# Confirmation on the Effect of Solar Irradiance

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1. May 14, 2015

### ecastro

The Sun emits approximately 1361 watts per square meter on Earth. I read in a book that since the Earth is approximately a sphere and only half of it is being illuminated by the Sun, the solar irradiance at the lit side of Earth is reduced by half (so around 680 watts per square meter). Is this correct? I thought the solar irradiance is a constant per square meter. So if the Earth is to be fully illuminated, the total power received by Earth from the Sun would be 1361 watts per square meter times the area of the Earth, I think this value should be the one to be reduced to half when only half of Earth is illuminated (i.e. 1361 watts per square meter times the area of the lit side of Earth)

2. May 14, 2015

### Bandersnatch

You are right. The area of Earth should be divided by two instead.
However, do notice that the result will be the same as if you kept the area and divided the irradiance. It matters only insofar as understanding what you're doing with the numbers.

3. May 14, 2015

### ecastro

@Bandersnatch - I'm still confused. Here, I quote the actual words:

"Finally, the Earth is approximately a sphere. Only half is being illuminated at any instant. For that half, the average sunlight incident per unit area is given by $\frac{Q_0}{2} = 680 \text{ W} \cdot \text{ m}^{-2}$."

I understand what you mean if I divide the irradiance by two, but given that the whole Earth is considered. However, the book said that "For that half", so it means that for that "half" of the Earth only.

4. May 14, 2015

### Staff: Mentor

Where did you find this? It indeed doesn't make any sense. The power per square meter will not change. If you wanted to find the total power of sunlight falling on the Earth, you would just use half of the Earth's surface area for your calculations.

5. May 14, 2015

### ecastro

It's from "Atmospheric Radiation - A Primer with Illustrative Solutions" by Jim Coakley and Ping Yang. Throughout the discussion, the book used the same analogy to calculate the effective temperature of the Earth (around 250 - 270 Kelvins; the blackbody temperature that fits the Earth's spectra). So the correct calculation should be, for half of the Earth, $1361 \text{ W} \cdot \text{ m}^{-2} \times \frac{4 \pi R^2_\oplus}{2}$ or $\frac{1361 \text{ W} \cdot \text{ m}^{-2}}{2} \times 4 \pi R^2_\oplus$. It's either consider half of the Earth and the whole solar irradiance, or consider half of the solar irradiance and the whole surface area of the Earth (the latter being an average). I don't know if I should continue reading this book, though.

6. May 14, 2015

### Staff: Mentor

Well, the math comes out fine either way you do it, but that's seems like a sketch way of doing it in my opinion. However, I don't calculate these things for a living, so there may be a very good reason for doing it this way.

7. May 15, 2015

### D H

Staff Emeritus
No, it's not correct. The Earth is approximately a sphere, so the half that is illuminated by the Sun is approximately a hemisphere. The surface area of a hemisphere with radius $R$ is $2\pi R^2$, twice the area of a circle with the same radius. This reduces the amount of sunlight averaged over the surface of the Earth by another factor of two. Combining these two effects (nighttime and latitude), the energy received by the Earth from the Sun averaged over the surface of the Earth is 1/4 of the solar constant.

8. May 15, 2015

### Bandersnatch

Looking up the passage in the book, it goes like this:
First the book introduces a figure illustrating the energy balance, including the equation with 1/4th factor of the solar constant $Q_0$, as well as albedo. Then it proceeds to explain where does this equation come from step by step. The very next sentence after the one you quoted tells you where the factor of two comes from ('the fractional factor of "1/2" represents the ratio of the area of the circle that blocks the sunlight to the area of the hemisphere having the same radius'). Then the thought process goes on to include the unlit hemisphere for the factor of 1/4.

In other words, the sentence looks like a cock-up if quoted out of context, but I see no problem with it in the context in which it appears. no, scratch that. The only cock-up here was my earlier response in post #2. The issue here is that the solar constant is, as explained by D H and in the book, not the energy per unit area of the surface of the Earth, but per unit area of the surface of a sphere of R=1AU centred on the Sun, so it's the wattage per square metre of a circle of Earth's radius, not a hemisphere.
You need to spread it across the hemisphere, which is where the 1/2 comes from.

Last edited: May 15, 2015