# Homework Help: Black Body Radiation

1. Jan 22, 2009

### __JR__

1. The problem statement, all variables and given/known data
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 distance from the Sun to Earth is 1.5e11 m. Assume the total power output of the sun is 4.47 x 1026 W.

2. Relevant equations
Stefan Boltzmann forumla - E(T) = $$\sigma$$T4

3. The attempt at a solution

Where I'm struggling is with the role the distance between the two bodies plays. Since the radiation is emitted radially outward, the distance between the two is relatively large, and the area on earth is relatively small, does this mean I need to observe an infinitesimally small "window" on the surface of the sun whose radiated waves will strike this area on earth? Or, since the 1m2 on the curved surface of the earth is relatively small I can assume that it, and a corresponding surface on the sun, is flat?

The power output of the sun can be determined by a surface temp of 6000K and a radius of 6.95x108 m (we solved this in an earlier problem).

2. Jan 22, 2009

### G01

At the distance of the earth from the sun, the light is impinging on a spherical surface centered at the sun with the radius equal to the earth's orbit. This surface area of this sphere is the area that the power is "spread" over.

Can you use this information to find the power per unit area at the earth's distance from the sun? If so, the answer should then follow trivially.

3. Jan 22, 2009

Imagine the radiation leaving the sun as the surface of an expanding sphere, calculate the surface area of this sphere when it reaches the earth's surface (i.e a sphere with a radius equal to the distance between the sun and the earth). By considering what proportion of the total surface area 1m^2 is, you can work out how much of the total power strikes that area.

4. Jan 22, 2009

### __JR__

Thanks for the replies, I think I've got the right answer.

So the total power output of the sun is 4.47e26 W, and the SA of the new sphere is 4*pi*(1.5e11)^2 m^2 = 2.82743e23 m^2.

Dividing gives 1580.94 w*m^(-2). Sound right?

5. Jan 22, 2009

looks good to me!

6. Jan 22, 2009

### G01

Looks good to me as well.

(BTW, that was perfect timing with our previous posts, Prosthetic Head!)

7. Jan 22, 2009

### __JR__

Thanks again!