I read this question about feasibility of capturing hydrogen from the Solar wind a while ago. Hydrogen in solar wind was exploited in that discussion. Here, does cosmic ray often close in towards a planet (Venus, Mars) in predictable orientation? If so, can the energy and atoms in the cosmic ray be at least partially captured and exploited?

Thank @Elmore and @PcMan for their comments. Lightning occurs when positively charged clouds connect with the negatively charge ground. If theirs stand, can an artificial atmosphere with dense clouds and frequent lightning absorb or adsorb the positively charged atomic nuclei? The electrons of the nuclei will still arrive at the planet to make overall charge neutral; meanwhile lightning brings foreign electrons to those bare atomic nuclei. Per Elmore, I would think a planet as a whole can absorb and transform the electric energy by a medium sized nuclear reactor outputs through lightning into thermal energy: raising the atmospheric temperature. If cosmic rays consist mostly of protons and electrons, hydrogen-1s after losing their energy can chemically make water vapor with oxygen in the atmosphere.

  • $\begingroup$ Solar panels on solar sails? $\endgroup$
    – nzaman
    May 14, 2021 at 4:39
  • 4
    $\begingroup$ There is very very little energy in cosmic rays, incomparably less energy than in solar light. Those high energy atoms / ions / photons are "high energy" for an atom or a photon; but they are sufficiently rare that they don't add up to much. $\endgroup$
    – AlexP
    May 14, 2021 at 6:49

2 Answers 2


The energy content of cosmic rays really is tiny. Quick glance at the numbers tells that the frequency of particles by energy drops faster than the energy increases. So 100 times more energetic particles are more than 100 times rarer. Thus the lower energy spectrum carries ~all of the total energy.

In the low end, 1GeV particles arrive at a rate of 10^4 particles per square meter per second. This would give a power output of 1.6 microwatts (µW), which is about a billion times less than the solar radiation gives as light at Earth's distance. (Solar output is about 1300W per square meter.)

Total (kinetic) power of cosmic rays hitting an Earth sized target comes out as roughly a gigawatt, which is what a medium sized nuclear reactor outputs as electricity today.

  • $\begingroup$ Thank you for the comment. $\endgroup$
    – Kav
    May 14, 2021 at 17:26

Cosmic rays come from all directions, they possess an enormous spread of possible energies
(from under 1 ev up to 10^20 ev, although we only really call those from 10^9 ev upwards by this name)..

So any device to harvest them will need to be quite versatile, able to handle energy levels from the subultraminimicroscopic, up to some serious stuff.

And they are very few and far between. The most common ones are only about 10000 per second, per square meter. At about 1Gev each, that's a total energy flux of 1/650000th of a watt per square meter.

A cosmic-ray-panel strong enough to power a Tesla charge station will be the size of Vermont.

If your purpose is to gather mass, not energy, then it is even worse.
You will collect the mass of one grain of pollen every 4.77years, per square meter of collector surface. (pollen taken as 2.5e-9 grams)

  • $\begingroup$ So instead of a collector, what does the targeted celestial body have? $\endgroup$
    – Kav
    May 16, 2021 at 4:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .