There are two things we need to know to work out how much illumination the primary is giving the moon:
- How bright and far away is the sun?
- How reflective is the primary?
EDIT: I've added in some comparisons with light bulbs thanks to Patty Jansen pointing out that the human eye can adapt to see in lighting conditions much dimmer than the sun
Star light, star bright?
The amount of light that reaches your planet-moon system from its sun will depend on what kind of star it is. Stars come in different sizes and different temperatures. Most stars lie on what is known as the Main Sequence. Two notable exceptions are red giants and white dwarfs. The main sequence refers to the band of stars running diagonally through the Hertzsprung-Russell Diagram (HR diagram, previous links to two different images). Is basically a plot of how much light a star gives out (it's magnitude or luminosity) against it's colour or temperature. Stars are then divided into types (O, B, A, F, G, K, M) based on colour/temperature. Giant stars (other than blue giants) lie above the main sequence and white dwarfs lie below it. The sun is a G type star with temperature 5800 K (on the surface, that is; it's much hoter on the inside). K means Kelvin and is the standard unit of temperature. To convert from Kelvin and Celcius, you need to subtract 273, so the sun is 5500ºC (with rounding).
Using a star's temperature we can work out how much energy, in the form of light, reaches our planet. The first step is to assume that the star is a black body. This might sound conter-intuitive since the last word you're likely to use to describe the sun is "black", but from a physics perspective, a black body is something that absorbs all incident light and emits light based on its temperature. Well, I say "light", but really I mean electromagnetic radiation.
The Stefan-Boltzmann law tells us how much energy a black body emits based on its temperature. When we're talking about stars, this is called the luminosity. The formula for calculating luminosity is:
|A stars luminosity, given it's radius, R, and temperature, T. σ = 5.67 × 10-8 is the Stefan-Boltzman constant and π = 3.14|
The temperature has to be in Kelvin and the radius in meters to give luminosity in units of Watts (yes, like your light-bulbs) which is a measure of energy emitted per second. Radius and temperature are slightly trickier to come up with numbers for. If you're using a real star, you can just look it up on Wiki or Wolfram Alpha (Wiki even has a page listing the nearest stars to Earth). Otherwise you can make up a star with the characteristics you want such as temperature or class, then go to the second HR diagram I linked and look at the diagonal lines of radius. Whether you want a main sequence star, white dwarf or giant, this should give you an idea of radius (in units of the radius of the sun).
That's all well and good, but what we actually want to find is the light reaching a planet, not the total light emitted. Because stars emit light in all directions at once, their total energy output end up being diluted over an expanding sphere of light. Basically, not all the energy the sun produces hits our planet. It depends on how far away the planet is. This next formula will tell us how much energy hits the planet:
|P is the energy per second hitting each square meter of the planet and D is the distance from the planet to its sun.|
So P is the energy from the sun that hits a square meter of a planet which is D meters away from the sun. We're not quite there yet, but let's take a break and calculate some numbers. I'm going to work out the energy from the sun that hits the Earth/moon and Jupiter each second.
- Earth/moon are about 1.5 × 1011 m from the sun. The sun's radius is 6.955 × 108 m and its temperature is 5800 K. The energy hitting a square meter of the Earth or moon each second is 1400 Joules.
- Jupiter is 7.8 × 1011 m from the sun. The energy hitting a square meter of Jupiter each second is 50 Joules, which is about 3.6% of the energy hitting the Earth. Jupiter's greater distance from the sun means that the sun's energy is about 30 times more spread out by the time it gets there. (As I calculate below, this is still about 14000 times brighter than the full moon as viewed from Earth.)
Light doesn't get completely absorbed by the planet, however. Some of it reflects back out into space and can illuminate other nearby objects. The property which determines how reflective something is (in this context) is called albedo. The average albedo of a planet is a number between 0 (non-reflective) and 1 (absolutely reflective), which represents the percentage of incident light that will be reflected.
In practice, it's fairly easy to implement albedo. The reflected energy is the incident energy multiplied by the albedo. Just multiply P above by albedo, A, and you get the power reflected off each square meter of planet. You can look up albedos for different planets/moons on Wiki and elsewhere. (But we all know Wiki's the easiest. It lists albedos in the summary box on the right of the relevant page. If more than one is given it's the Bond albedo, not the geometric albedo, that you want.)
What we actually care about, however, is how much of that reflected light goes on to reach the moon our colony is built on. In a way, we just reuse the equations I've already included above. Instead of putting L into the equation for P, use the P from the sun multiplied by albedo, radius becomes the radius of the planet, and distance is now the distance between planet and moon:
OK, so this is getting increasingly more complicated looking, but remember that you only really have to do the last step. There rest are only there by way of explanation.
Now, one last thing before I calculate some more numbers. That last equation assumes that the primary planet appears full in the sky. If it's half full, you have to halve that number, if it's a quarter full you have to divide by four. Honestly? Just approximate.
- The moon has an albedo of 0.136. The energy the full moon is reflecting at the earth is 0.0037 W/m2.
- For the purposes of comparison, a 100 Watt light bulb from 10 meters away has a brightness of 0.02 W/m2.
- The Earth has an albedo of 0.306. The energy Earth reflects at the moon is 0.12 W/m2. So because it's bigger and more reflective, the Earth as seen from the moon gives off about 32 times more energy per second. That means the full Earth in the lunar sky is roughly 32 times brighter than the full moon in Earth's sky and six times brighter than a 100 W light bulb.
- Jupiter has an albedo of 0.343. Ganymede is 1.1 × 109 m away. The brightness of full Jupiter in Ganymedean sky is 0.07 W/m2. That means Jupiter is almost twenty times brighter than the full moon. Not surprising given how big it is in the Ganymedean sky. A half-full Jupiter would be 10 times brighter than the full moon, a quarter-Jupiter about 5 times as bright and so-forth. The varying quantity here is what fraction of Jupiter's disk is illuminated (and that we're working under the assumption that Jupiter reflects evenly in all directions). A quarter-full Jupiter would be about as bright as a light bulb and a full Jupiter would be as bright as three and a half light bulbs 10 meters away.
- Io is 4.2 × 108 m from Jupiter. The brightness of full Jupiter in Io's sky is 0.48. So Jupiter is shining a whopping 130 times brighter than the full moon. By comparison, the sun as seen from Io is only about 100 times brighter than Jupiter. Light-bulb-wise, Jupiter would be as bright as 24 100W light bulbs 10 meters away.
- For a bit of fun, the brightness of full Io (albedo 0.63) as seen from Ganymede varies from 1.2 × 10-4 W/m2 when it is at its closest point to Ganymede to 2.4 × 10-5 W/m2 when it is at its furthest. Neither of those are very bright, but it would still definitely be visible. It's about 0.6–3% the brightness of the full moon.
- And finally, let's say we put Jupiter at the same distance from the sun as Earth is. Now Ganymede would get around the same amount of energy from the sun per square meter as Earth does and Jupiter would be a lot, lot brighter. How bright? 1.9 W/m2, which is 500 times more light that Earth gets from the moon and as bright as almost 100 light bulbs from a distance of 10 meters.
And there you have it. A method for approximating how much light you'd get reflected from a gas giant planet (or whatever planet/moon/asteroid you like). Unfortunately this post ended up being a little bit more complicated than I had initially anticipated (where complicated really means more maths), but it's a small price to pay for painstaking accuracy... Well, some semblance of accuracy, at any rate. There are a lot of approximations in the above (for example, the albedo varies for different types of terrain; so Earth's albedo is higher over clouds than over forest), but on average, it's close enough. Phew!
One last thing I came across after writing this post. I was looking for something else and came across this photo of Jupiter and Io. Notice how the line between Io's sun side and dark side (called the terminator) is very distinct and solid, whereas Jupiter has a bit more of a gradient going from light to dark? This is because Jupiter has an atmosphere (a very thick one, but the effect applies to Earth's atmosphere too) whereas Io's atmosphere is whispy and not really much to write home about. The atoms/molecules/particles in the atmosphere reflect light in all directions, allowing it to diffuse through a bit, giving us that gradient from light to dark. Io, on the other hand, only reflects light off its surface, leading to the solid terminator you can see in that image. Just something to think about when writing those realistic descriptive passages. ;-)
Update: I photoshopped some Jupiters into skies to give a size comparison with the full moon. You can see them here.