One of the staples of SF art is images of alien worlds with satellites or planetary twins hanging low and huge in the daylight sky. This blog post brings he trope home by simulating what the Earth’s Moon would look like if it orbited the Earth at the distance of the International Space Station.
The author correctly notes that a Moon that close would play hell with the Earth’s tides. I can’t be the only SF fan who looks at images like that and thinks “But what about Roche’s limit”…in fact I know I’m not because Instapundit linked to it with the line “Calling Mr. Roche! Mr. Roche to the white courtesy phone!”
Roche’s limit is a constraint on how close a primary and satellite can be before the satellite is actually torn apart by tidal forces. The rigid-body version, applying to planets and moons but not rubble piles like comets, is
d = 1.26 * R1 * (d1 / d2)**(1/3)
where R1 is the radius of the primary (larger) body, d1 is its density, and d2 is the secondary’s density (derivation at Wikipedia).
And, in fact, the 254-mile orbit of the ISS is well inside the Roche limit for the Earth-moon system, which is 5932.5 miles.
The question for today is: just how large can your satellite loom in the sky before either your viewpoint planet or the satellite goes kablooie? To put it more precisely, what is the maximum angle a satellite can reasonably subtend?
We need to constrain the conditions a bit more, actually. By holding the mass of the satellite constant but decreasing its density we can push up its angular diameter. It’s not easy, mind you; because volume increases with the cube of radius, it will take roughly an eightfold fall in density to double the diameter.
But the artistically interesting cases are terrestroid rocky worlds or moons orbiting each other and it turns out their densities don’t vary by a lot. Earth’s is 5.51g/cm**3, the Moon’s is 3.43g/cm**3 and Mars’s is 3.93g/cm**3. For comparison the Sun’s density is 1.41 and Jupiter’s is 1.33.
Let’s start by setting the primary and secondary densities equal, then. This is convenient, because in the formula above we actually minimize d (and thus maximize subtended angle) when the density ratio is 1 (if it goes below 1 the primary and secondary switch roles). Our simplest case for minimizing d – two worlds of equal mass/density/radius – is actually pretty plausible.
OK, computation time. By hypothesis our worlds have the same radius R. The minimum distance from a point on the surface of the primary to a point on the surface of the secondary has to be 1.26 R. We must add R to get to the secondary’s center, because the visual angle subtended by the secondary will be that of a disk passing through the center of the secondary at right angle to the line of vision to it [[Edit: This turns out not to be quite right – see the comment thread – but close enough]].
The implied triangle has one corner at the surface of the primary world, another corner at the center of the secondary, and a third on the secondary’s limb subtending the maximum angle. x = 2.26R, y = R. Elementary trigonometry gives us the following formula:
a = 2 * arctan(R / 2.26 * R) = 2 * 0.41 radians = 46 degrees.
That’s actually pretty dramatic – 80 times the size of a full moon. More importantly the human visual field is about 150 degrees; our loomingest possible satellite could take up almost a fifth of it. Looks like those SF illos are fairly plausible after all.
Oh, and why did I title this “The Roche motel”? Because of the destructive effects when a hapless celestial body wanders inside the limit. “Satellites check in…but they don’t check out.”
(Two mistakes in the original posting have been corrected.)