“You know moons are tidally locked, don’t you?”
Once upon a time I was talking to a friend about the setting for Bluehammer, and I mentioned that it was set on a moon orbiting a gas giant. I had already figured out that there would be a certain amount of time when the parent planet would occlude the sun from the moon, and in the interests of minimizing the amount of time in the day that this occlusion would last I put the moon in an orbit that would take a couple of weeks to complete,
But my friend was right: every moon in our solar system – even the asteroid captures like Phobos and Deimos around Mars – are tidally locked: the same face always points towards the parent.
For the orbit that my moon was in, that meant that its day/night cycle was a week of darkness followed by a week of daylight. The planet-ward side would always at least have the parent planet in the sky to light the way, except during the period of occlusion.
Would that glowing ball in the sky be enough for earthly crops to grow?
The answer to that required two pieces of information:
- how big would the parent planet appear in the sky?
- how much of the local star’s light would be reflected?
How Big In The Sky
The interesting thing about gas giants is that Jupiter is about as large as they get, in terms of volume. They can be more massive, but the actual diameter of the body won’t increase much because the gas will just compress. So, I used the radius of Jupiter and the orbital dynamics of my moon (particularly the orbit’s radius) to figure out the angle subtended by the parent planet in the sky. I put my moon in about the same spot as Callisto, and those are the numbers I use here.
Orbital radius: 1.88×106 km 
Diameter of planet: 1.4×105 km
The angle we want is at the pointy end of a very extended isosceles triangle. An isosceles triangle is two right angled triangles back to back, so if we calculate the small angle for one of those and double then that will be subtended angle of the gas giant in the sky.
Formula: tan(θ) = Opposite / Adjacent sides
Adjacent side = orbital radius = 1.88×106
Opposite side = radius of planet = diameter / 2 = 7×104 km
tan(θ) = 7×104 / 1.88×106 = 3.723×10-2 = 0.03723
θ = tan-1(0.03723) = 2.13˚
parent planet subtended angle = 2θ = 4.26˚
That is about eight times bigger than the moon’s subtended angle of 0.52˚, which of course happens to be about the same as our Sun.
How Bright On The Ground
The amount of energy reflected from a planetary surface is the albedo of that surface.
The obvious comparison is with our own moon, which seems nice and bright in the sky, at least at night when it is full. What is surprising is that the Moon’s albedo is really low at 0.12 – it’s comparable to that of coal. By contrast, Jupiter has a much higher albedo of 0.52, so that much more of the Sun’s light is reflected from that surface.
So, let’s say that 50% of the local sun’s radiation is reflected back from the parent planet.
That’s going to work out, right? A body in the sky eight times as wide as our Sun reflecting five times as much light as the moon will light the sky enough to grow crops, yes?
The difference is between emitted and reflected light: our sun emits a lot of energy, and we are primary consumers of it – the surface of Earth absorbs the energy from the Sun directly.
By contrast, a planet is a convex reflector so the relevant calculation is not just what angle the parent planet subtends in our sky, but what angle the child moon subtends in the gas giant’s sky. That will tell us what minuscule proportion of the local star’s energy received by the planet will be reflected back onto the moon.
Callisto’s radius = 2.4×103 km
Orbital radius = 1.88×106 km
tan(θ) = 2.4×103 / 1.88×106 = 1.277×10-3 = 0.001277
θ = tan-1(0.001277) = 0.073˚
Callisto subtended angle on Jupiter = 0.146˚
Let’s say that the sky is 180˚ wide:
fraction of sky angle = 0.146˚ / 180˚ = 0.000813 = 8.13×10-4 = ~0.08%
… except the sky is not a line, so we need to square that proportion to get the actual fraction of energy reflected from the parent planet.
fraction of sky area = 8.13×10-4 = 6.6×10-7
Oh, and halve that again to account for the albedo
proportion of energy received by Callisto = 3.3×10-7
Wow – that’s not a lot. But what does it mean?
Well, let’s say that the amount of energy collected by this moon is 1. The amount of energy collected by the parent planet is more than that, related to the relative surface area of the planetary disc.
Callisto radius = 2.4×103 km
Jupiter radius = 1.4×105 km
Area of circle = πr2
Jupiter : Callisto disc ratio = Jupiter disc area / Callisto disc area = (1.4×105)2 / (2.4×103)2  = 1.96×1010 / 5.76×106 ~= 2×1010 / 6×106 = 0.3×104 = 3×103
In other words, Jupiter collects about 3,000 times the amount of solar energy that Callisto does, but Jupiter reflects back to Callisto 3.3×10-7 of that: 3×103 x 3.3×10-7 = 10×10-4 = 1×10-3
Jupiter’s reflected light is about a thousandth of a Sun.
Which is a lot more than the moon gives us, but not likely to be enough for healthy crop growth.
So, I had to think of something else, but that is another story.
I mentioned Fermi estimates in my last post. This TED Ed video is a good introduction to these kinds of order of magnitude estimates.
 I also wanted to put it that far out for other reasons: so that the moon could have its own satellites, and so that the tidal forces on the moon would not make it unsurvivably volcanic. Living on Io would be tough in part for that reason.
 and also because Io is a toxic ball of sulphur.
 Pluto and Charon are mutually tidally locked, another point against Pluto being considered a planet in my book.
 a “week” here meaning seven Earth days. These words get slippery when not dealing with Earthly rhythms.
 my moon is a little larger than Callisto and so orbits a little further out.
 using kilometres rather metres since these are large scale distances, but the units will also cancel out for these calculations
 this brightness is partly because the moon’s surface has a strong opposition effect: it looks brighter when the light is behind the observer. That’s not relevant for the reflectivity of a gas giant, however.
 the π factor cancels out