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In a book, I've to read for a class at my university, called 'Duurzame Energietechniek' (Dutch, translated: renewable energy) they gave a formula that looks like:

$$\text{G}_\text{sc}\left(\text{n}\right)=1367\cdot\left(1+0.03\cos\left(\frac{360}{365}\cdot\text{n}\right)\right)\space\space\space\space\space\space\space\space\space\left[\text{W}/\text{m}^2\right]\tag1$$

Where $\text{n}$ is a day in a year.

That formula is called the sun constant formula, but I do not know where it comes from.

Question: Can someone help me derive this formula?


EDIT:

I understand that the energy flux is given by:

$$\text{G}=\mathcal{k}\cdot\text{T}^4\cdot\left(\frac{\text{R}}{\text{D}}\right)^2\tag2$$

Where $\mathcal{k}$ is the Boltzmann constant, $\text{T}$ is the surface temprature of the sun, $\text{R}$ is the radius of the sun and $\text{D}$ is the distance from the sun to the earth.

Now, for $\text{D}$ we know that is changes over a year because the earth makes a elliptical orbit around the sun.

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This answer first of all gives a simple-minded approach which has some numbers, and then a more hairy one which doesn't.

The simple approach, with numbers

So, first of all, $1367\,\mathrm{W/m^2}$ is the solar constant: it's the measured flux of energy from the Sun at the top of atmosphere (TOA), averaged over a year. So this flux is, the TOA flux for the point on the planet where the Sun is directly overhead (all other points get less). I'll call this $G_0$.

But the Earth's orbit has some eccentricity in it, so in fact sometimes this flux is a bit higher, and sometimes it's a bit lower. To first order we could model this by saying that the flux looks like

$$G = G_0\times(1 + E\cos (2\pi y))$$

where $E$ is some fudge factor based on the known eccentricity of the planet's orbit, $y$ is the time in years ($y$ is not constrained to be an integer), with $y=0$ being chosen as the point where the Earth is closest to the Sun. Observationally, $E \approx 0.03$.

Well, perhaps we want the constant in terms of day of the year, rather than year, which would be more useful. This would then look like

$$G = G_0 \times\left(1 + E\cos \left(\frac{2\pi}{365} d\right)\right)$$

Where $d$ is the day number, and $365$ is an approximate thing here: this will be OK for a while, but it will drift.

And now you have to realise that climate scientists talk to people who build spacecraft more than they talk to people like me: they work in degrees like engineers do. And $2\pi$ radians is $360$ degrees. So, finally, we get:

$$G = G_0 \times \left(1 + E\cos \left(\frac{360}{365} d\right)\right)$$

Which is your formula, and where things are working in degrees.

(I'm actually really disappointed now: I started this reply thinking 'aha, this is because you are using a model with a 360-day year (which many climate models have done historically, and many still do in fact) and I can explain this bit of obcsurity'. But no, sadly.)

A more hairy approach, without numbers

First of all we know the Sun looks quite like a black body at some temperature $T$, so the flux leaving the Sun is $\sigma T^4$ in the normal way. The total power passing through any surface surrounding the Sun is constant, so the flux at a radius $R$ is given by

$$\sigma T^4 \left(\frac{R_0}{R}\right)^2$$

Where $R_0$ is the radius of the Sun.

We could plug numbers for $T$, $R_0$ and $R$, the average radius of the Earth's orbit, into this and we will get $1367\,\mathrm{W/m^2}$. But the thing to know is how this varies with $R$, since Earth's orbit has some eccentricity. So we want to expand

$$\sigma T^4 \left(\frac{R_0}{R + \delta R}\right)^2$$ in terms of $\delta R$:

$$\begin{align} \sigma T^4 \left(\frac{R_0}{R + \delta R}\right)^2 &= \sigma T^4\left(\frac{R_0}{R}\right)^2 \times \left(1 - 2\frac{\delta R}{R} + O(\delta R^2)\right)\\ &\approx \sigma T^4\left(\frac{R_0}{R}\right)^2 \times \left(1 - 2\frac{\delta R}{R}\right) \end{align}$$

And now, well, we know that $\delta R$ is a periodic function of time, with the period being a year (to a good approximation & ignoring orbital variation), and there's no constant term. So, expressing time, $t$ in years, we can write $\delta R$ as

$$\delta R = \sum\limits_{n=1}^{\infty} a_n \sin(2\pi n t) + b_n \cos(2\pi n t)$$

And to first order, and adjusting the zero of $t$ suitably, we get

$$\delta R \approx a \cos(2\pi t)$$

where $a= a_1$. So plugging this into the above, we get

$$\sigma T^4\left(\frac{R_0}{R}\right)^2 \times \left(1 - 2\frac{a \cos (2\pi t)}{R}\right)$$

Whic is what we wanted.

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  • $\begingroup$ First of all, thanks for your answer. But I do not see where the cosine term comes from? Why can we model it that way? $\endgroup$ – Jan Feb 5 '17 at 12:42
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    $\begingroup$ @JanEerland Think of it as the first term in a Fourier series approximation: the variation is periodic, and the variation is small. You could also model the mechanics to come up with a Newtonian expression for $1/r^2$ as a function of time and with the orbital eccentricity in it: it would show that the $\cos$ term would be a good approximation when the eccentricity is small. $\endgroup$ – WetSavannaAnimal Feb 5 '17 at 12:48
  • $\begingroup$ @WetSavannaAnimalakaRodVance Is the formula in my question right? About the energy flux. $\endgroup$ – Jan Feb 5 '17 at 13:51
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    $\begingroup$ @JanEerland: I've added a bunch more detail: really, a whole new part to the answer with a derivation in terms of (the first terms of) power & Fourier series. I hope that makes sense. It may have errors as I typed it straight in rather than thinking on paper. $\endgroup$ – tfb Feb 6 '17 at 0:13
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And the value of $1367$ W/m$^2$ essentially comes from the radiation of a black body at 5700 K with a radius of 696 000 km located 150 000 000 km away.

Note that this value is indeed that on the top of the atmosphere. The average value on Earth should take into account the absorption through the atmosphere, reducing the intensity to roughly $1000$ W/m$^2$.

This value of course depends on your latitude on Earth - the closer to the pole, the thicker the atmosphere layer. We talk about "air mass". This value is that of the standard AM 1.5, used for most solar cell characterization (considering the name of your book, I guess this is what you will soon focus on).

For the complete spectrum, see for instance here.

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