1. Mixture ratio for Oxygen/Methane is 3.6 to 1
  2. Liquid oxygen temperature of -183°C
  3. 16 Psyche (in shade) -113.15°C (https://simple.wikipedia.org/wiki/16_Psyche)

Looks like the idea of bringing enough Methane for the return trip, while making LOX from water locally - has some legs to it?

Question - how much of LOX will have to be produced locally? 20-30 ton will be enough for the return trip?

When Starship becomes operational, a mission to 16 Psyche is almost guaranteed to be considered. The propellant depot will be the main issue to address for this mission.


1 Answer 1


Question - how much of LOX you will have to make? 20-30 ton will do?

20-30 tons will not be enough. Add another zero.

I happen to have a delta-v estimate for a 16 Psyche return, which is 4050 m/s for a Hohmann transfer.

With the exhaust velocity of the raptor engine of 3.56km/s, that's a mass ratio of:

$\frac{m_0}{m_1} = e^{\frac{4050m/s}{3560m/s}} = 3.12$

With the caveat that Raptor operates at a mixture ratio of 3.6:1 and not 4.5:1. I guess running oxygen rich would make sense if the oxygen is "free", but it's going to impact the exhaust velocity negatively having significant amounts of unreacted oxygen in the exhaust mix. So a mass ratio of at least 3.12

I'm not sure there are better numbers for the final dry mass of Starship than "around 100 tons", so with the above mass ratio and your mixture ratio, that's around 250 tons of oxygen.

  • $\begingroup$ "SE - stop firing the good guys" - thank you very much! As always, extremely useful answer. This is what I have suspected, but you have confirmed with numbers. $\endgroup$
    – anon
    Commented Feb 3, 2023 at 17:46
  • $\begingroup$ "SE - stop firing the good guys" - do you think 4050 m/s for a Hohmann (final burn from 16 Psyche) - is a final verdict? Possible gravity assist from Mars would not reduce this number? $\endgroup$
    – anon
    Commented Feb 3, 2023 at 17:57
  • 1
    $\begingroup$ @TheMatrixEquation-balance The linked answer has a Jupiter flyby at 2400m/s. From a back of the envelope calculation, a Mars flyby would be about the same. $\endgroup$ Commented Feb 3, 2023 at 18:25

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