During a solar eclipse, some of the visible area of the sun is obscured by the moon. The exact percentage varies with time and with the observer's location. My question is, does the sky darken by this same percentage?

In other words, suppose the sun is 80% covered. Does sky light diminish by 80%? Are objects illuminated by 20% of their normal amount?


Assuming the intensity of the sunlight is proportional to the area of the Sun which is showing, we can roughly calculate the intensity of sunlight, $I(d)$ as:

$$I(d) = 1 - \frac{\mathrm{Area}_{\mathrm{overlap}}}{\mathrm{Area}_{\mathrm{Sun}}}$$

Where $\mathrm{Area}_{\mathrm{overlap}}$ is the overlap between the Moon and the Sun (the green area).

(source: uga.edu)

Some calculation finds the overlap to be:

$$\mathrm{Area}_{\mathrm{overlap}} = 2r^2\cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2}\sqrt{4r^2 - d^2}$$

Where $d$ is the distance between the centres of the Moon and the Sun. Similarly, the area of the Sun is:

$$\mathrm{Area}_{\mathrm{Sun}} = \pi r^2$$

Putting this together, you get the graph:

Intensity changing with time

Where the Y axis is brightness (ranging from 100% to 0%) and the X axis is $d$ (ranging from $2r$ to $0$).

I hope this answers your question: the brightness does not change proportionally to coverage/time.


This assumes that $d$ increases linearly with time (the moon doesn't speed up or slow down as it moves across the Sun).

  • 5
    $\begingroup$ Yes, but sky brightness is caused by an overall scattering of solar light, not be the light only at your current location - this is why the sky does not go pitch black as soon as the Sun sets ... so while the level of brightness is likely proportional to your formula, that won't be the only factor. $\endgroup$ Aug 23 '17 at 15:33
  • $\begingroup$ @adrianmcmenamin Agreed, but without experimental data, this is the best I can do $\endgroup$
    – Beta Decay
    Aug 23 '17 at 17:46

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