Short on time? This page is the recap for the calculus-based course (separate from Physics 1). For derivations, graphs and worked examples, use the full study guide.
Recap
- System: central body M plus satellite m, interacting only by gravity. If m ≪ M, the central body’s motion is negligible (its speed is m/M times the satellite’s).
- Potential energy: integrate the force from infinity, U(r) = −∫ F_r dr, to get U_g = −GMm/r. It belongs to the system and is negative for all finite r.
- Energy is conserved because gravity is the only force and it is conservative.
- Angular momentum about the central body is conserved because gravity points along r, so it exerts no torque about the centre.
- Circular orbit: r, v, K, U, E and L are all constant.
- Elliptical orbit: E and L constant; K is largest and U lowest at the closest point.
- Escape: a body with E = 0 just reaches infinity, with its speed falling to zero there.
Key relationships
| Relationship | When it applies |
|---|---|
| U_g = −GMm/r | any separation r (centre to centre), zero at r = ∞ |
| E = ½mv² − GMm/r = constant | gravity the only force |
| L = mvr sin φ = constant | gravity the only force; φ is the angle between r and v |
| v = √(GM/r) | circular orbit |
| K = GMm/(2r) = −½U; E = −GMm/(2r) = ½U | circular orbit |
| r_p v_p = r_a v_a | closest and farthest points of an ellipse (v ⟂ r) |
| v_esc = √(2GM/r) = √2 × v_circular | escape from distance r |
Assumptions behind the numbers
- m ≪ M, so the central body is treated as fixed.
- No drag and no thrust. Firing a thruster is an outside force and changes E and L.
- r is measured from the centre of the central body, not from its surface. G = 6.67 × 10⁻¹¹ N·m²/kg².
Mistakes to avoid
- Altitude used as r. Add the central body’s radius.
- Positive U. With the zero at infinity, U_g is negative.
- “Higher orbit is faster.” v = √(GM/r) falls as r rises.
- K constant in an ellipse. Only E and L are.
- L = mvr at every point of an ellipse. Only at the two ends.
- Mass in v_esc. The satellite’s mass cancels.
Quick self-check
- A satellite in a circular orbit has U = −6.0 × 10⁹ J. Find K and E. (K = 3.0 × 10⁹ J; E = −3.0 × 10⁹ J)
- A circular orbit’s radius doubles. By what factor does the speed change? (1/√2 ≈ 0.71)
- A fictional asteroid has mass 6.0 × 10¹⁵ kg and radius 5.0 × 10³ m. Find the escape speed from its surface. (√(2GM/R) ≈ 13 m/s)
Next: practice questions.