If the moon had a mirror surface, would the earth be equally illuminated as by the sun during full moon, or would it require a different mirror shape?

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.


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.


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.