Evidently you mean via the energy of insolation and some assumed physics supporting e = mc^2? Not much, for two reasons. 1) The earth is in rough thermodynamic balance. The incoming energy from the sun is balanced by the outgoing energy of the earth's infrared radiation. 2) e = mc^2 doesn't just happen. You have to have a specific event, such as a photon producing an electron and a positron. But in order to do that, the photon has to have enough energy to cover the masses of the two particles, 1022 electron volts. And the only photons with that kind of energy are up in the gamma ray range. Very few solar gamma rays (there are a few), make it down to the surface of the earth.
Particles from the recent solar flare would have made it to Earth... but their mass would have been absolutely minute.
It would contribute thermal energy, though. While the mass of any individual particle doesn't increase, the mass of the system does, doesn't it?
Solar wind or not, our atmosphere does "evaporate," which is why it doesn't have much in the way of lighter elements - like hydrogen.
Correction Hi selfAdjoint, I think you mean for "enough energy", 1.02 x 10^6 electron volts. Also, in the context of this string, isn't the absorption of gamma rays by the stratospheric environmental gases manifest as the ionization of those gases that end up rising into the ionosphere? Any Earthbound high voltage (>1.02 MeV) accelerator is capable of producing copious "pair-production" events. Cheers, Jim
According to my geology books, the atmosphere is not "evaporating." That is except for hydrogen. Helium is to big to escape the earth.
Another factor to take into account when computing how much mass and radiation the Earth gets from the Sun is the extremely small solid angle we cover. Sun's wind and radiation escapes in all directions; Earth's radius is roughly 1/100th of the Sun's, and is located at about 100 Sun diameters from it. Edit: Using Earth's orbit as unit of length (say, u), Sun's radiation is distributed on a spherical surface with an area 4[pi] u2, while Earth "catches" only an area of [pi]r2, with r ~ (10-4u)/2: Earth's receiving area = [pi](0.5*10-4u)2 = [pi]0.25*10-8u2 So, our fraction of the total is: f = (Earth's area)/(Area of the sphere) ~ 0.25*10-8/4 = 6.25*10-10 Now that I think of it, it is impressive the ammount of power the Sun is putting out, given the fact that we get warm at all!