5
$\begingroup$

Could it be possible ( using a reasonable amount of energy and ressources ) to capture protons from solar wind on (ex. the moon) in order to produce water for a long term self-sufficient mission. The amount of water should compensate for system losses.

You are free to make assumptions and improve my question as english is not my native language

EDIT: Assuming a proton flux of 2e12 m^-2 s as provided here (https://astronomy.stackexchange.com/questions/18829/composition-and-proton-flux-from-the-solar-wind) we would need to capture an area of 10^6 m^2 in order to get 0.11kg of protons per year ( the mass required to make 1 liter of water )

$\endgroup$
  • 3
    $\begingroup$ To make hydrogen of protons you need electrons too. To make water of hydrogen you need oxygen too. $\endgroup$ – Uwe Mar 3 at 20:00
  • 1
    $\begingroup$ @Uwe there’s plenty of oxygen on the moon, but what could be a source of electrons? Could we get it from solar wind too ? $\endgroup$ – Simon Talbot Mar 3 at 22:00
1
$\begingroup$

Protons are positively charged so it would be quite easy to capture them using a magnetic field. With that said, the density of solar wind is incredibly tenuous by the time it reaches the moon so you'd have trouble accumulating significant mass of hydrogen over a reasonable amount of time. Additionally, you'd still need to liberate oxygen from whatever oxides it is currently residing in on the moon which would be energy intensive.

Finally, there is already an abundance of frozen water in permanently shadowed craters on the moon. In 2009, the Moon Mineralogy Mapper discovered significant evidence for lunar water which was re-confirmed by the LCROSS probe later that year.

$\endgroup$
1
$\begingroup$

There may be a simple way of capturing the protons from the solar wind. It would involve using an electric field to decelerate the protons in order to capture them.

Please see Could protons in the Sun's solar wind be used to create a photonic laser thruster for a spacecraft?

$\endgroup$

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.