astronomyorbital-motionsunweather
In the diagram, it shows that the fundamental reason for different seasons is when the northern hemisphere is titled towards the sun there's summer in northern hemisphere and winter in southern hemisphere and vice-versa after the six months.
But because the earth spins around its axis, why doesn't the seasons change during day and night? is there something missing in the diagram to prove that?
Degrees and radians are just different units for the same quantity, angular displacement. So your question is fundamentally the same issue as whether you should use, say, meters or feet to represent a distance. You just have to convert the quantity in each case to the unit that your code expects.
It's a little confusing in this case because in some cases, they've inserted the conversion factors in the formulas for you. You can recognize these conversion factors by the appearance of $\frac{180}{\pi}$ or $\frac{\pi}{180}$, which pretty much never appears in a formula except to facilitate a conversion from radians to degrees or vice-versa (respectively). For example, in
$$\Lambda = \frac{180\lambda}{\pi}$$
$\Lambda$ and $\lambda$ are actually the same quantity, but $\Lambda$ is in degrees and $\lambda$ is in radians. So when you calculate
$$\tan^{-1}\biggl[\frac{\Omega}{\sqrt{1 - \Omega^2}}\biggr]$$
if your calculator is in radian mode or you are using a function that outputs in radians, you will get an answer in radians, i.e. $\lambda$. But if your calculator is in degree mode or you are using a function that outputs in degrees, it does the conversion (the multiplication by $\frac{180}{\pi}$) for you, and you will get an answer in degrees, i.e. $\Lambda$. You don't multiply by $\frac{180}{\pi}$ again.
For simplicity, I would suggest always using trig functions that accept arguments in radians. Then when you have a value in degrees, you can convert as necessary.
In
$$\Omega = \sin\biggl(\frac{\pi\theta}{180}\biggr)\sin(\delta) + \cos\biggl(\frac{\pi\theta}{180}\biggr)\cos(\delta)\cos(\omega)$$
the formula is written with the conversion factors from degrees to radians built in. In other words, as written, it expects $\theta$ to be in degrees, but it assumes you are using radian mode (i.e. trig functions that take arguments in radians). $\theta$ needs to be converted from degrees to radians, hence the factors of $\frac{\pi}{180}$, but $\delta$ and $\omega$ are assumed to already be in radians.
In $$\delta = \frac{23.45\pi}{180}\cos\biggl(\frac{2\pi}{365}(172-J)\biggr)$$
you will notice that the factor of $\frac{\pi}{180}$ is already there to convert degrees to radians. But what it is converting is not the output of the cosine, which is a pure number (not an angle); rather, it is the $23.45^\circ$, which is the tilt of the Earth's axis relative to its orbital plane.
For a definition of the UV index and a rough discussion of influencing factors see e.g. https://en.wikipedia.org/wiki/Ultraviolet_index.
You can compare the model shown there against actually measured data for the US: http://www.cpc.ncep.noaa.gov/products/stratosphere/uv_index/uv_annual.shtml.
Both sources will support your statement that UV exposure in summer is significantly higher (by up to almost an order of magnitude) than in winter.
This graph
shows the difference in irradiance between the top and the bottom of the atmosphere. For the UV-B range of 280-315nm over 50% of the radiation is being absorbed while going trough the atmosphere at right angle, i.e. the extinction will be significantly larger than that when the sun is low over the horizon. I would agree that the geometric angle makes a significant difference in the available irradiance. Combine that with the time we spend outside (or, do not spend outside) and the fact that we are almost completely covered up in European latitudes, and our actual UV exposure can only be a tiny fraction of that in summer. Having said that, unfiltered summer exposure of more than half an hour is considered dangerous.
Best Answer
In the summer, the sun stays up all day, it comes up early and stays up late, while in the winter it sets early and comes up late. Also, the sun is lower down in the sky in winter, so that like a flashlight at an angle, the illumination per square meter is decreased by the cosine of the angle.