The second conservation law of the Two-Body Problem (Curtis section 2.4).

Conservation of Angular Momentum

  • Central force: the force and the acceleration are both along ̂𝐫
    • 𝐅∝̂𝐫,̈𝐫∝𝐅
  • Cross with 𝐫:
    • 𝐫×̈𝐫∝𝐫×𝐫=0
  • The left side is a total derivative (because ̇𝐫×̇𝐫=0):
    • dd𝑡(𝐫×̇𝐫)=0

Angular Momentum

  • 𝐡=𝐫×̇𝐫=const
  • 𝐡 is constant, so 𝐫 and ̇𝐫 always lie in the same plane: the orbit is planar

Kepler’s Second Law

  • Area swept out by the position vector in a short time (from Dr. Healy's notes)
    • figure from go.umd.edu/ENAE601-figures
  • In a short time 𝛿𝑡 the position vector sweeps out a thin triangle with base 𝑟 and height ℓ=𝛿𝑟sin𝜃:
    • 𝛿𝐴=𝑟ℓ2=12𝑟𝛿𝑟sin𝜃=12|𝐫×𝛿𝐫|
  • Rate of sweeping out area:
    • ̇𝐴=lim𝛿𝑡→0𝛿𝐴𝛿𝑡=lim𝛿𝑡→012|𝐫×𝛿𝐫|𝛿𝑡=12|𝐫×̇𝐫⏟𝐡|=ℎ2
  • Area swept out over a finite time interval from 𝑡1 to 𝑡2:
    • 𝐴𝑡1,𝑡2=∫𝑡2𝑡112ℎd𝑡=12ℎ(𝑡2−𝑡1)
  • Equal times ⇒ equal areas: Kepler’s second law proved

Transverse Velocity

  • Split the velocity into a radial part 𝑣𝑟 and a transverse part 𝑣⟂ (perpendicular to 𝐫) (Curtis figure 2.9)
    • 𝜙: angle between 𝐫 and ̇𝐫
  • In time 𝛿𝑡 the radius turns through ̇𝜃𝛿𝑡, so the transverse displacement is 𝑟̇𝜃𝛿𝑡≈𝑣⟂𝛿𝑡
    • 𝑣⟂=𝑟̇𝜃in the limit𝛿𝑡→0
  • Magnitude of the angular momentum:
    • ℎ=𝑟𝑣sin𝜙=𝑟𝑣⟂=𝑟2̇𝜃
  • Only the transverse velocity contributes to ℎ