How much of the surface of other planets is lit by the sun?

Since the sun is much larger than the earth, it should ideally light up a considerable portion of the earth rather than only half of it.

However the sun lighting only half of the earth can be explained using the fact that the sun is at a considerable distance away from the earth. As the distance of a light source from the object increases the portion of the object lit up reduces.

Applying the same logic to other planets (especially the ones beyond the earth), I would like to know how much of the surface of these other planets is lit up by the sun?

Going by the distance thing, a portion less than half should be lit up, which would mean a fraction of the surface (mainly around the centre from north to south) would always be in darkness.

Would be nice if diagrams could be used to explain

• I cannot follow your logic that distance decreases the lit surface to below 50%. The sun is bigger than any planet, so always 50% and a tiny bit more are illuminated. The tiny bit is very tiny as the sun is very distant Commented Oct 15, 2023 at 6:19
• Well, there are two different effects at play here. Assuming a point source, closer distance is less illumination. For an extended source, closer distance is more illumination. For far away planets since the sun is effectively a point, they have effectively 50% illumination, whereas the earth is a fraction of a percent above 50% illuminated directly by the sun Commented Oct 15, 2023 at 7:15
• @Adiyarkku why do you think that the Earth is exactly 50% illuminated? (This sounds like a pedagogical simplification, like π=22/7, that's close to the truth, but not perfectly true.) Even discounting atmospheric effects, orbital eccentricity, and asphericity, it's clear that the sun illuminates measurably more than 50%: full night is on average 4 minutes shorter than daytime. As for the poles being illuminated for 6 months each, no that's not true either: each equinox there's a period of about 31 hours when both poles are directly illuminated (plus a bit longer because of refraction). Commented Oct 16, 2023 at 9:29
• @Adiyarkku Ahh, from elementary school text books? So yes, they're approximations for children. If you put a diagram showing exactly half next to a diagram showing the correct proportion, they look the same, and most people could not tell the difference. If you used a ruler to measure the diagram, the difference would be less than the thickness of a line. Commented Oct 18, 2023 at 3:07
• Consider a big circle and a small circle right next to each other, along with the 2 outer lines that are tangential to both circles (representing how much of the small circle is illuminated, which is > 50%). When the circles move apart, the lines become closer to parallel, but they cannot ever become parallel, because this can only happen if the 2 circles are the same size. Parallel means 50% illuminated, so it will tend to 50% illuminated, and always be more than 50% illuminated. (I would've posted an answer, but for that I probably need diagrams, which I didn't really feel like drawing.) Commented Oct 18, 2023 at 8:04

1 Answer

OK, let start with some assumptions: spherical objects and no atmospheric effects.

Here's the relevant geometry with the object on the left representing a planet and the object on the right representing the sun:

Where d is the distance between the two spheres, r1 and r2 are the radii of the spheres. $$\theta_1$$ is the angle showing the portion of the sphere illuminated. We'll use it later to compute what fraction of the sphere is lit.

We can solve for $$\theta_1$$ easily by noticing that the solid line quadrilateral can be drawn as below. We're flipping it so that the side labeled L is on the bottom.

Noting that the upper triangle is a a right-triangle with sides known allows us to solve for the unknown angle $$\theta_1$$ as follows:

$$d \sin\left(\frac{\pi}{2}-\theta_1\right) = r_2-r_1$$

$$\sin\left(\frac{\pi}{2}-\theta_1\right) = \frac{r_2-r_1}{d}$$

$$\frac{\pi}{2}-\theta_1 = \arcsin\left(\frac{r_2-r_1}{d}\right)$$

$$\theta_1 = \arcsin\left(\frac{r_2-r_1}{d}\right) + \frac{\pi}{2}$$

Or, with a small angle approximation for when the distance is much larger than the difference in radii:

$$\theta_1 \approx \frac{r_2-r_1}{d} + \frac{\pi}{2}$$

The solid angle subtended by a cone with apex angle $$2\theta$$ is given by (see figure below for the definition of \theta relative to the apex angle):

$$\Omega = 2\pi(1-\cos\theta)$$

Image from Wikipedia.

Substituting $$\theta_1$$ for $$\theta$$:

$$\Omega = 2\pi\left(1-\cos\left(\arcsin\left(\frac{r_2-r_1}{d}\right) + \frac{\pi}{2}\right)\right)$$

$$\Omega = 2\pi\left(1+\sin\left(\arcsin\left(\frac{r_2-r_1}{d}\right)\right)\right)$$

$$\Omega = 2\pi\left(1+\frac{r_2-r_1}{d}\right)$$

Given there are $$4\pi$$ steradians in a sphere, the fraction, $$f$$, of the planet illuminated is given by:

$$f = \frac{\Omega}{4\pi}$$

$$f = \frac{2\pi\left(1+\frac{r_2-r_1}{d}\right)}{4\pi}$$

$$f = \frac{1}{2}\left(1+\frac{r_2-r_1}{d}\right)$$

As the distance between the objects gets larger, the term on the right becomes small and the fraction illuminated tends to 50%. As the term on the right is always positive, the fraction illuminated will always be more than half so long as the planet is smaller than the sun.

Using values for the mean radius and distance to the sun (from Planetary Fact Sheet - Metric) and with the Sun's mean radius of 695,700 (from Sun Fact Sheet):

Planet Mean Radius (km) Mean Solar Distance (km) Fraction Illuminated
Mercury 2,439.5 5.79x107 50.60%
Venus 6,502.0 1.08x108 50.32%
Earth 6,378.0 1.50x108 50.23%
Mars 3,396.0 2.28x108 50.15%
Jupiter 71,492.0 7.79x108 50.04%
Saturn 60,268.0 1.43x109 50.02%
Uranus 25,559.0 2.87x109 50.01%
Neptune 24,764.0 4.52x109 50.01%
Pluto 1,188.0 5.90x109 50.01%

If atmospheric effects are considered then there is the possibility of extinction occurring reducing the effective fraction of the planet that is illuminated. This would need to have a definition of how much light counts as illuminated before an assessment could be made for this effect.

Another effect, which may be more significant, is refraction by the atmosphere which causes a larger portion of the surface to be illuminated than if there is not atmosphere present. This same effect on Earth means that geometrically the sun is below the horizon while we can still see it rising or setting.

• Limiting the "fraction illuminated" to only 4 digits hides the details near the bottom of the table. I get 50.022218 for Saturn, 50.011675 for Uranus, 50.007422 for Neptune, and 50.005886 for Pluto. (My figures probably have spurious precision because I've ignored eccentricity and asphericity, especially for Pluto; but the 50% additive term should be ignored when educing the available precision from the precision of the solar system dimensions, so clearly more precision is available than in the table above.) Commented Oct 16, 2023 at 6:00
• The important point in all cases is: The stellar radius is always bigger than the planetary radius. Thus by simple geometric logic as illustrated in the sketches here, alway an area >= 50% must be illuminated. That is even true for any convex-shaped object orbiting the star. Commented Oct 16, 2023 at 7:33
• Well, technically one could have a gas giant orbiting a white dwarf, and being less than half illuminated by it. Commented Oct 16, 2023 at 7:38
• @DrPhil that's not as far-fetched as you might suppose. Even our moon is slightly egg-shaped, as it's tidally locked to the earth. So it's quite plausible that a planet tidally locked to a star could likewise become egg-shaped with its "fat" end facing the star. Commented Oct 16, 2023 at 9:53
• @Adiyarkku the bottom is simply a mirror image of the top, with the same internal proportion, and therefore the same overall proportion. (If A is 51% of B then 2A is 51% of 2B.) Commented Oct 18, 2023 at 3:12