In most diagrams of the Apollo Lunar Module descent profile (example at the start of this video), it looks like the LM turns over and points some of its thrust downward relatively early in the process.

I always wondered exactly how much delta-V they expend downward (how long at what angles and throttle levels), and what that meant for their gravity losses. I'm sure a fuel-ideal trajectory wouldn't have been practical for many reasons (hard to do a suicide burn when you're not sure if it ends on top of a boulder). But I'm just wondering how much fuel they sacrificed for those considerations?

  • $\begingroup$ Related: space.stackexchange.com/q/2493/26446 $\endgroup$ – DrSheldon Nov 18 '18 at 9:00
  • $\begingroup$ @DrSheldon Thanks for the pointer. Good info on the difference between the fuel they carried and the minimum required for their planned trajectory. I'm wondering about the difference between their planned trajectory and an ideal one with no gravity losses. $\endgroup$ – Nick S Nov 19 '18 at 0:50
  • $\begingroup$ @NickS Ideal trajectory with no gravity losses meaning: assuming the ship can do instantaneous burns of any magnitude? Like, do a small burn to drop perilune to 0 altitude then cancel the entire surface-relative velocity all at once when you get there? $\endgroup$ – Russell Borogove Nov 19 '18 at 18:18

The limiting case assumes a perfectly spherical moon and a lander that can do instantaneous burns of any magnitude.

In this case, starting from a 110km circular orbit, the lander does a brief burn to drop periapsis to 0 altitude, then burns to cancel its entire surface-relative velocity all at once when it gets there.

The initial burn is about 25 m/s; the terminal burn is about 1711 m/s (26 m/s to circularize; 1680 to cancel circular orbit velocity, 5 more to match the moon's rotation speed since the Apollo flight plan approaches the moon in retrograde orbit), for a total of 1736 m/s descent ∆v.

The nominal full-automatic descent profile for Apollo was 6827 feet per second, or 2081 m/s plus another 145 fps (44 m/s) budgeted for manual approach and hover time: 20%-22% more than the theoretical limit.

Assuming a fully loaded mass at undocking of 15200 kg, and a 311 s specific impulse for the descent propulsion system, the instant-burn solution requires 6596 kg of fuel; the Apollo profile (including approach and hover time, but not other contingencies) requires 7627 kg -- about 16% difference in fuel tankage, due to the nonlinearity in the rocket equation.

  • $\begingroup$ Oh, I assumed the optimal plan would start with a long burn to cancel orbital velocity, then a suicide burn once approaching the surface to bring vertical motion to 0 right at 0 altitude. I'm curious why that's not best? Also, is it possible to translate the delta v into actual fuel quantities? $\endgroup$ – Nick S Nov 19 '18 at 19:55
  • 2
    $\begingroup$ That's 1634 m/s to come to a dead stop at 110km, then falling for 368 seconds, in which time you gain another 597 m/s that you have to cancel, for a total of 2231 m/s. In my proposed elliptical-descent approach, gravity is working mostly perpendicular to the velocity we're trying to shed, mainly changing the direction of the velocity vector rather than simply increasing it. $\endgroup$ – Russell Borogove Nov 19 '18 at 21:08
  • $\begingroup$ ...in fact, after the 25m/s burn, the periapsis will be exactly half an orbit away on the opposite side of the moon. It takes about an hour to get there, and gravity pulls your velocity vector around from 1629 m/s in one direction to 1680 m/s in the opposite direction! Your proposal is certainly faster, though. $\endgroup$ – Russell Borogove Nov 20 '18 at 22:48

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.