Orbital Transfer, Impulse, and Elliptic Orbits
Between impulses, specific orbital energy epsilon=v^2/2-GM/r and angular momentum are conserved; for an ellipse epsilon=-GM/(2a), giving the vis-viva equation v^2=GM(2/r-1/a).
Why this shows up in the exam
Finding apogee after a burn · Stopping or ejecting from an orbit · Comparing elliptical-orbit states
Learn the idea
An instantaneous tangential speed change moves a satellite onto a new conic while position stays fixed during the impulse. A forward burn in a circular orbit normally makes the burn point the nearest point of a larger ellipse before any circularisation.
🧠 Memory hook: At an instant burn, position stays; energy and angular momentum jump.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- v² = GM(2/r - 1/a) — vis-viva relation for a Kepler orbit
- epsilon = v²/2 - GM/r = -GM/(2a) — specific orbital energy
- r_p v_p = r_a v_a — apsis angular-momentum relation
How to approach it
- 1Record r and post-burn v at the impulse
- 2Compute new energy and angular momentum
- 3Infer semi-major axis and turning points
Common slip-ups that cost marks
- •Treating the new path as immediately circular
- •Changing radius during an instantaneous impulse
- •Conserving mechanical energy across a powered burn
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A satellite moves in a circular orbit where GM = 4 x 10^14 m^3/s^2 and orbital radius is 2 x 10^6 m. Find its orbital speed.
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