# Homework Help: Energy in EM Waves problem (finding sun power output from EM wave intensity)

1. Mar 26, 2006

### confusedbyphysics

Hi, if someone could help me figure this out I'd really appreciate it.

Estimate the average power output of the Sun, given that about 1350 W/m^2 reaches the upper atmosphere of the Earth.

I know the answer is 3.8 X 10^26 W but I don't know how to get there. I've tried different equations and putting random numbers together but it's been waaayyy off. Can someone help point me in the right direction?? THANKS!!

2. Mar 26, 2006

### nrqed

The Sun emits uniformly....so the energy emitted per second gets spread over the surface of an expanding sphere. Does that give you an idea?

(do you know how the intensity of light, in W/m^2 is related to the power of a source of light?)

Patrick

3. Mar 26, 2006

### Staff: Mentor

Assume that the sun radiates its energy into space uniformly in all directions. Now imagine a hollow sphere the size of the Earth's orbit, with the sun at the center. 1350 W of solar power strikes each m^2 of the inside surface of that sphere. Can you take it from there?

4. Mar 26, 2006

### confusedbyphysics

No, I dont...This book is confusing, there isnt anything about power in the chapter in relation to the intensity, thats why I was looking through old chapters to see if there is a formula or something that could convert intensity into power outage. Just looking at the units, the intensity is the power of the wave over an area (m^2), right?

Do I use the surface area of the sun in the calculation somewhere?

5. Mar 26, 2006

### nrqed

There is a similar formula in sound waves. Intensity = power of the source / area over which the energy is spread (here, the energy is spread over the surface area of a sphere expanding away from the Sun). By the time the energy has reached the Earth, what is the area of the sphere over which the energy has spread out? And you won`t need the radius of the Sun (basically, the Sun can be treated as a point source here)

6. Mar 26, 2006

### confusedbyphysics

the area over which the energy has spread out is over the Earth, right? surface area = 4(pi)r^2 = 5.11 X 10^8..but only part of the energy is hitting the earth, the rest is going out into the space, right?? im lost

7. Mar 26, 2006

### confusedbyphysics

ok if i think of the orbit of the earth around the sun

the radius between the earth and sun is 1.496 X10^11 m

the surface area is 4pi(r^2) which comes out to 2.81 X10^23

then if I use the equation you gave me above 1350 X (2.81 X 10^23) = 3.8 X 10^26, THE RIGHT ANSWER!!!!!!

thank you both so much for your help it makes sense now

Last edited: Mar 26, 2006