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Say you have two spacecraft in identical orbits. Say at a point after apoapsis and before periapsis one of the spacecraft burns retrograde to lower its periapsis, and the other spacecraft does nothing and stays on the original orbit.

Which of the two craft would reach their respective periapsis first?

The craft that burned has less distance to travel than the one that didn't. But then the one that burned reduced that distance by accelerating away from the periapsis. Does one of these always outweigh the other, or does it depend?

DTCantReachSpace
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    This could be answered by the Kepler 2nd law. The one with lower orbit has less area to sweep before its periapsis, hence less time. – user3528438 Sep 12 '23 at 04:18
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    @user3528438 I'm not convinced, perhaps just because I can't really picture it. Do you think you could post that answer as an answer, showing how the areas swept are established from the question? Thanks! – uhoh Sep 12 '23 at 06:29
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    "It depends" -- the closer to periapsis when the burn is made, the more periapsis is moved away from the craft. The effect is especially pronounced on an almost circular orbit. – JCRM Sep 12 '23 at 12:56
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    I think this question cannot be answered unless you specify the phase angle (between apo- and periapsis) the burn occurs. If the burn occurs at a small phase angle, the "burner" wins. If the burn occurs close to periapsis, the non-burner wins. Do you want the phase angle at which it is a tie? – Woody Sep 13 '23 at 19:32
  • @Woody "I think this question cannot be answered unless..." and yet it has just been quite nicely answered. However there's still no expression for "the phase angle at which it is a tie" and I think that would be great! Since $t(\theta)$ is analytic (even though $\theta(t)$ isn't) it seems pretty straightforward to work out. – uhoh Sep 14 '23 at 01:19
  • Here's an example of a tie. https://i.stack.imgur.com/kxWXp.png The eccentricity of the initial orbit is 1/3, the retrograde burn occurs at 3/4 of the orbit period, reducing the current speed by a factor of 0.8391384. I used a 4th order Leapfrog integrator. The burner wins by a tiny fraction at that step size (~1/3475 of the period). – PM 2Ring Sep 16 '23 at 19:13

1 Answers1

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It depends on where you are in your orbit.

Specifically, it depends on where you are in the orbit relative to your periapsis when you start the burn.

Visualizing orbits is difficult at first, but a rule of thumb is that prograde or retrograde thrust raises or lowers the exact opposite point on the orbit. If you fire your thrusters retrograde, the point you're lowering the most will be the point directly across the center of the orbit from your current location.

The easy case is if you're at or nearly at your apoapsis -- at the furthest point in your orbit from the periapsis. In that case, the periapsis doesn't change, but just gets lower, you shorten your orbital period, and consequently you'll reach the periapsis faster than the ship that didn't decelerate. (For example, if the original period was 90 minutes and you're at the apoapsis, then your partner ship will take 45 minutes to get to periapsis. If you fire long enough to bring your period down to 80 minutes, you'll reach the far end in 40 and beat them by a whole five minutes.)

But if you're anywhere else in the orbit, decelerating will lower the point precisely opposite you more than any other point on the orbit, so given enough thrust, that point will become the new periapsis, leaving you nearly 180 degrees away from it. Depending on where you are in the orbit, this might mean you get there faster or slower.

For instance, assume your orbit is around 90 minutes long. Suppose you're just two minutes from periapsis when you fire: in that case, the point you're lowering will be the opposite side of the orbit, the point which was (almost) your old apoapsis. If you lower it enough, it will become your new periapsis and you won't reach it for another half-orbit, while your partner will reach their unchanged periapsis in just a couple minutes.

By contrast, if you passed periapsis twenty minutes ago, then firing retrograde will pull the periapsis towards you as it moves closer to directly across the orbit from your current point.

Now, if you consider whether you'll reach the radial where the old periapsis was before (i.e. whether a ground observer looking straight up as the "finish line" would say you passed overhead faster than your partner), then the answer is usually that you'll get there first. Dropping the orbital altitude shortens your path, so you orbit faster even though you're starting out at a lower speed. However, I believe in the aforementioned scenario where you're nearly at the periapsis already when you fire, slowing your orbital speed means the other ship gets there first (and then you'll catch up and get ahead once you start down your newly eccentric orbit and get the benefit of the shorter distance). Over a a short enough distance, the curvature of the orbits is negligible, so it would function more like earthly physics and slowing down means you get there slower.

(If you're interested, I can recommend Kerbal Space Program as a fantastic way to get a handle on the physics of movement in orbit. While it's tuned more for fun than realism, the orbital physics is good enough, and you can set up scenarios like this to try them out for yourself.)

Darth Pseudonym
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  • Excellent answer, but it would be more correct to use "delta-v" rather than "thrust" in paragraph 4. – Russell Borogove Sep 13 '23 at 22:47
  • "Thrust" is fine in para 4. Sure,""impulse" is the correct term,but the lay-person's usage of "enough thrust" will encompass duration. Personally I'd avoid delta-v – JCRM Sep 15 '23 at 08:57
  • I considered phrasing that differently, but I decided in the end that being more clear to a lay-person would be better than being more technically correct, at least in this context. – Darth Pseudonym Sep 15 '23 at 13:28