Short on time? This page is the recap. For diagrams, derivations and worked examples, use the full study guide. This is the algebra-based course: no calculus is needed.
Recap
- System: a central body of mass M and a satellite of mass m, interacting only by gravity. If m ≪ M, the central body’s motion is negligible: treat it as fixed.
- Gravity on the satellite points at the central body’s centre, so it exerts no torque about that centre: the satellite’s angular momentum is constant.
- Gravity is the only force, so the system’s mechanical energy is constant.
- U_g = 0 at infinite separation, so U_g is negative at every finite r, and every bound orbit has E < 0.
- Circular orbit: r and v fixed, so K, U_g, E and L are all constant.
- Elliptical orbit: E and L constant; K and U_g trade. Fastest at the closest point, slowest at the farthest.
- Escape speed: the speed that makes E = 0. The satellite then slows towards zero speed only as r grows without limit.
Key relationships
| Relationship | When it applies | Use it to |
|---|---|---|
| U_g = −GMm/r | any separation r, centre to centre | find potential energy (zero at infinity) |
| E = ½mv² − GMm/r = constant | gravity the only force | link speed and distance anywhere on an orbit |
| L = mvr sin θ = constant | gravity the only force | compare speeds; L = mvr at closest and farthest points |
| v² = GM/r | circular orbit | orbital speed |
| K = GMm/(2r), E = −GMm/(2r) | circular orbit | energies; E = −K and U_g = 2E |
| v_esc = √(2GM/r) | escape from distance r | minimum speed to escape (√2 × circular speed) |
Assumptions behind the numbers
- The satellite’s mass is negligible compared with the central body’s.
- No atmosphere, no other bodies, no rocket thrust: gravity from one central body is the only force.
- Central bodies are spherical, so r is measured from their centre. G = 6.67 × 10⁻¹¹ N·m²/kg².
Mistakes to avoid
- Measuring r from the surface. Add the radius of the central body.
- Positive potential energy. U_g = −GMm/r is always negative.
- “Higher orbit, more kinetic energy.” Higher circular orbit: less K, more (less negative) E.
- Using L = mvr at a general point on an ellipse. Only where v ⟂ r.
- Making escape speed depend on m. It cancels.
- Using mgh for large changes in height. g changes with r; use −GMm/r.
Quick self-check
- A satellite in a circular orbit has K = 5.0 × 10⁹ J. What are E and U_g? (E = −5.0 × 10⁹ J; U_g = −1.0 × 10¹⁰ J)
- A probe moves at 6.0 km/s at its closest point, 1.0 × 10⁷ m from a planet’s centre. Its farthest point is 3.0 × 10⁷ m away. How fast is it there? (2.0 km/s)
- Planet B has four times the mass of planet A and the same radius. Compare their surface escape speeds. (B’s is twice A’s.)
Next: practice questions.