Total power entering earth's atmoshpere

  • #1
Assuming the EarthÕs radius is 6378 km, the solar constant is 1370 W/m2 and our planetary albedo is 0.31 then

a) Determine the total power (from solar energy) entering the EarthÕs atmosphere

b) What power would be absorbed by the EarthÕs surface in the absence of an atmosphere?

c) What would the surface temperature be assuming it radiated into space all the energy it absorbed?

Anyone know what to do here? I havn't a clue what equations to use and having trouble finding them...
 

Answers and Replies

  • #2
My attempt is
a. Total power entering atmosphere = (4pier^2)/2(solar constant) = 350 X 10^9 watts

b. 31% would be absorbed my the earth if there was no atmosphere, 108.5 watts absorbed.

c. The temperature of Earth if 108.5 watts absorbed and rest is sent back to space would be i have no idea even if that number is right...
 
  • #3
Dick
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Well, you've got power per area. How much area does the earth present to the sun? What does albedo do to the problem?
 
  • #4
Well thats why i divided by 2, and albedo is the pecentage of energy the earth absorbs...
 
  • #5
Dick
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Good in principle, but you want a cross-sectional area, like a disk. And you are orders of magnitude off. Did you convert km to m?
 
  • #6
no.. am I suppose, ha yah i am! ok so make it X 10^12, and should I be dividing the 108.5 ten to the twelve by have the area of the earth to get, 424.5 watts/ meter squared?
 
  • #8
no wait thats wrong still, should be 3.5 times ten to the 17th. then take 31 % of that...
 
  • #9
so this is what i have now
a. Total power entering atmosphere = (4pier^2)/2(solar constant) = 3.5 X 10^17 watts

b. 31% would be absorbed my the earth if there was no atmosphere, 1.09 X 10^17 watts absorbed.

c. The temperature of Earth if 1.09 X 10^17 absorbed and rest is sent back to space would be 1.09 X 10^17/ 2pier^2 = 424.7 watts/m^2 on the Earth, giving a temp of 20 degrees Celsius?
still just guessing on the temp tho...
 
  • #10
Dick
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I want pi*r^2 for the area of the earth. Sunlight doesn't strike the polar regions directly. And albedo is the percentage reflected, not absorbed. For temperature check out the Stefan-Boltzmann law.
 
  • #11
Ok, so pi r^2 for all equations, so working more in 2d then.. thanks alot!
 
  • #12
Dick
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No, pi*r^2 just for power received. 4*pi*r^2 for the area that radiates heat back!!!
 
  • #13
a. Total power entering atmosphere = (pi r^2)(solar constant) = 1.75 X 10^17 watts

b. 31% would be absorbed my the earth if there was no atmosphere, 5.4 X 10^16 watts absorbed.

c. Earth absorbs 5.4 X 10^16/ 4pi r^2 = 105.5 watts/m^2 = (boltzmann's constant)T^4, Thus T = this will be in Kelvins... lower than 1???
 
  • #14
may bad, using internet calculators.. so I got T = 207.7 K
 
  • #15
Dick
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Once more - albedo is the percentage reflected - not absorbed.
 
  • #16
ok thanks got it now!
 

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