Let's say that we have a space station orbiting the Earth around the equator at a speed of 360 degrees per 24 hours, always being roughly on the line between the centers of the Earth and the sun. From the point of view of those on the station, the Earth would always look fully illuminated. (In case it makes a huge difference, let's say that it orbits at the same height as the ISS.)

Now, let's say that the half of the station facing the Earth is a large, dome-shaped window, allowing the entire planet (well, the half of it facing the station) to be seen at the same time. It doesn't have any windows facing the sun, however.

If the power were to go out on the station (thus disabling all artificial illumination and leaving the light reflected off of the Earth the only source of light), how well-illuminated would the station be? Would there be enough light to find one's way around the station, perhaps even doing some repair work?

Here on Earth, we can still see somewhat with a full moon as our only source of light. Now, the Earth doesn't seem to reflect quite as much light, but it's several orders of magnitude closer to the station than the Earth is to the moon, so the question is whether or not it's enough to be a passable source of light.

  • $\begingroup$ Something to consider is that if the space station is to orbit at a point where it always sees the fully illuminated earth, it would have to be at the Lagrange point $L1$, which is outside the orbit of the moon. So the light reflected from the earth would then again be falling off as $r^2$ from the earth. $\endgroup$ – tmwilson26 Nov 20 '15 at 21:16

This depends on the reflectivity of the objects and the size in the sky.

The visual albedo of the moon when near (but not at) full, is about $0.12$. Earth's is around $0.39$. Being in low-earth orbit, the actual amount will vary based on the terrain and atmosphere. If it's overcast below, the value could be much higher. You can assume that a given area of the earth would appear 2 to 3 times brighter than the same area of the moon.

The full moon has an angular diameter from earth of about 30 arcminutes or one-half a degree, so the angular radius is about a quarter of a degree.

From an altitude of $400km$, the angular radius of the earth would be $\sin^{-1}(\frac{6371km}{6771km})$ or about 70 degrees. This means the ratio of the apparent size of the earth versus the moon would be about $(\frac{70}{0.25})^2$ or more than 78,000 times larger. Along with the likely increase in albedo, it would be 100,000 to 200,000 times brighter.


This Q boils down to

How bright is Earthshine?

According to the usual source

Oceans reflect the least amount of light, roughly 10%. Land reflects anywhere from 10–25% of the Sun's light, and clouds reflect around 50%.

So the amount of sunlight reflected (i.e. albedo) depends on what part of the earth is facing your observer and how cloudy it is.


  • I can easily walk around at night in the light from the full moon.

  • The earth is a much larger object - in terns of subtended angle - when viewed from earth orbit, than the moon is from earth.

Therefore a "full earth" would be a bright enough source of Earthlight to perform a great many tasks.

Another interesting note about the ISS:

What's perhaps not best known is that the station's solar panel arrays are double-sided to also collect some of the sunlight reflected off the Earth's albedo (what our planet reflects from the Sun). They generate roughly up to 120 kW of power (on average about 84 kW)

- https://space.stackexchange.com/a/5596/3101

So Earthlight is strong enough to power (at least parts of) a space station.


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