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The question Has anything ever executed an orbit change such as between ISS and Hubble? has a in interesting answer. It made me wonder, would something like the (patented) Lunar Flyby technique use less detla-v than a direct, propulsive inclination change of about 23 degrees?

edit 1: Let's say it's not a delicate communications satellite, so aerobraking is allowed for part of the circularization at the end, and there is no rush, a transfer time of maybe six months would be OK.

edit 2: Hubble to ISS is OK too. Starting at the lower of the two inclinations is probably going to be necessary if this is at all possible.

note: something like this was used to correct the orbit of PAS-22, making it the first commercial spacecraft to reach near-lunar space if I understand the Wikipedia article correctly. It says:

Using on-board propellant and lunar gravity, the orbit's apogee was gradually increased with several manoeuvres at perigee until it flew by the Moon (9) at a distance of 6,200 km from its surface in May 1998, becoming in a sense the first commercial lunar spacecraft. Another lunar fly-by was performed later that month at a distance of 34,300 km to further improve the orbital inclination.

These operations consumed most of the satellite's propellant, but still much less than it would take to remove the inclination without the Moon assist manoeuvres.

(9) http://www.spacedaily.com/news/Book_Reveals_How_Hughes_Saved_ComSat_In_1997.html

Would a maneuver involving the moon also be lest costly in delta-v than a direct plane change if transferring from the Hubble to the ISS?

See also: New Book Reveals How Engineers Saved Hughes Satellite On Christmas Day 1997

For reference, here is Figure 2 of US6116545

enter image description here

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  • $\begingroup$ nice questions feel free to advise me on mine. $\endgroup$ – Muze the good Troll. May 7 '18 at 2:31
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The trans lunar injection of the Apollo missions took a Δv of about 3000 - 3200 m/s. Let us assume all inclination change is for free (quite reasonable with careful planning and possibly multiple fly-bys) and lowering the orbit to LEO again is free as well due to the possibility of aero-braking. Compared to that, the direct inclination change of 23 degrees in LEO takes about 3100 m/s Δv. That means we need about the same Δv for both paths.

Neither Hubble nor the ISS is in the low 200km LEO of the Apollo missions but significantly above at 600 km and 400 km, respectively. This gives us an advantage of about 200 m/s in favor of the lunar fly-by.

Why was the maneuver so successful in case of PAS-22 then? It was already in a GTO with an apogee of more than 36000 km. From there it was much less than 1000 m/s of Δv to get to the moon.

One thing I did not take into account: I don't know if it is actually possible to do the required orbital change using the moon due to orbital geometry.

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  • $\begingroup$ Thanks for taking a look at this! Would you say then that it's close, and can't be ruled out right away, and deserves a closer look? Without perishable astronauts on board, this doesn't have to be done quickly, and so a delta-v well below 3000 m/s can still get the satellite close enough to the moon so that it's gravity can be used in various ways. The Apollo missions didn't have this luxury. $\endgroup$ – uhoh Jul 3 '16 at 12:19
  • $\begingroup$ Apollo didn't waste much energy for a faster travel - 3000 m/s is what you need to raise your apogee from 200km to 380.000km. You need a precise calculation of how much of the needed Δv you can get from Moons gravity to tell if it's feasible or not. $\endgroup$ – asdfex Jul 3 '16 at 13:13
  • $\begingroup$ So are you answering "yes" or "no" here? "Would lunar flyby be less costly in delta-v than direct change from ISS to Hubble orbits?" $\endgroup$ – uhoh Jul 3 '16 at 13:23
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    $\begingroup$ I'm not sure. Most likely not. In any case, the difference will be small for this particular case. For higher inclination changes definitely yes, for lower changes definitely no. $\endgroup$ – asdfex Jul 3 '16 at 13:45
  • $\begingroup$ OK great - I appreciate a second opinion here. I also felt that it is intriguingly close. $\endgroup$ – uhoh Jul 3 '16 at 14:03

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