# Energy carried by electromagnetic waves question

1. Mar 3, 2015

### Nyxious

1. The power radiated by the sun is 3.9 x 10^26 W. The earth orbits the sun in a nearly circular orbit of radius 1.5 x 10^11 m. The earth's axis of rotation is tilted by 27 degrees relative to the plane of orbit, so sunlight does not strike the equator perpendicularly. What power strikes a 0.75-m^2 patch of flat land at the equator?

2. Relevant equations
S(Intensity) = Power/Area
S2 = S1 x cos^2 (delta theta)
Surface area = 4pi x radius^2 (maybe?)
3. The attempt at a solution
Surface area using the radius 1.5 x 10^11m and got A=2.83 x 10^23
Next, i used S = Power/Area using the power of the sun and the area and got 1379W/m^2
Then, I used S2=S1 x cos^2( theta) and got 118W/m^2
Finally, I used S=P/A again but with the patch of flat land area and got a final answer of 88W

The answer is 920W so if someone could go over this with me I'd greatly appreciate it!

Last edited: Mar 3, 2015
2. Mar 3, 2015

### Staff: Mentor

Okay, looks good so far.
What's the reasoning behind using cos(θ)2 rather than, say, just cos(θ)? You should suspect that something is amiss when the resulting available power for a whole square meter is smaller than the proposed correct result.

3. Mar 3, 2015

### Bystander

For starters, let's take a look at this piece of your calculation: Where you got the particular form of the equation, what numbers you plugged into it, and for what purpose? Just as an aside, the axial tilt is ~ 23 degrees.

4. Mar 3, 2015

### Nyxious

Hmm, both of you have stated a possible problem about the equation S2=S1 x cos^2 (theta) but I can't seem to find the problem. Doesn't this equation show how the intensity is decreased due to the angle of the axial tilt? I believe i'm missing something very obvious haha

5. Mar 3, 2015

### Staff: Mentor

Where does the equation come from? Did you derive it yourself? To me it doesn't look like a correct equation for this application.

6. Mar 3, 2015

### Nyxious

It's an equation given by my text book. Malus's Law I believe? Any suggestions on other equations I could use?

7. Mar 3, 2015

### Staff: Mentor

Malus's Law applies to polarized filters. There are no filters involved here.

You might just have to derive your own relationship. Light from the Sun at our distance from it is essentially parallel light rays. Consider a flat surface that's tipped at an angle to the oncoming flux of light. The light that falls on that surface is the light that passes through some cross section that is perpendicular to the oncoming light:

How does the area of that cross section compare to the area of the tilted surface?

8. Mar 3, 2015

### Nyxious

Would the cross sections area be smaller than the area of the tilted surface?

9. Mar 3, 2015

### Staff: Mentor

You should be able to deduce that from the geometry in the image that I drew. What's the trigonometric relationship between the height of the cross section and the edge length of the surface shown?

10. Mar 3, 2015

### Nyxious

Cos (theta) = adj (Height of cross section) / hyp (length of surface), correct?

11. Mar 3, 2015

### Staff: Mentor

Correct. So given that the width of the surface presented to the light is not affected by the tilt angle (the width being the dimension of the surface that is going into the page in the drawing, and not seen edge-on), how does the area of the cross section compare to that of the surface?

12. Mar 3, 2015

### Nyxious

Would the area be the same?

13. Mar 3, 2015

### Staff: Mentor

The same as what?

14. Mar 3, 2015

### Nyxious

The cross sections area being the same as the area of the tilted surfaces width face. Was that the question you were asking? I may have read your question wrong.

15. Mar 3, 2015

### Staff: Mentor

No, I'm asking how the area of the flat surface compares to the area of the cross section. For a rectangular surface the area is L x W (length by width). The width of the flat surface presented to the light is unchanged by the tilt, but the length that is presented in cross section has changed by the amount that you worked out a few minutes ago.

In other words, what is the area of the cross section?