Vai's Uni Notes

Unbound Orbits

2 min read

Parabolic and hyperbolic trajectories: conic sections with 𝑒≥1 that never return.

Energy of the Conic Section

  • The sign of the specific mechanical energy classifies the orbit:
    • ℰ︀<0: bound orbits (circle, ellipse)
    • ℰ︀=0: parabolic
    • ℰ︀>0: hyperbolic
  • ℰ︀=−𝜇2𝑎 applies to the bound orbits, not to the parabola (𝑎 is not defined)
  • For the hyperbola, ℰ︀=𝜇2𝑎 (with 𝑎 taken positive; equivalently −𝜇2𝑎 with 𝑎<0)

Parabolic Orbits

  • Curtis section 2.8
  • Eccentricity 𝑒=1; semimajor axis 𝑎 not defined
  • Orbit equation:
    • 𝑟=𝑝1+cos𝜃
  • Perigee radius, with 𝑝 the semi-latus rectum:
    • 𝑟𝑝=𝑝2

No semimajor axis

  • The ellipse formula 𝑟𝑝=𝑎(1−𝑒) does not work for a parabola (crossed out in the notes): 𝑎 is undefined and 1−𝑒=0; use 𝑟𝑝=𝑝2
  • Energy is zero:
    • ℰ︀=𝑣22−𝜇𝑟=0
    • as 𝑟→∞, 𝑣∞=0: the parabola is the trajectory that just barely escapes
  • So the speed at radius 𝑟 on a parabola is the escape speed:
    • 𝑣2esc2−𝜇𝑟=0

Escape Speed

  • 𝑣esc=√2𝜇𝑟
  • Compare the circular speed 𝑣circ=√𝜇𝑟: 𝑣esc=√2𝑣circ

Hyperbolic Orbits

  • Curtis section 2.9
  • Hyperbolic escape trajectory with the true anomaly of the asymptote theta_inf (from Dr. Healy's notes)
  • 𝑒>1, so 𝑎𝑒>𝑎: the focus (Earth) is farther from the center than the vertex is
  • Solve the orbit equation for the true anomaly:
    • cos𝜃=1𝑒(𝑝𝑟−1)
  • As 𝑟→∞ the trajectory approaches its asymptote at the true anomaly of the asymptote 𝜃∞:
    • cos𝜃∞=−1𝑒
    • the true anomaly is limited to −𝜃∞<𝜃<𝜃∞

Hyperbolic Excess Speed

  • Far from the Earth (𝑟→∞) the potential energy vanishes and the speed tends to 𝑣∞:
    • ℰ︀=𝑣22−𝜇𝑟=𝑣2∞2

Hyperbolic Excess Speed

  • 𝑣2∞=𝑣2−2𝜇𝑟=𝑣2−𝑣2esc
  • The speed left over after escaping

Graph View

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