# Conceptual Doubt regarding the calculation of the Solar Constant

The Solar constant is the intensity of the solar radiation in the upper atmosphere. It's value is about $1400$ $Wm^{-2}$.

Now we begin by stating that the Power radiated by the sun is about $3.9*10^{26}$W. Then we using the Formula $I=\frac{P}{4\pi d^2}$, we compute the Intensity. (Here d is $1.5*10^{11}$m).

I understand the mathematics but have trouble in interpreting the meaning of the solar constant. The Intensity formula used above is generally depicted by authors as follows (as per my understanding): There is a sphere around the Sun of radius $d$ and if the power is radiated equally in all directions then each point on the Sphere receives an Intensity $I=\frac{P}{4\pi d^2}$. But the Earth is not a point object, so is this assumption valid?

Next, what do you mean by the upper atmosphere? I've come across another version:

The Solar constant is the intensity of the solar radiation in when the sun is directly overhead. It's value is about $1400$ $Wm^{-2}$.

I guess my confusion is due to a lack of an intuitive understanding of this concept and I would be grateful if anyone could provide me with one.

• Strictly speaking, $I$ is not intensity but irradiance: en.wikipedia.org/wiki/Irradiance . "Intensity" is specially inappropriate here, since this word has a specific, different meaning in photometry and radiometry. – L. Levrel Apr 13 '16 at 20:12
• Just want to check, but I assume you are aware that we have been measuring this to be ~1360 W $m^{2}$ since the 1970s... – honeste_vivere Apr 14 '16 at 15:48

## 1 Answer

First of all the Earth's radius is about $6.4\times 10^6\mathrm{m}$, while its orbital radius is about $1.5\times 10^{11}\mathrm{m}$: about $2.3\times 10^4$ times its radius: yes, it is a good approximation that the whole Earth is at the same distance from the Sun.

Secondly, this is the flux passing per unit area of the imaginary shell: the flux per unit area at the top of the atmosphere is equal to this when the top of the atmosphere is parallel to the imaginary shell. This is, obviously, when the Sun is directly overhead. At sunset the flux at the top of the atmosphere is zero, of course.

You can get the total power reaching the Earth by just integrating over the disk the Earth presents to the Sun, or simply by multiplying by $\pi r^2$ where $r$ is the Earth's radius.