I was thinking how to solve this problem. $1\,\mathrm{AU}$ is roughly the distance from the Earth to the Sun, $1.4960 \times 10^{11}\,\mathrm{m}$.

The radius of Earth is approximately $6.4 \times 10^{6}\,\mathrm{m}$, and the radius of the Sun is approximately $6.96 \times 10^{8}\,\mathrm{m}$.

How could we estimate the percent of radiation which the Earth receives, ignoring astrophysical "noise" like dust?

The radiation emitted by the Sun roughly follows the Stefan-Boltzmann law and the radiation emitted is roughly $\propto T^4$, and the surface area of the Earth is roughly $\propto r^2$.

Would you simply take the ratio between the Sun's surface area divided by the Earth's surface area?

## closed as off-topic by Rob Jeffries, Sebastian Riese, CuriousOne, Kyle Kanos, rob♦Sep 21 '15 at 2:27

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• Consider the solid angle subtended by the sun with the earth. – SchrodingersCat Sep 20 '15 at 16:21
• If you mean total radiation at the Earth's surface, would our atmosphere be more of an aspect you could consider than any dust between here and the sun? I am not sure how accurate you want to be, but you could of course check it on Wikipedia. – user81619 Sep 20 '15 at 16:22
• The Sun radiates isotopically (roughly). So you work out what fraction of a $4\pi$ solid angle is occupied by the Earth as seen from the Sun. To first order it is the projected surface area divided by the area of a sphere at the Earth's orbit. – Rob Jeffries Sep 20 '15 at 16:41

All the Sun's power $P$ passes uniformly through a sphere with radius of 1 AU. Calculate the total surface area of this sphere and call it $S$.
The Earth's disc also has a surface area that can be calculated from its radius. Call this surface $S_E=\pi R_E^2$.
$f=P\frac{S_E}{S}$.