Heating during spiraling would depend on the density too.

Let's estimate it. Dissipating power is N = Fv = v3DS, so power per Earth surface area is N/S=v3D = 3.8x10^9 watt/m2


In the first model (dense atmosphere)Earth surface would receive about half that flux (another half will be radiated in space), or about 2x10^9 watt/m2 which is about million times more that what we receive from Sun per square meter with a daylight.

In the second more realistic model, it is 10-100 times less, depending strongly on the actual density of Sun's atmosphere.

It seems to be extremely brignt and hot shock wave, but just how quickly it may evaporize Earth - I do not know at this point. The think is that this radiation (with maximum likely in near UV range) is strongly absorbed by air, and little gets through. If the air will be gone (as being hot beyoud escape velocity), then Earth would hit solar hydrogen atmosphere directly, which will result in heating of the surface to about 5300 K in the case of 1/10 density. You can get this number equating incoming energy to black-body heat dissipation: sigmaT4/sigmaTcurrent4=105.

Now, how quickly will Earth evaporize if its surface is constantly heated to 5300 K? Again, I do not know now, (but probably can estimate too).

• What about the forces? - doctordick - March 19, 2001 - 22:39 UTC
  • yes, pressure is Dv^2 - alexander - March 19, 2001 - 22:57 UTC
    • duh -yep-yep-yep - doctordick - March 20, 2001 - 01:58 UTC