So I know that the Sun's luminosity is $3.839 \cdot 10^{26}$.
What I want is the luminosity per square meter at the Earth's surface. So this is what I have got so far:
I know the distance to the Earth from the Sun is $149.60 \cdot 10^9$ m. So imagining the Sun creates a sphere with a radius of $149.60 \cdot 10^9$ m, the surface area of that sphere would be $2.812 \cdot 10^{23} \mathrm{m}^2$. If I divide the luminosity of the Sun by that surface area, I get 1356 W/m$^2$.
Assuming this is right (if not, please correct me), is this actually the amount of Watts per square meter that would reach a solar panel on Earth under ideal conditions? I didn't account for the atmosphere of the Earth, and the amount of light it blocks - as far as I understand.
So what calculations am I missing here, and how would I go about finding what I want (that is, how many Watts per square meter hit a solar panel on Earth due to the Sun under ideal conditions)?
Thanks!