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Say I'm in a circular orbit around the Earth. I give my motor a burn in the tangental direction. My understanding is that my trajectory now becomes an ellipse, and if I want to enter a new, higher circular orbit, at the apogee of the ellipse I have to fire my motor again to attain the velocity needed for that orbit. If I don't, my spacecraft will fall back. So here's the question: I'd expect it to fall back on a mirror image trajectory -- mirroring the elliptical path that got me to the apogee -- and here's the real issue: once it has arrived at perigee, directly opposite the point where I first fired my engines, I expect it to continue past that point and draw a mirror image trajectory on the other side of it's orbit ... and that trajectory should be a mirror image of the first half of the orbit ... which would put the Earth in the center of this ellipse rather than at one of the foci.

If the first quarter of this orbit (from power on to apogee) is elliptical then the second quarter (apogee to perigee) 'should' be a mirror image of the first quarter and then the second half of the orbit (perigee back to the initial position where i first fired my motor) should be a mirror image of the first half ... which again puts the Earth at the center of an elliptical orbit ... but orbits don't work like that. It's paradoxical. What to do about it? Somehow I've got to end up with an orbit around a focus, no?

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    $\begingroup$ "once it has arrived at perigee, directly opposite the point where I first fired my engines". No, the perigee is at the point where you made the burn. $\endgroup$
    – PM 2Ring
    Commented Sep 17 at 12:51
  • $\begingroup$ But I'm expecting a symmetrical perigee on the opposite side. Of course I'm making a mistake somewhere, but my entire thinking is based on the idea of symmetry. $\endgroup$
    – Ray Andrews
    Commented Sep 17 at 13:49
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    $\begingroup$ open KSP, put a rocket in orbit, circularise it, then fire prograde and see how your orbit changes $\endgroup$
    – njzk2
    Commented Sep 17 at 20:53
  • $\begingroup$ KSP? Sounds interesting but never heard of it. $\endgroup$
    – Ray Andrews
    Commented Sep 18 at 13:58
  • $\begingroup$ KSP = Kerbal Space Program, a very fun spaceflight simulation game. (Tip: I'd recommend trying the original game, not the sequel, which is currently stuck in early access while its development has been halted.) $\endgroup$
    – Ilmari Karonen
    Commented Sep 18 at 15:24

2 Answers 2

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Your instinct that the four quarters of the orbit would be symmetrical isn't correct, because the spacecraft doesn't start each quarter the same way.

Here's the start of the scenario (The spacecraft is travelling counter-clockwise from this POV): A green planet, an orange spacecraft, and a magenta circular orbit line.

Then, the spacecraft thrusts prograde, changing the orbit into an elliptical one: The spacecraft now has a highly elliptical orbit.

As you can see, when the spacecraft has traveled 90° forward in its orbit, its in a very different state than it was when the scenario started. It's further away from the planet, moving more slowly, and moving away from the planet at an angle. A diagram of where the space ship would be after 90°, compared to where it would be if it hadn't thrusted.

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    $\begingroup$ Nice diagram and I chafe at nothing, but in my thought experiment I'm doing my burn ... Awwww, damn ... yes I get it. Perigee to apogee is 180deg not 90deg. The way I visualized it, I'd be doing my burn at the grey point but still getting to the same apogee. Yup, it's clear now. Many thanks. So .... this means that in a real world situation you want to do your burn 180deg away from the intended apogee -- good to know just in case I'm ever driving a spaceship. $\endgroup$
    – Ray Andrews
    Commented Sep 18 at 14:05
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Let's consider the elliptical orbit (following the initial burn). The foci of the orbit (including Earth) lie on a line connecting the perigee and apogee (the line of apsides). This is indeed a line of symmetry for the ellipse. However, that does not mean that the Earth is at the center of the ellipse. See the image attached (from http://dx.doi.org/10.1109/MILCOM.2008.4753110).

Elliptical orbit diagram

The source of confusion may be indicated by the incorrect statement in the question that the apogee to perigee is a quarter of the orbit. This is not the case, that is always going to be half the orbit (both in terms of distance travelled and time).

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  • $\begingroup$ Nice diagram, although its use of mean anomaly is a little puzzling, and possibly misleading. $\endgroup$
    – PM 2Ring
    Commented Sep 17 at 13:26
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    $\begingroup$ Good catch @PM2Ring, that should be true anomaly! $\endgroup$
    – FTT
    Commented Sep 17 at 13:48
  • $\begingroup$ Ok, that's something to chew on. When I say that the apogee to perigee is a quarter of the orbit, that's only because my sense of symmetry tells me that the whole orbit will resolve into four identical mirrored quarters. I'm wondering at what point my 'mirroring' goes wrong. My first quarter starts with a tangental burn and it ends with apogee at (I think) a tangental point, so it's 90 degrees of rotation. From that point, I'm supposing that the next quarters all have the same shape. $\endgroup$
    – Ray Andrews
    Commented Sep 17 at 13:53
  • $\begingroup$ Hmmm ... Ok, seems the 'first quarter' ... don't quite get it yet, but maybe I can't accelerate at exactly a tangental direction from my circular orbit. If I want to become the ellipse, my burn must vector slightly inward from the circular path. The first quarter is more than 90 degrees of rotation. Ok, that's it. Or I'm on the scent anyway. $\endgroup$
    – Ray Andrews
    Commented Sep 17 at 14:04
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    $\begingroup$ @RayAndrews FWIW, the diagram in this answer also seems to be "not to scale" (or, to use a less polite term, incorrect). In particular, for a real elliptical orbit, the perigee is always (as the name indicates) the point along the orbit that's closest to the Earth. Here the Earth seems to have been drawn too far to the left, and thus too far from the perigee point. The diagrams in HiddenWindshield's answer are more accurate in this respect. $\endgroup$
    – Ilmari Karonen
    Commented Sep 18 at 6:47

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