How bright is the full moon as compared to the Sun (as a ratio).?

Question

I realize that moonlight is reflected sunlight, and the moon is not a perfect reflector. What is the ratio between the brightness levels (or luminous intensity) of the sun and the full moon? If percentage is more convenient expression, go ahead and use that.

Answer 1

When referring to light reflected by moons or planets in this solar system, astronomers use the term "albedo" to denote the ratio of reflected light intensity to sunlight intensity.

("Albedo" is a Spanish word that means "whiteness.")

Unfortunately, albedo can be measured and calculated in several different ways, each of which yields a different answer.

According to different measures, the albedo of the moon ranges from seven (7 percent) to about 12-1/2 percent. I personally prefer the simple comparison of luminous flux, which yields the lowest of those numbers. That is, in my opinion, the albedo of the moon is seven percent.

Answer 2

The apparent magnitude (m) of a star, planet or other celestial body is a measure of its apparent brightness as seen by an observer on Earth. The brighter the object appears, the lower the numerical value of its magnitude.

The Sun is 14 magnitudes brighter than the full moon, Obviously its more than that for a crescent moon or a new moon,

A magnitude difference of 5 means a difference in brightness of 100 and of 15 means a difference in brightness of 1,000,000

A magnitude difference of 1 means a difference in brightness of the fifth root of 100 or 2.512 (the Pogson ratio)

So a magnitude difference of 14 means a difference in brightness of 1,000,000 / 2.512 = in round figures 400,000 and the ratio is 1: 400,000.

If the moon is at or close to perigee it will look larger and brighter. The earth being close to aphelion will make the sun look smaller and less bright. So these figures will vary a bit.

Answer 3

Last night, when I could not stay asleep (a full moon seems to have that effect on me) I asked myself the same question, and with a few numbers I remembered, I argued as follows: per area about the same amount of light falls onto earth as onto moon (both are about the same distance from sun). The fraction of light then remitted from that area of the moons soil is about 10% and goes in all direction of a half sphere. This half sphere has a radius (for an observer on earth) of the distance Moon-earth. (I remembered that this distance is about 1.3 light seconds, or 1.3 times 300,000 km, that is about 400,000 km). Now the size of the moon matters: If the apparent area of the moon was 2 times as big we would have twice as much moonlight, right? And further, if the moon was twice as far from us its apparent area would be one quarter. (Not one half, if you think about it.) So the amount of light is proportional to the area and the area is proportional to the square of the diameter. I remembered that the diameter of the moon is about one third of the diameter of the earth (12,000 km) which makes the diameter D of the moon about 4,000 km, and it makes the area of the circular moon 0.25*pi*4000*4000square kilometer. (but I did not have to worry about the 0.25pi as we'll see). The light from that area is diluted over 0.25*pi*400,000*400,000 square kilometer (at the earthly observers location. Result: the full moonlight shining on earth is a fraction 0.1 (10%) * 4000*4000/(400,000*400,000)=0.1*4*4/(400*400) of the sun's, or 1/100,000.

This morning I found this on the Internet:
“The actual ratio of full-moon light to sunlight is approximately 1/400 000. [See Allen, Astrophysical Quantities, 3rd ed., Athlone Press, p. 144 (1973).] It would be considerably smaller were it not for an extraordinary property of the moon's surface which like glass beads (but for quite a differ-ent reason) concentrates the scattered light in the backward direction. In fact, we receive from the full moon about ten times as much light as from the half-moon, whereas for a matte Lambert Law sphere the ratio would be just pi.”
So my 10% remission value was an oversimplification, and the explained optical properties of the moon surface seem to explain nicely (almost quantitatively) the factor of four error of my estimate.

Answer 4

The full moon has an apparent magnitude of -12.6. The sun has an apparent magnitude of -26.8. The sun's apparent magnitude is 449,032 times greater than the moon's.

Answer 5

Approximate illumination of horizontal plane (sun overhead in clear sky = 1), with no moon, Venus, or city lights unless stated otherwise. Disclaimer: This is NOT official or certified data and is NOT for use for legal purposes. Angle of sun above horizon in degrees(- is below) sun (or moon) and sky light of clear sky, but not sun (or moon) thin clouds (sun can marginally be seen faintly through clouds) Solid clouds, maybe light rain Storm clouds Very dark storm clouds
90 1 1/8 1/8 1/32 1/128 1/512
45 0.7 1/8 1/11 1/44 1/176 1/704
14 1/5 1/16 1/40 1/160 1/640 1/2560
5 Unkown 1/256 Unknown Unknown Unknown Unknown
0 1/512 1/512 1/4096 1/16384 1/65536 1/262144
-2 1/5120 1/5120 1/40960 1/163840 1/655360 1/2621440
-4 1/51200 1/51200 1/409600 1/1638400 1/6553600 1/26214400
-6 1/512000 1/512000 1/4096000 1/16384000 1/65536000 1/262144000
-8 1/5120000 1/5120000 1/40960000 1/163840000 1/655360000 1/2621440000
-10 1/51200000 1/51200000 1/409600000 1/1638400000 1/6553600000 1/26214400000
-12 1/512000000 1/512000000 1/4096000000 1/16384000000 1/65536000000 1/262144000000
-14 1/2560000000 1/2560000000 1/20480000000 1/81920000000 1/327680000000 1/1310720000000
-90 (starlight, airglow) 1/5120000000 1/5120000000 1/40960000000 1/163840000000 1/655360000000 1/2621440000000
Conditions - - - - - -
Full moon at zenith 1/500000 1/4000000 1/4000000 1/16000000 1/64000000 1/256000000
Half moon at zenith 1/5500000 1/44000000 1/44000000 1/176000000 1/704000000 1/2816000000
Very bright city lights N/A 1/500000 1/100000 1/100000 1/100000 1/100000
light of medium-sized town N/A 1/5500000 1/1100000 1/1100000 1/1100000 1/1100000
Sun 6 degrees below horizon - This is when the theory says you begin needing outdoor light in the evening (or stop needing it in morning)
Sun 12 degrees below horizon - Horizon at sea at the limit of usability
Sun 18 degrees below horizon - Astronomers can consider it full night for their purposes.
MYTH: The process of it getting light in the morning is complete at sunrise, and darkening in the evening begins at sunset.
FACT: The moment before sunset or the moment after sunrise, the sky is 64 X as dim as noon. If skylight at noon is equated with a chessboard, at sunrise/set it is a mere one square on that chessboard. If unshaded ground illumination is taken, the direct sun at noon is about 8 X as bright as diffuse sky light, so the noon:sunset illumination ratio is 512:1, like 8 chessboards to one square on one!

The first problem with this is the limits as a whole. When the Sun is less than 14 degrees ABOVE the horizon, the sky is significantly dimmed because the lower atmosphere is receiving sunlight that has passed through more air than at noon ('Self-shading'). For SAD-relieving purposes, for example, just having the sun above the horizon isn't good enough. It needs to be at least 14 degrees above the horizon. When the Sun is on the horizon, ambient outdoor light is approximately equal to that of a school classroom or an office, whereas at noon, it's about 64 times as bright in the shade, and 512 times as bright in the sun, or about that of a camera flash about 2 feet away(sun) or 6 feet away(shade). By the time the sun has dropped to 6 degrees below the horizon, its light is dimmed by another factor of 1000 compared to sunset. Ambient outdoor light is now approximately equal to that in a TV-lit room. Not bright enough for most things, as the theory goes. By the time the Sun has reached 8.5 degrees below the horizon, its light is reduced to be about the same as a full moon directly overhead. At 12 degrees below, light pollution in most medium-to-large cities makes it invisible. In the absence of the Moon, or city lights, it's about the same as a room illuminated by a small LED (like the light on a surge protector to indicate that it's on). Sunlight becomes less than half of total illumination, away from city lights, in the absence of the Moon and Venus, at 14-15 degrees below the horizon, not 18. At 18 degrees below, its light contribution, now a paltry 1 percent of total sky illumination away from city lights, in the absence of the Moon and Venus. It's about the same as that of a dark room at night with the door closed, and LED in the neighboring room, and its reflected light coming through the crack under the door!

Answer 6

about one tenth (1/10)