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Luna orbits Earth which in turn orbits around Sol. The period of the Lunar rotation is around 28 Earth days; i.e. one lunar day + one lunar night.

Unlike Earth, Luna is not possessed of an atmosphere. This leads to extremely sharp temperature, and illumination gradients. The side of Luna facing away from the Sun is as dark as the other is bright.

On a clear winter new-moon night starlight alone is enough to make out the general landscape, flora & fauna here on Earth.

What is the average illumination incident on the side of the moon facing away from the Sun?

p.s. I seek the approximate average amount of light incident per unit area on the dark side in terms of a ratio/percentage/fraction of the sun-illuminated side

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When the sun goes down on the Moon, Earth is waxing and at half full as seen from its surface on the near side. At lunar dawn Earth is a waning 'half-Earth', if you will. This is noticeable to us in the way you can see the full disc of a young waxing crescent Moon in the early evening, because Earth is lighting it up. This is known as earthshine.

Isidoro Martinez wrote an overview of the space environment that has this to say on page 20:

The Earth has a larger albedo than the Moon $(\rho_E/\rho_M=0.30:0.12=2.5)$, and larger size $((R_E/R_M)2=(6370/1740)2=13.4)$. An observer on the Moon will see ‘full Earth’ (at new moon) 4 times wider than the Sun disc, and 100 times brighter than we see full moon, corresponding to an apparent magnitude (if the observer reference point is exchanged in the definition of apparent magnitude) of $m= -17.7$ (as can be checked from the full moon magnitude, $m= -12.7$, and the ‘brightness’ ratio, $I_E/I_M=2.512^{(mEmM)}=2.512^5=100)$; notice that the illuminance it creates on the lunar surface is around $20 lx$, a good value for ‘ambient light’ in a living room, and for outdoor night lighting (Apollo 8 astronauts described relief features flying over the dark side of the Moon); full moon shine yields around $0.25 lx$ on the Earth surface (although it may reach $1 lx$ at great altitudes near the equator); good reading light is about $200 lx$ (up to $2000 lx$ for precision work, matching natural diffuse light; maximum Sunlit yields nearly $100 000 lx$). Notice also that the brightness ratio $(100:1)$ does not coincide with the product of Bond albedo ratio $(0.30:0.12=2.5)$ times area ratio $(13.4)$, which is $2.5·13.4 =34$, because of atmospheric absorption and directional effects.

Monitoring of earthshine from here on Earth has been used to monitor changes in Earth's albedo. Such study reveals that its albedo is rather variable depending on the part of the planet that is visible. The graph is from Earthshine: not just for romantics. It is based on a mathematical model which takes into account the difference in albedo of land versus ocean, but ignores clouds because modeling for them is much more complex. Ocean is much darker than most land (except where the sun is glinting off it towards the observer). Modeled like this, when the Pacific is facing the Moon the planet reflects half as much light as it does over Eurasia and Africa. Differences are especially big when the Moon is crossing its ascending or descending node, aligned with Earth's equator so the polar ice caps aren't very visible. The clouds distributed over the Earth dampen this variation a lot though. At any rate, the graph shows that the range of results over a month is from near 0 to 0.06 W/m2.

earthshine intensity over a month

Peter Thejll, Chris Flynn, Hans Gleisner and Andrew Mattingly

The image below shows the albedo of all of Earth's land masses, with data taken by the Moderate Resolution Imaging Spectroradiometer.

albedo of earth land masses

Credit: Image courtesy Crystal Schaaf, Boston University, based upon data processed by the MODIS Land Science Team

albedeo of different substances From Encyclopedia of the Earth: Albedo

Unlike the values in lux mentioned in the Martinez overview, these earthshine studies measure the energy of the light reaching the Moon. The studies seek to understand how climate and albedo interact, and so need to see all the energy in earthshine. The camera used in the study quoted above extends into the near infra-red range, and the energy from that invisible range would accounts for about a third of the values it obtains. The peak result in that graph is closely matched by simply multiplying Earth's average albedo, its projected area, and the energy in sunlight at this distance from the sun, and dividing that by the area of a hemisphere at the average distance of the Moon's orbit. That takes the energy reflected off the sunlit half of the planet and spreads it out to what it would be at some spot at the distance of the Moon.

$$\frac {(0.3) \pi (6.37 * 10^6)^2\,m^2)(1366\, W/m^2)}{\pi (3.85 * 10^8)^2\,m^2} = 0.112\;W/m^2$$

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The moon is tidal-bound to the earth. As a consequence, one (visible) half always faces earth. The other (invisible) half never faces earth. The invisible hemisphere was unknown to us until space travel.

If the invisible side of the moon is not lit by the sun, it experiences an "earthless starlit night". So the illumination would be the same as "a moonless starlit night, which is measured at 0.0001 lux" to quote OnoSendai's comment.

On the visible side of the moon, the incident illumination would be fully determined by light radiated from earth (if the pertinent location is not getting any sunlight.) But I do not know how much light we emit.

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