Two-body orbit propagation without leaving Cartesian coordinates: “the 𝑓 and 𝑔 functions”.

Two-Body Initial and Boundary Value Problems

  • Initial value problem: propagation
  • Boundary value problem: the Lambert problem
  • Orbit propagation: given an orbital state vector at one time, what is it at a future time?
    • [𝐫0̇𝐫0]⟶[𝐫(𝑡)̇𝐫(𝑡)],𝐫0=𝐫(𝑡0)
    • 𝑡0: the epoch time

The f and g Functions

  • Method of Lagrange coefficients (Curtis section 2.11, but here using the change in eccentric anomaly rather than the change in true anomaly)
  • The motion stays in the plane of 𝐫0 and ̇𝐫0, so the new state is a linear combination of them:
    • 𝐫(𝑡)=𝑓(𝑡,𝑡0)𝐫0+𝑔(𝑡,𝑡0)̇𝐫0̇𝐫(𝑡)=̇𝑓(𝑡,𝑡0)𝐫0+̇𝑔(𝑡,𝑡0)̇𝐫0
  • At the epoch:
    • 𝑓(𝑡0,𝑡0)=1=̇𝑔(𝑡0,𝑡0),̇𝑓(𝑡0,𝑡0)=0=𝑔(𝑡0,𝑡0)
  • Solve for 𝑓 by crossing with ̇𝐫0:
    • 𝐫×̇𝐫0=(𝑓𝐫0+𝑔̇𝐫0)×̇𝐫0=𝑓(𝐫0×̇𝐫0)=𝑓𝐡
    • In in-plane (perifocal) components:
      • 𝑓=|𝐫×̇𝐫0|ℎ=𝑥̇𝑦0−̇𝑥0𝑦ℎ,̇𝑓=̇𝑥̇𝑦0−̇𝑥0̇𝑦ℎ
  • Solve for 𝑔 by crossing 𝐫0 with 𝐫:
    • 𝐫0×𝐫=𝐫0×(𝑓𝐫0+𝑔̇𝐫0)=𝑔𝐡
    • 𝑔=𝑥0𝑦−𝑥𝑦0ℎ,̇𝑔=𝑥0̇𝑦−̇𝑥𝑦0ℎ

The f and g in Terms of Delta E

  • Perifocal coordinates in terms of the eccentric anomaly:
    • cos𝐸=𝑒+𝑟cos𝜃𝑎=𝑒+𝑥𝑎,𝑎√1−𝑒2sin𝐸=𝑟sin𝜃=𝑦
    • 𝑥=𝑎(cos𝐸−𝑒)𝑦=𝑎√1−𝑒2sin𝐸̇𝑥=−𝑎̇𝐸sin𝐸̇𝑦=𝑎̇𝐸√1−𝑒2cos𝐸
  • Rate of the eccentric anomaly, from Kepler’s equation and 𝑟=𝑎(1−𝑒cos𝐸):
    • 𝑛=d𝑀d𝑡=̇𝐸−𝑒̇𝐸cos𝐸=̇𝐸(1−𝑒cos𝐸)⏟𝑟𝑎=̇𝐸𝑟𝑎
    • ̇𝐸=𝑎𝑛𝑟=1𝑟√𝜇𝑎
  • Compute ℎ𝑓:
    • ℎ𝑓=𝑥̇𝑦0−̇𝑥0𝑦=𝑎(cos𝐸−𝑒)𝑎̇𝐸0√1−𝑒2cos𝐸0+𝑎̇𝐸0sin𝐸0𝑎√1−𝑒2sin𝐸=𝑎2√1−𝑒2̇𝐸0[cos𝐸0(cos𝐸−𝑒)+sin𝐸0sin𝐸]
  • Use cos(𝐸−𝐸0)=cos𝐸cos𝐸0+sin𝐸sin𝐸0 and ℎ=√𝜇𝑝=√𝜇𝑎(1−𝑒2):
    • 𝑓=𝑎2√1−𝑒2ℎ1𝑟0√𝜇𝑎[cos(𝐸−𝐸0)⏟Δ𝐸−𝑒cos𝐸0]=𝑎𝑟0[cosΔ𝐸−(1−𝑟0𝑎)]=1−𝑎𝑟0(1−cosΔ𝐸)
    • using 𝑒cos𝐸0=1−𝑟0𝑎

Lagrange Coefficients

  • 𝑓=1−𝑎𝑟0(1−cosΔ𝐸)𝑔=Δ𝑡−√𝑎3𝜇(Δ𝐸−sinΔ𝐸)̇𝑓=−√𝜇𝑎sinΔ𝐸𝑟𝑟0̇𝑔=1−𝑎𝑟(1−cosΔ𝐸)
  • Δ𝐸=𝐸−𝐸0 comes from solving Kepler’s equation for the time step Δ𝑡
  • Worked example: Curtis Example 2.13

In Universal Variables

  • With the universal anomaly 𝜒 and the Stumpff functions 𝐶, 𝑆:
    • 𝛼=1𝑎=2𝑟0−𝑣20𝜇

Universal Lagrange Coefficients

  • 𝑓=1−𝜒2𝑟0𝐶(𝛼𝜒2)(3.69a)𝑔=Δ𝑡−1√𝜇𝜒3𝑆(𝛼𝜒2)(3.69b)̇𝑓=√𝜇𝑟𝑟0[𝛼𝜒3𝑆(𝛼𝜒2)−𝜒](3.69c)̇𝑔=1−𝜒2𝑟𝐶(𝛼𝜒2)(3.69d)
  • Note the 𝑟 (the radius at the new time) in the denominator of ̇𝑓 (circled in the notes)

Propagation with Lagrange Coefficients

  • Block-matrix form of the propagation:
    • [𝐫(𝑡)̇𝐫(𝑡)]=[𝑓(𝑡,𝑡0)𝐼̇𝑓(𝑡,𝑡0)𝐼𝑔(𝑡,𝑡0)𝐼̇𝑔(𝑡,𝑡0)𝐼][𝐫0̇𝐫0]
    • 𝐼: the 3×3 identity matrix
  • Careful: this is not the state transition matrix, because 𝑓, 𝑔 themselves depend on the initial state (see State Transition Matrix)