1
$\begingroup$

I've been reading about black body radiation and I came across the topic of solar irradiance.

If we consider the sun to be a perfect blackbody, then the intensity of the solar radiation at a distance of $1\space AU$ is roughly equal to around $1360 \space Wm^{-2}$. This is the flux density of solar radiation and is known as the solar constant.

However, I want to calculate the amount of energy received by the earth. Note that I'm ignoring the reflected energy and albedo for now.

According to most of the sources that I've come across, including this Wikipedia article, this article and this youtube video, among others, the energy that the Earth receives is found by considering the Earth to be a solid disc, instead of a sphere. Then we multiply the solar constant to the area of this disc. I'm not sure about the intuition behind this. Shouldn't we consider the earth to be a sphere and not a disc while calculating the total flux ?

After they have found out this total energy, they divide it by $4$ as the area of a sphere is four times the area of a disc. This gives us the average energy over the surface of the Earth. However, I'm still unable to understand the motivation behind doing so.

Why would we consider the Earth to be a solid disc to find the total energy received, and then divide this by the total area of a sphere, to find the energy received per unit area ?

$\endgroup$

1 Answer 1

2
$\begingroup$

Imagine a very long cylinder. The circular part faces the Sun. The long part does not. Clearly, the long part of the area does not help the cylinder absorb extra sunlight from the Sun.

The reason you consider the Earth to be a disk while calculating the received flux is similar. I'll explain it in a different way which may be more intuitive. The Sun's light is spread across a full sphere of area $4\pi D^{2}$ where D = 1AU, the distance between the Sun and the Earth. The Earth casts a disk-shaped shadow that covers $\pi R^{2}$ area of that Sphere. The shadow covers some fraction of the Sun's light. The fraction of the Sun's total luminosity that Earth receives is then

$\frac{\pi R^{2}}{4\pi{D^{2}}}$

The third way to explain why it ends up being a disk, is somewhat less intuitive, but I'll include it regardless. The angle between the incident sunlight and the ground affects the power/(area of the ground). This is one of the reasons why the poles are cold whereas the equator is hot, and one of the factors behind the seasons. It turns out that taking the angle into account, the area that matters is just the component of the area that is perpendicular to the Sunlight (or the component of the normal vector of the area which is parallel to the Sunlight).

In short, the Earth receives energy from 1 direction (the Sun) and only shows the Sun a disk-worth of area, but the Sun is emitting energy in all directions over the area of a sphere

The part below does not directly apply to your question but I include it in case there's any confusion about multiple instances of spherical areas.

A confusing thing you will then find, is that the total spherical radius of the Earth is used again $4\pi R^{2}$ in these calculations. This is often because you might be interested in calculating the temperature of the Earth, by balancing Energy In and Energy Out.

Energy In will depend on the Disk Area of the Earth / Spherical Area of Earth's orbit and the Sun's Luminosity L: $L\left(\frac{\pi R^{2}}{4\pi D^{2}}\right)$

Energy Out will depend on the total spherical area of the Earth and its temperature, cooling according to the Stefan-Boltzmann Law. $\sigma T^{4} 4\pi{}R^{2}$

So we can balance the two:

$L\left(\frac{\pi R^{2}}{4\pi D^{2}}\right) = \sigma T^{4} 4\pi{}R^{2}$

In a sense, $\sigma T^{4}$ is the average flux (energy per area) received over the Earth.

$\endgroup$
7
  • 1
    $\begingroup$ Thank you so much. I suppose the biggest confusion arose here, due to the concept of 'flux'. Normally when we calculate (say) electric flux through a hemisphere or a disc due to a an external point source, we have a rather complex integral that we need to solve. We can't say that electric flux through a sphere is this 'shadow' of the disc and something like that. That was the source of my confusion. $\endgroup$ Oct 1, 2021 at 19:02
  • 1
    $\begingroup$ Frequently in Gauss's law applications of flux, the surface area normal vector is designed to be parallel with the flux, so you often end up taking the whole surface area. In the case of the Earth, the area normal vector is not parallel with the flux, so the parallel component must be taken, which ends up being the disk. $\endgroup$
    – Alwin
    Oct 1, 2021 at 19:03
  • 1
    $\begingroup$ Yeah, but say I choose to calculate the complicated integral, where I integrate the value of the flux density over each area element, I'd get an exact answer. Comparing the earth to a disc, is just an excellent approximation, isn't it ? The real answer would be found out by integrating the different values of flux, over the earth's surface with different area vectors. Is that correct ? $\endgroup$ Oct 1, 2021 at 19:09
  • 2
    $\begingroup$ If you calculate the complicated integral while assuming the Earth is a perfect sphere and the light is parallel, you will find that it all works out perfectly so that you get the same answer. In short, $\int \vec{F}\cdot \vec{dA} = F \pi R^{2}$ because $\pi R^{2}$ is the perpendicular area, if $\vec{F} = F \hat{r}$. The approximation made here is that the total angular area of the Earth as seen from the Sun is small so that $\hat{r}$ is just 1 direction. $\endgroup$
    – Alwin
    Oct 1, 2021 at 19:12
  • 1
    $\begingroup$ If I assume that the light is parallel and constant, then the area must be that of a disc, and we would get the integral that you mentioned. But generally, the light isn't parallel, and it varies ever so slightly, doesn't it ? And the area element of each region is not parallel, so we have to integrate over all these elements ? $\endgroup$ Oct 1, 2021 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.