# Solar irradiation on the Earth as a function of latitude.

1. Sep 18, 2011

### Lavabug

1. The problem statement, all variables and given/known data
Given the following info:

Solar output(modulus of the Poynting vector): $1.4kW/m^2$
Earth radius: 6370km
Average Earth orbit radius: $1.49*10^11 m$

Find the total power output of the Sun, power received on the Earth, and the power received in Spain if its surface area is 5*10^2 km^2 and is at a latitude of 40 N.

2. Relevant equations
Poynting vector formula?

3. The attempt at a solution
I managed to get the right answer for the power received on the Earth, but I'm not sure how to proceed with the other 2 questions. Wouldn't the power received at a given latitude depend on the time of the year? I found this applet that calculates it for you and apparently at 40 N, the solar output never reaches 1.4kW at any time of the year:
http://pvcdrom.pveducation.org/SUNLIGHT/SHCALC.HTM

... so I'm guessing I'll have to resort to spherical trigonometry some way or another? I am given no other info and this is a course purely on electromagnetic optics.

2. Sep 18, 2011

### Spinnor

The time of year is important as is the time of day as is the Earth's atmosphere.

Say you have a black square, 1 meter by 1 meter that absorbs all light that falls on it. Say you are at the equator and hold the square perpendicular to the suns light. What is the power absorbed? Now hold the same black square so that the suns light and the unit normal of the black plane make and angle theta. The power now absorbed is rescued by a factor of cos(theta)?

3. Sep 18, 2011

### Lavabug

I think I see it now, drawing a circle (earth) with normal surfaces on the equator and at 60ยบ from the N pole, using trigonometry I get P(at 60 N) = Power(known)*(cos60), which is a halve of the known power output at the given orbital distance (which I got right: 3.9*10^26W).

However, what you're saying assumes the equator is aligned with the sun/the Earth isn't tilted, but since the problem says nothing to that avail so I guess its reasonable? Thanks.

How would you solve the first part? Wouldn't I need the radius or surface area of the Sun?

Last edited: Sep 18, 2011
4. Sep 18, 2011

### Spinnor

You are given what is needed, the power/m^2 at the distance the Earth is from the sun and the distance to the sun.

power per area times area = power

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