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.
Best Answer
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.