I am having trouble understanding why the current location you are at in an orbit remains unaffected when performing an orbital maneuver.

As an example, if I want to raise the apoapsis of my orbit (most efficiently) I would burn prograde at the periapsis. But why doesn't the periapsis point change as well?

  • $\begingroup$ Because you are roughly still at the periphrasis when you finish the burn. In an ideal orbit, & no more delta-v changes, you must return to your starting point. $\endgroup$
    – johnDanger
    Dec 27, 2022 at 21:42

3 Answers 3


When calculating propulsive manoeuvres, we usually consider them as impulses. That is, happening in a very short time. This is usually a good enough model for chemical rocket engines which deplete their propellant in minutes or even seconds.

So, usually: Place the burn started $\approx$ place the burn ended.

This can be combined with a fundamental property of orbits: They repeat.

So exactly one orbit later, we are back in the original location, regardless of what propulsive impulse was performed there.

If, on the other hand, your burn takes a while, like with an ion engine or with a small motor needing to burn through a lot of propellant, the original location does as a matter of fact not stay unaffected.

For the apoapsis raising manoeuvre specifically

You are at the periapsis, so your orbit has only tangential velocity and no radial velocity. By adding more, you still only have tangential velocity.

There are only two points in an elliptic orbit with tangential velocity only, the periapsis and the apoapsis. So if what we are raising is the apoapsis, the current point must by necessity be the periapsis, and thus remain unchanged. (you could flip them though, by doing a retrograde burn to take the tangential velocity below the circular orbital velocity, making the current point the apoapsis). By angling the burn a little, so it also adds radial velocity, would indeed be changing the periapsis location.

  • 16
    $\begingroup$ Note that even for long burns, the spacecraft will still return to the point where the burn ended. The instant you've stopped thrusting, you are in an elliptical orbit that will return you to that location (provided you haven't burned to escape, set up a collision course, or getting close enough to a third body that your orbit is non-Keplerian). There's no period of adjustment or settling as the orbit reacts to your actions (a misconception I've seen more than once). $\endgroup$ Dec 26, 2022 at 1:36
  • 3
    $\begingroup$ @ChristopherJamesHuff Your comment seems a more general and better answer to the question, and perhaps might be better as a proper answer that expands on this slightly, with more detail on the misconceptions and how they come about. $\endgroup$
    – cjs
    Dec 26, 2022 at 7:38
  • $\begingroup$ I guess one could do a burn in not exactly tangential direction, but e.g. radially (or diagonally). Then the "you still only have tangential velocity" would be wrong, but it still doesn't help to increase the periapsis. $\endgroup$ Dec 26, 2022 at 20:41
  • $\begingroup$ @PaŭloEbermann But you could lower the periapsis that way. It would just be suboptimal way to change the periapsis altitude. Since the orbit has to pass through your current point, the constraint is that you can never raise the periapsis higher than you currently are, or bring the apoapsis lower than you currently are. But you can absolutely change their altitudes within those constraints at any point around the orbit by burning in some direction (it just won't usually be optimal way to adjust the apses or orbit geometry.) $\endgroup$
    – tylisirn
    Dec 27, 2022 at 7:17

Simply because the point of burn is part of both orbits.

When you fire your engine pro-/retrograde, all you do is change the spacecraft's velocity. The spacecraft does not magically jump to a new height that would correspond to a circular orbit. Instead, it simply becomes too fast/slow for its old orbit in the place of the burn. And the new orbit will be whatever orbit happens to have the spacecraft at the place of burn with the new velocity.

As such, the orbits before and after the burn will not just coincide in location, they will also coincide tangentially: The new orbit has the spacecraft moving at the same place in the same direction, but at a different speed. It is only when you integrate the effect of the changed speed over time that it accumulates a difference is place.


The key point is that the spacecraft will return to the point where the burn ended. Whether the burn is long or short: the instant the spacecraft stops thrusting, it is in a closed elliptical orbit that will bring it back to that location after a full orbit (provided it hasn't burned to escape or set up a collision course, and there's no additional bodies or other perturbations that make it inaccurate to simplify things to Keplerian orbits).

For short burns, the spacecraft will often not move a substantial fraction of an orbit over the course of the burn, and the entire burn can be simplified to a single point. However, even for a long burn, the orbit is only actively changing while the spacecraft is under thrust (ignoring outside factors like atmospheric drag). Contrary to some depictions of orbits in media, there's no period of adjustment or settling as the orbit "reacts" to your actions: the spacecraft won't spiral in or out or take multiple orbits to "stabilize", it will instantaneously be following an elliptical trajectory that brings it back to the same location.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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