Suppose the apparent diameters of the sun and the moon are exactly the same (which in fact very close to the real situation). If the moon had a perfect mirror surface, would the reflected visible light of a full moon (at night) illuminate the earth with the same intensity as the visible light of the sun would do?

Or would this only happen if we place a giant flat perfect mirror which reflects the light of the sun during the night so that every person on the night side of the earth could see the sun?

  • 9
    $\begingroup$ This video has a related visualization: youtu.be/w8I25H3bnNw?t=131 This video considers the moon as a Disco ball, not a spherical mirror, but the effects would be similar. $\endgroup$ – Seth Feb 15 '17 at 20:44
  • $\begingroup$ Great video! It would be a nice idea to make a video of a circular mirror with a diameter equal to that of the earth, rotating around the earth at the same distance as the moon does, to see how the reflected sunlight illuminates the dark side of the earth. $\endgroup$ – Deschele Schilder Feb 16 '17 at 7:56
  • 5
    $\begingroup$ @Seth Worth clarifying that the portion you linked to (131 seconds in) shows the moon as a disco ball if it were at the distance of the ISS. $\endgroup$ – owjburnham Feb 16 '17 at 21:42

No, because of the sizes of their surfaces. Let's make these simplified assumptions:

  1. The Earth and the Moon are both spheres 1 AU from the Sun.
  2. The total amount of sunlight an object receives is proportional to the solid angle it takes up from the Sun's point of view.
  3. The Sun and the Moon are each visible from a hemisphere of the Earth.

Then the total amount of sunlight received by the sunlit hemisphere of Earth is proportional to the square of the Earth's radius, while the total amount of sunlight received by the sunlit hemisphere of the Moon is proportional to the square of the Moon's radius. Since the Moon is ≈1/3.67 the radius of Earth, it receives ~1/13.5 the total amount of sunlight.

Certainly, even a perfectly reflective Moon can't reflect more sunlight than it receives, so even if all of the light bouncing off of the Moon reached the Earth it would only provide brightness comparable to a cloudy day.

Of course, owing to the geometry, most of the light bouncing off of the Moon doesn't land on Earth; it goes off into space in directions that miss the Earth completely. Making another simplifying assumption, I think we can say that the fraction of it that reaches Earth is proportional to the fraction of the Moon's sky taken up by the Earth. The Earth has an apparent size of about 2 degrees as seen from the Moon, so its angular size is $2\pi\left(1 - \cos\frac{2^\circ}{2}\right) \approx 0.00096$ steradians. A hemisphere is $2\pi$ steradians, so the Earth occupies about 0.00015 hemispheres (about 0.015% of the Moon's sky). Now we have that geometrically, a perfectly reflective Moon should illuminate the Earth at about $\frac{0.00015}{13.5} \approx \frac{1}{90,000}$ the intensity of the Sun.

In real life, the light from a full Moon is about 1/480,000 the brightness of the noon Sun. Given the Moon's albedo is somewhere between 0.1 and 0.2 depending on the angle of incidence, and given the huge simplifications made in the above math, I think this indicates that we're in the right ballpark.

| cite | improve this answer | |

You seem to be asking if the reflection of the sun from a spherical mirror, a convex surface would be the same as the reflection from a flat mirror.

A convex mirror is dispersive

enter image description here

The image in the diagram above is a virtual image. Light does not actually pass through the image location. It only appears to observers as though all the reflected light from each part of the object is diverging from this virtual image location

In addition to absorption from the surface that both a flat and a convex mirror would have and so the energy would be lost, a lot of the rays will be dispersed to space and not reach earth. A flat mirror (always at full moon) has fewer losses.

From conservation of energy, a flat mirror could only give as much energy as it got from the sun, and the cross section of the earth is much larger than the cross section of the moon, so there will be proportionally that much less energy from this image compared to the midday sun energy.

| cite | improve this answer | |
  • 3
    $\begingroup$ @annav-But if we make the flat mirror as great as the diameter of the earth it can illuminate the night side of the earth to turn the night into day. $\endgroup$ – Deschele Schilder Feb 15 '17 at 17:01
  • 12
    $\begingroup$ Well, this is another problem, not the moon problem. $\endgroup$ – anna v Feb 15 '17 at 17:37
  • 5
    $\begingroup$ @descheleschilder even if you had a mirror that was exactly the diameter of the earth, the density of light hitting it is less because it must be 'behind' the earth and therefore farther away and because the reflected light is again dispersed where some of it will miss the earth. Only if the mirror were somewhat larger than the earth and were it able to focus all reflected rays on the earth would you upper bound of light density possibly be the same as what the earth gets from the sun directly. $\endgroup$ – JimmyJames Feb 15 '17 at 18:34
  • 4
    $\begingroup$ @JimmyJames It wouldn't have to be very far "behind" Earth compared to the distance from the Sun to Earth, so the effect of being further away could be marginal. A flat mirror would not disperse the light and if a flat mirror the size of Earth was properly aligned, pretty much all of the reflected light would hit Earth (again assuming that it is much closer to Earth than the Sun). Hence descheleschilder's idea of a flat mirror the size of Earth to turn night into day is theoretically possible (as far as the optics is concerned). $\endgroup$ – jkej Feb 15 '17 at 19:15
  • 2
    $\begingroup$ @annav-The example of the total solar eclipse is related to this problem, in the sense that it is the opposite of the problem in question. In the case of a total eclipse by an earth-sized moon, you turn the day side of the earth into night (by absorption of light), and in the case of the reflection of the sun by a flat, circular shaped mirror with the diameter of the earth, you turn the night into day by the reflection (instead of absorption) of light. $\endgroup$ – Deschele Schilder Feb 15 '17 at 20:24

I expect details of the mirror are not the point of the question. We will assume a perfectly reflecting mirror, and curve it so all the light is more or less spread evenly over the surface of the Earth.

No this would not be equivalent to a second Sun. As Anna says, the mirror would need to be bigger.

enter image description here

Edit: I have updated the sketch to show an annular mirror. To light up the Earth like a second Sun, it would have to intercept about as much light as the Earth and reflect it evenly on the night side. That means if the mirror was a flat disk, it would have to have about the same surface area as a flat disk the size of the Earth.

Since it is a few thousand miles farther from Earth, the Sun would be slightly dimmer. The area would have to be proportionally larger. Given sunlight intensity follows an inverse square law, the ratio would be $d_{Earth}^2/(d_{Earth} - \Delta d)^2$.

Given $d_{Earth} = 93,000,000$ miles, this is pretty close to $1.00$.

But both the Earth and the mirror would be curved. The area of the mirror would depend on how the mirror is curved and angled. Without those details, all we can really say is that it would be about the size of the Earth.

If this was a difuse reflector, there would be another inverse square law loss that Anna has covered.

I am assuming a mirror surface. It would reflect all the light on the Earth, and look like an annular sun.

| cite | improve this answer | |
  • 1
    $\begingroup$ @mmesser314-If the mirror is big enough we would see a perfect copy of the sun, regarding the visible light, everywhere on earth. I think indeed in the case of the spherical moon, the sun appears to be smaller and so the image won't be able to illuminate the whole earth. $\endgroup$ – Deschele Schilder Feb 15 '17 at 15:05
  • 1
    $\begingroup$ Yes. A mirror the size of the moon would be too small. From the sketch, it would have to be about the size of the Earth. $\endgroup$ – mmesser314 Feb 16 '17 at 0:08
  • $\begingroup$ @mmesser314 But, if the mirror were about the size of the earth, a large fraction of it would be in the earth's shadow during the full mirror-moon. $\endgroup$ – David Richerby Feb 16 '17 at 9:21
  • $\begingroup$ @DavidRicherby - I updated the sketch to show how it would work. $\endgroup$ – mmesser314 Feb 16 '17 at 14:42

Even if a circular perfect mirror has the diameter of the earth, and assuming the distance from the sun of the earth is equal to the distance from the sun to the mirror, this mirror will never be able to illuminate the dark side of the earth completely.

If the mirror's inclination is 0 degrees, then the light from the sun (if the mirror is on the opposite side as the sun) won't be able to reach the surface of the mirror because the earth prevents all light from the sun to reach the mirror.

If the inclination is such that all light from the sun will be able to reach the mirror, and the orientation of the mirror is such that all the reflected light reaches the earth, only a part (though a big part) of the light will reach the dark side of the earth. The other part will reach the daylight side of the earth, so a big part of the light will reach the nightside, while the other part will make a little part of the dayside brighter. A relatively small piece of the dark will stay in the dark.

When the mirror rotates in an orbit perpendicular to the plane in which the earth rotates then [again with a proper orientation, which is, in this case, a 45 degrees inclination with respect to the plane in which the earth rotates around the sun (the ecliptic)] half part of the dark side will be illuminated, while half part of the dayside will receive twice as much visible light.

In fact, after giving it a second thought, the circular mirror, due to its inclination with respect to the ecliptic, can't illuminate the earth as much as the sun does. In the previous example of the rotating circular mirror, you have to change the circular form into an elliptic form.

| cite | improve this answer | |

Looking at a perfect flat mirror, is like looking through a hole of the same size with a sun in the other side. Therefore, you will see a second sun only if you are aligned with it so you can see the whole sun through the hole. Since the sizes and distances are the same sizes an distances involved in a solar eclipse, the light of the mirrored sun will follow the same pattern as the shadow in a usual solar eclipse. Therefore:

  • The total light received by Earth will be small. It will be the same amount as the average decreasing in light in a solar eclipse. That is, as small as other answers have shown by different reasoning.
  • The area from where you can get the same light from the mirror than from the Sun will be as big as the area from where you can see a total solar eclipse - very small.
  • Nearby areas will get part of that light, just as some areas can see partial solar eclipses.

In summary, only an small area of Earth will get another sun and the average effect will be tiny.

| cite | improve this answer | |
  • $\begingroup$ @Pere-I understand your way of thinking because I was thinking along similar lines. This would answer would be true tough if there was a plane, circular mirror with the same diameter as the moon (and with the right orientation) circling the earth. Because a sphere shaped mirror will diminish the amount of light reaching the earth (see the first answer by anna v), the amount of light reaching the earth will be less than the "amount" of shadow in a full solar eclipse. $\endgroup$ – Deschele Schilder Feb 17 '17 at 14:01
  • $\begingroup$ @descheleschilder Yes, this answer covers the case of a perfect flat mirror with the same apparent diameter of the Moon. If the mirror were spherical the Sun could be seen by anyone able to see the mirror from the Earth night side, but this case is already covered by the accepted answer. $\endgroup$ – Pere Feb 17 '17 at 14:23

As you ask it, no. Any mirror, spherical or not, would rob the light of some energy (change its frequency) as the mirror would be slightly heated and pushed in the process of reflecting he light.

It would make nights less dark, but not provide nearly as much light or energy as a second sun.

| cite | improve this answer | |
  • 1
    $\begingroup$ -I made an edit in my question to make more clear what I mean.With the observed diameter, I meant the apparent diameter, and I was referring to a perfect mirror, so no energy is lost in the reflection. $\endgroup$ – Deschele Schilder Feb 15 '17 at 16:57
  • $\begingroup$ The question says that the observed [apparent] diameter of the moon and sun are the same. $\endgroup$ – David Richerby Feb 16 '17 at 9:23
  • $\begingroup$ He changed it. It used to say earth. $\endgroup$ – Jake Watrous Feb 16 '17 at 12:41
  • $\begingroup$ ... so amend your answer to reflect the (now corrected) question. $\endgroup$ – Emilio Pisanty Feb 16 '17 at 14:50
  • $\begingroup$ Also, the proposition that the mirror would 'rob the light of some energy', particularly by changing its frequency, is pretty much completely wrong, particularly if you want to blame the heating for it. Heating would diminish the amplitude and is explicitly ruled out as the question posits a perfect mirror. The frequency would only change via a Doppler shift, which would be ridiculously minuscule. $\endgroup$ – Emilio Pisanty Feb 16 '17 at 14:52

Not the answer you're looking for? Browse other questions tagged or ask your own question.