 10⁷ | ? |
Hypothetical data. A 500 kg satellite moves in an elliptical orbit around a planet with much greater mass. At the closest and farthest points, the satellite's velocity is perpendicular to the line joining it to the planet's center.
What is the satellite's speed at the farthest point?
What is the change in the system's gravitational potential energy as the satellite moves from closest approach to the farthest point?
Which statement correctly compares the closest approach and the farthest point?
0 of 4 answered