Suppose you wanted to achieve a polar orbit around the Moon, starting from LEO. Would it be better to start from a polar orbit around the Earth? "Better" meaning using less fuel, having more frequent launch windows, etc.

  • 1
    $\begingroup$ The are no launch windows when going to the moon as they would be for Mars. Being on an orbit around Earth, some launch times might be a bit more convenient than others, but that's about it. $\endgroup$ Feb 26, 2018 at 8:10
  • 4
    $\begingroup$ In an equatorial orbit your optimal transfer window happens once per orbit. A polar orbit will have to wait for the moon to line up with its own orbit, so transfer windows would only happen twice a month. $\endgroup$
    – Vincent B
    Feb 26, 2018 at 9:49
  • $\begingroup$ Whether you save's fuel will be a balance of how much fuel you use adjusting orbital inclination once you reach the moon against how much fuel you save by launching with Earth's spin rather than perpendicular to it $\endgroup$
    – bendl
    Feb 26, 2018 at 17:17
  • $\begingroup$ @vincentb, could you elaborate on that? In not sure I understand. $\endgroup$ Feb 27, 2018 at 10:53

1 Answer 1


If you are already in low earth orbit the fuel gains are negligible when you start from a polar orbit. The possible transfer windows however are greatly reduced, to two per month (one moon orbit), compared to every 90 minutes. (one spacecraft orbit)

Fuel efficiency


One might assume that this is where the big gain could be made by starting in a polar orbit. But that doesn't have as big of an impact as one might expect.

Because LEO and the orbit of the Moon are very far apart, <2,000km and 385,000km respectively, a small change in the inclination of the starting orbit will translate to a big change at the Moon. This means you don't need to be in a polar orbit around Earth to arrive in a polar orbit at the Moon.

I'm not going to do the math on that, but I'd wager the difference in required delta-v when you start in polar orbit or an orbit that's on the same plane as the Moon is less than 1%. (The Moon orbits at an inclination of 5.15°)


While this was not part of the question I do think this is important to mention. Launching into a polar orbit takes considerably more fuel than an equatorial orbit. This is because when you launch into an equatorial orbit you launch into the same direction as the rotation of the Earth, effectively getting a small boost in your velocity.

When you launch into a polar orbit you don't get this benefit, even the opposite. You are already rotating with the earth, so you need to cancel this velocity to get into a polar orbit.

Transfer windows

To transfer to the moon you need to time your transfer burn just right. A bit too early in your orbit and you arrive behind the Moon, a bit too late and you'll be too far ahead.

trans-lunar injection https://en.wikipedia.org/wiki/Trans-lunar_injection

In an orbit that is on the same plane as the Moon, the optimal transfer window would happen once per orbit of the spacecraft.

In a polar orbit you can't choose to launch behind or ahead of the Moon by starting the transfer earlier or later in your orbit, because your orbital plane is perpendicular to the Moon. This means transfer windows would only happen twice a month. Once on the "upswing" of your orbit and once on the "downswing".

  • $\begingroup$ +1 I think this is a great answer! Being a non-KSP-enlightened individual, I found your explanation and image particularly helpful to picture the situation in an Aha! sort of way, thanks. $\endgroup$
    – uhoh
    Feb 28, 2018 at 6:42
  • 1
    $\begingroup$ I think it's worth mentioning that launching into a polar orbit takes considerably more fuel than an equatorial orbit only if you have free choice of launch sites and launch from the equator. It is prohibitively expensive to get into an equatorial-ish orbit from a higher latitude launch site, so rockets launched from say Russia are stuck with "polar-ish" orbits. This is why a polar transfer might be used anyway. $\endgroup$ Feb 28, 2018 at 10:20

Your Answer

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

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