Any Kerbal Space Program player will know that burns prograde and retrograde to the velocity vector are most efficient closest to the body being orbited, while burns normal and anti-normal are most efficient furthest from the body being orbited, where efficiency is defined by how much delta-v is required to go from a given starting orbit to a given target orbit.

But energy can neither be created nor destroyed, so when burning in the right direction at a suboptimal point in the orbit (suboptimal true anomaly, to be technical), where does all the additional energy expended go? It's not cosine loss, as this effect, which has a few different names based on the position in the orbit but is most familiar as the Oberth Effect, is encountered even when the burn is instantaneous, so it's not that one part of the burn is cancelling out another.

So where is the extra energy going?

  • 2
    $\begingroup$ After drafting an answer I ended up bailing on, I think this is fundamentally a terminology question. The "energy of an orbit" means nothing, physically, and thus is not subject to the familiar conservation laws of physics... begging the question "how is the energy of an orbit determined"? Maybe the kinetic+gravitational potential of the vehicle integrated over an orbital period? What about precession? $\endgroup$ Sep 23 '21 at 19:35
  • 2
    $\begingroup$ As for physical kinetic energy, it's always going to be the same if you sum the chemical potential of the vehicle, its kinetic energy, and the kinetic energy of the exhaust. Doesn't matter if you're trying to make a poorly-advised plane change burn at perigee or a better-advised ejection burn at the same. $\endgroup$ Sep 23 '21 at 19:36
  • 1
    $\begingroup$ Actually, the "energy of an orbit" is probably just the invariant kinetic energy (in an inertial frame)+gravitational potential. Would not be hard to define a "momentum of an orbit" (of a massive body). $\endgroup$ Sep 23 '21 at 19:57
  • 2
    $\begingroup$ The energy end up more in the exhaust gases, and less in the rocket $\endgroup$ Sep 24 '21 at 0:48
  • 2
    $\begingroup$ Efficiency depends on what you want to achieve. If you want a circular orbit you will need a burn at the apogee. $\endgroup$
    – Florian F
    Sep 24 '21 at 9:12

After writing my comments, I started writing a new answer. That got long, so here's a shorter one.

The "energy of an orbit" may be poorly defined and depending on the definition, is not subject to energy conservation laws. Let's ignore it and focus on the kinetic energy.

Kinetic energy is not necessarily conserved. Momentum is. Momentum (derived from velocity) is a vector; kinetic energy is derived from the magnitude of velocity and is a scalar. You can exchange momentum without changing your kinetic energy (this is called changing directions). If you want to do an energy conservation calculation, you must include the stored chemical energy of your unburned propellant, the gravitational potential of your vehicle, and both the gravitational potential and the kinetic energy of your exhaust. For simplicity, it's probably best to do this in the inertial reference frame of the body you're orbiting.

Here's a shorter answer: the energy goes into the energy of your exhaust's orbit. When you burn prograde, the energy of your exhaust's orbit goes into your own.

  • 30
    $\begingroup$ the energy goes into the energy of your exhaust's orbit. This is the crux of the answer right here. An inefficient burn will put your exhaust's gases into a more energetic orbit more efficiently. $\endgroup$
    – Ingolifs
    Sep 23 '21 at 23:48
  • 3
    $\begingroup$ I'm not sure why you insist that "energy of an orbit" may be poorly defined. It is, as you suggest yourself, the sum of the kinetic and potential energy which are entirely well defined. Differences in that energy between two different orbits determine how much energy you need to transfer to the rocket to get it from one to the other orbit. And why do you think that this energy "is not subject to energy conservation laws"?? $\endgroup$ Sep 24 '21 at 12:19
  • $\begingroup$ "I'm not sure why you insist that "energy of an orbit" may be poorly defined" You're right, it's really just because it took me a while to figure out the definition & I hadn't seen it stated before. Even so, I'm still not sure if the "energy of an orbit" is a property of the orbit or a property of the body orbting. For example, is the energy of the orbit the same for a 1kg object as a 1000kg object if they're in the same orbit? If so, is the "energy of the orbit" somehow normalized by the mass? These are questions I don't know the answer to so I'm going to weasel word my way out of it as I did $\endgroup$ Sep 24 '21 at 19:32
  • $\begingroup$ As for energy conservation laws, energy isn't conserved in any one form except for certain types of interactions defined as such. Case in point: burning chemical propellant introduces new energy to the system of orbits. Thus, the energy of the orbit is nonconservative except in certain types of interactions (like drifting along a geodesic) $\endgroup$ Sep 24 '21 at 19:33
  • $\begingroup$ There’s a logical hole in this answer that I’m not sure how to fill. A prograde burn’s optimal position in the orbit matches that of a retrograde burn. Same goes for a normal/antinormal pair. Yet what is for a rocket a prograde burn is for the exhaust a retrograde burn, if you want to imagine the burn from the perspective of the exhaust as throwing a rocket backward. So an efficient position for the rocket’s burn is also an efficient position for the exhaust’s burn, and vice versa, which seems to run counter to this argument $\endgroup$ Oct 9 '21 at 0:55

The sum of all mechanical energy will be the same after your ideal burns. The difference will be that the portion given to your exhaust will be greater for higher burns.

For a burn farther away from the planet:

  • The craft will be higher, so the PE of the exhuast will be greater.
  • The craft will (usually) be slower, so the KE of the exhaust will be greater.
  • 1
    $\begingroup$ Actually, at least in low- or medium Earth orbits, the orbital velocity is higher than the exhaust velocity of chemical engines, meaning that the exhaust will still be in a prograde orbit and thus higher original speed of the craft correlates with higher KE of the exhaust. $\endgroup$ Sep 25 '21 at 19:15
  • $\begingroup$ @leftaroundabout Correct -- what really matters is the change in KE of the exhaust (versus the KE when it was traveling with the craft as propellant). This change is greater (less negative) the slower the craft. Likewise, the first bullet in the answer is not relevant because there is no change in PE of the exhaust. $\endgroup$
    – nanoman
    Sep 26 '21 at 12:31
  • $\begingroup$ @nanoman actually no, the change in exhaust-KE is greater the faster the craft. Basically, $v^2 - (v-v_\mathrm{E})^2 \approx 2\cdot v\cdot v_\mathrm{E}$ for .$v_\mathrm{E}\ll v$ And the first bullet does matter too: if you burn at apogee, it means you had to carry all the fuel with you higher up in the gravity well and only there get rid of it. Better to burn at perigee and have a lighter craft rising first to the original apogee and then even higher. $\endgroup$ Sep 26 '21 at 14:02
  • $\begingroup$ @leftaroundabout I said "greater (less negative)". The exhaust speed before burn is $v$ and after burn is $v - v_{\mathrm{E}}$, so the change in KE of exhaust is proportional to minus $2vv_{\mathrm{E}}$ and is negative for $v_{\mathrm{E}} \ll v$. The slower the craft, the less the KE extracted from the exhaust -- or as an algebraic equivalent, the greater (less negative) the KE put into the exhaust. I used the latter phrasing as it is the closest correct and relevant statement to the answer's second bullet. More energy is wasted on the exhaust when the craft is slower. $\endgroup$
    – nanoman
    Sep 26 '21 at 20:12
  • $\begingroup$ @leftaroundabout Also, the first bullet is not an additional contribution. The burn does not change the PE of the exhaust (it is emitted at the same location of the craft). The propellant is already traveling with the craft and there is no cost "to carry all the fuel with you higher up in the gravity well". In the following formula, the "craft" before and after excludes the propellant (about to be) burned, i.e., no change in mass: (change in energy of craft) = (energy of burn) - (change in energy of exhaust) = (energy of burn) - (change in KE of exhaust). Only KE changes during the burn. $\endgroup$
    – nanoman
    Sep 26 '21 at 20:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.