# Back to oribt from Mars - how many km^2 yr of solar power to make enough propellant?

In this cool video of a presentation about SpaceX's GPU-based computational fluid dynamics, there is a slide early on about making rocket fuel on Mars - specifically using water from the ground and carbon dioxide from the atmosphere to make oxygen and methane:

2H2O + CO2 -> CH4 + 2O2


Going in that direction requires energy, and I'm guessing solar power in some form or other would certainly be one way.

There may be multiple ways to use solar power to drive endothermic chemical reactions. There is photocatalysis, concentrated solar thermal catalysis, photovoltaic for thermal catalaysis and/or electrolysis just for example. "What is the best method..." would be a great question, but it's not this question.

This question: Ballpark figures: if seven people wanted to get in some capsule (could be the same Dragon that got them there or not) now sitting on a methane/LOX burning booster, blast off and get back to Mars orbit to rendezvous with a waiting vehicle to go home, and for some reason they had to make the propellants from water and carbon dioxide, roughly how many square kilometer years of solar power on Mars would be necessary to synthesize those propellants?

Assume food and water and vehicle for the return is waiting for them in orbit, but they have to make the propellants to get to orbit on Mars as shown in the screen capture.

• This question and al of the answer may be helpful here. – uhoh Jul 28 '16 at 9:38

Don't have exact figures but let's estimate it a little...

Mars orbital velocity is roughly 3.3 km/s. Your dragon capsule has a mass of about 8 tons. So your entire rocket would be something like 50 tons to reach Mars orbit. The methane/oxygen needed in that is roughly 36 tons. Of that, about 12 tons is methane. To generate 1 kg of methane we need 50 MJ of energy.

Mars distance to the sun is roughly 1.5 AU on average. Thus the solar power generated per m² is about 44% of that of earth.

Near the equator you have roughly 600 W/m² of solar radiation at midday. Due to planetary rotation you can use half the day to generate power and it will increase from sunrise until midday and then decrease until sunset - you can increase that amount by spacing and rotating the panels though, so let's assume we end up with effective 150 W/m².

If we use really efficient solar panels, we can probably generate up to 50 W/m². You can probably convert 60% of that energy to fuel.

so 12,000 * 50 MJ / 60% ~ 1TJ. With 50 W/m² that would mean about 2 * 10^10 m² * s or 634 m² * year.

This does of course not account for losses storing or compressing the fuel yet... with all the assumptions, you might end up with 10-50 times that number though.

• OK that's exactly what I was hoping for - lucid ballparking. Do you think then that it can or can't be excluded - that at least from this calculation, it's not immediately impossible. I'm trying not to ask "can solar work" - only "can solar be immediately excluded?" So much for being lucid on my part. – uhoh Jul 28 '16 at 15:14
• I'd say solar can't be immediately excluded, but on a mass basis, nuclear power would be more practical and possibly more reliable. – Russell Borogove Jul 28 '16 at 15:20

From Wikipedia, I learned that 1 kg of methane was produced using 700 W of power in a day.

The Case for Mars estimates 82 tonnes of Methane/ LOX required for the ERV, which assumes launching directly to Earth. Of this, about 22 tons (20 metric tons) of it is methane. This is for a return directly to Earth vehicle, not just to Mars orbit.

The overall power requirement can then be computed. 20000/365=54.8 kg of fuel required per day. That means 38.3 kW of power on average per day. Given the same efficiency that Adwaenyth listed, that's about 767 m^2 of solar panels to produce the fuel in a year. Of course, a real mission would have 2 years to produce the fuel, at least, more likely 3 years. It would also require a certain amount of cooling.

The Mars Direct plan requires sending a small nuclear reactor, on the order of 80 kW, to supply this power. It also suggest some solar power for backups and remote sites. At the estimated 1 ton/ 5 kW of solar panels, that does make sense.

• Actually I want to ask what that reactor be like (e.g. weight, "remote control") but that's a different question so I won't here. Is it based on steam and a small turbine or Stirling engine? Porta-reactors have been though about for a very long time, I'm sure it would work fine, but I only know (a little bit) about the giant kind. – uhoh Jul 28 '16 at 15:21
• All I know is an estimate of 4 tons of weight. Beyond that... – PearsonArtPhoto Jul 28 '16 at 15:32
• OK thanks. At 20W/kg 24/7 it might be hard for solar to compete! – uhoh Jul 28 '16 at 15:46
• The thought I'd that for a settlement, solar made from Mars will be better, but until then, nuclear is the better choice. – PearsonArtPhoto Jul 28 '16 at 15:48
• 80kw of Nuclear power is already difficult enough, because you need that reactor cooled somehow... and it's not like you have an abundance of coolant like the water we use for the reactors here. – Adwaenyth Jul 29 '16 at 5:46