3 ms·
Wow, it's completely non-intuitive that this is more efficient! Neat.
by idiotsecant 2y ago
Wow, it's completely non-intuitive that this is more efficient! Neat.
- daeken 2y agoFor an intuitive example of where this is more efficient, consider plane changes. The delta-V for a plane change is 2 * sin(angular_change/2) * speed The higher an orbit is, the slower you're going at apogee. If you're doing a large plane change, going even slower will save you a *ton* of delta-V, so you can transfer up to a much higher orbit, plane change at apogee, and then transfer down to your desired final orbit. That's by far the most useful place for bielliptic transfers, not just going from one orbit to another on the same plane.
- avmich 2y agoIn the limit, going from first space velocity V1 (low circular orbit) to second space velocity (parabolic speed) V2 = sqrt(2) * V1 takes delta V = (sqrt(2)-1) * V1, then at infinity you do plane change to the opposite with expending zero delta V, then come back and return to the same orbit, only the opposite direction (largest speed change) spending again delta V = (sqrt(2)-1) * V1. Comparing this total of 0.82 * V1 to spending 2 * V1 if you do that on the spot - just reversing the orbital velocity in place - shows the benefit.