## [SOLVED] Amount of Energy emitted from the sun to the earth

1. The problem statement, all variables and given/known data

The Sun's surface temperature is 6000 K. Assume that the radiation emitted from the Sun moves radially outward from the Sun and that no radiation is absorbed between the Sun and Earth. How much energy in the form of radiation will fall per second on an area of 1 m2 on Earth, if that area is perpendicular to the straight-line path of the radiation? The radius of the Sun is 6.95×108 m and the distance from the Sun to Earth is 1.5 ×1011 m.

2. Relevant equations

I tried to use the dQ/dt= -k*A*1/L*deltaT, but I was unable to find a constant to use for k. After going through my notes, I couldn't find any other equation that looked relevant.

3. The attempt at a solution

I assumed that the change in temperature would be minimal, since the temperature of the sun is much greater than that of the earth -so I left the temperature at 6000K. I took the radius of the sun to find the area. And, plugged the rest of the numbers in accordingly.

dQ/dT= [-k((6.95E8)^2*pi) * 6000K]/1.5E11

However, I believe that this equation is not suited for this problem and more suited for heat transfer from one end of a rod to another (or something like that).
I don't know where to go from here- or since I don't think this is the correct equation, I don't think I know where to start in the beginning. I would appreciate any help in getting this problem started. Thank-you

-Meredith

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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 You could try using the Stefan-Boltzmann law: $$L=4{\pi}R^2{\sigma}T^4$$ Where L is luminosity, R is the radius of the stellar object, T is the temperature, and sigma is the Stefan-Boltzmann constant. (http://en.wikipedia.org/wiki/Stefan%...zmann_constant) and then the inverse square law http://en.wikipedia.org/wiki/Inverse_square_law
 I used the L=4*pi*R^2 (sigma)*(T^4) to find the amount of radiation the entire body is emitting and I got 4.46E26. I then said that L/SA=1m^2/X and solved for X; where SA is equal to 4*pi*(R^2+Distance): 4.46E26/(4*pi*(R+ 1.5E11)^2 I got that X equals 1.563E3 W, but I am still unsure of whether I did the problem correctly. Thank you!