Imagine all the extraterrestrial matter that earth absorbs over a given time frame (multiple of years, to account for the orbital path). If all of that is converted (perfectly) to energy at E=mc^2, is that figure greater than the energy that hit earth in the form of solar radiation, during that time?

an interesting question. i don’t think it has a simple answer, and certainly an accurate answer will require research, and not the surely bs answers you’ll get.

I would make a guess(and that’s all this is) that it would be from the extraterrestrial matter. I say this because a large percentage of the solar radiation that we would get is repelled by the earths magnetic field, and some that isn’t is trapped in that field eventuality becoming northern lights when it is converted into energy, but this is only a guess on my part from the somewhat limited physics I know.

Assume that the Earth is just a circular disc in space, with a diameter of 12,750 Km. The area of that disc is 127.7 trillion (10^12) square metres. Each square metre would receive 1.6 KW of energy from solar radiation, which makes up a total power of 204.3 PW (Peta Watts, 10^15 Watts.) Over the course of a 365.25 day year, that’s 6.45 * 10^24 J.

The equivalent of 71,700 metric tons of mass per year in solar radiation. We also receive a small amount of mass and energy from the solar wind, but it is insignificant in comparison to the energy from em radiation.

OK, so what about meteorites? Well, although the solar radiation varies slightly over time, the meteorite flux varies much more wildly, and can be between 10,000 – 1,000,000 metric tons per year.

Therefore it’s a little tricky to say which is greater. It would appear that most of the time it’s the meteor strikes, but every now and then when there’s a lull, solar radiation is king again.

The part about the sun is a pretty straightforward calculation.

Look up the sun’s luminosity.

Divide by 4 pi a^2, where a is one au, the distance from the earth to the sun, to get the luminous flux (or just look that up).

Multiply by pi r^2, where r is the earth’s radius to get the luminous power incident on the earth.

Multiply by the absorbtivity of the earth to get the power aborbed.

Divide by c^2 to convert the energy rate to an equivalent mass rate.

Multiply by some arbitrary time if you want to get a mass.

You can easily look up all that and calculate. I don’t have a good idea, though, how to do the ET part unless you can look it up somewhere. It may be that the sun’s equivalent mass per time is small enough or big enough that the answer is fairly obvious. You’ll have to work it out and see.

—- I found a source that said the meteorite material ranges from 37-78,000 tons (presumably imperial tons). That should give you a basis to compare.

Two estimates of mass influx quoted in the reference are 1.7 and 2.2 X 10^8 kg/year