We can indulge in a little thought experiment which helps illustrate the concept of meta-stable equilibria.
Earth orbits the sun at a distance of about 150 million kilometers, which provides a path around the sun conveniently close to about a billion kilometers, a path which the planet traverses each year. The planet travels thru space at about 31 kilometers per second to achieve this “round” trip, leapfrogging the whole way with its co-orbiting body Luna.
The maximum from Earth the the moon is about a little over 400,000 km. For the purposes of our thought experiment, do let us tweak the distances just a little and imagine the moon at a constant 400 megameters, rather than the slightly eccentric elliptic orbit, and let the earth be in a circular orbit around the sun with a path length of exactly one billion km. Neither of these assumptions would be even noticeable if nature could be adjusted to fit these easily-calculated numbers.
If we go with those numbers, then there are 2500 earth-moon distances which add up to the path around the sun.
In our thought experiment, imagine replacing the moon with a twin of earth, then having 2498 additional twin earths joining those two, equally spaced at 400,000 km, along the 1 billion km circular path around the sun. Instead of orbiting each other, the 2500 earths stay on the circular path in an orbit which is metastable: if nothing tweaks anything, they continue in their path in a curved path.
For the sake of this thought experiment, let the rotation stop completely, or rather almost: make day length exactly one year. The same face is toward the sun always, tide locked, analogous to how one face of the moon is always toward our home planet. Assume away all the other planets and the moon, for we do not want any external gravitational influences messing up our thought experiment.
For the next part of our thought experiment, we must think long long ago, in a galaxy far far away. Darth Vader is on his Death Star. He has already destroyed Alderaan, for no reason other than that’s what Dark Siders do: evil. But the rebel forces, led by Luke Skywalker, are about to whoop his Dark-Side butt. Since this is a thought-experiment, we get to imagine a different outcome. Vader knows the rebels are about to drop a torpedo down his Death Star’s exhaust port, and knows what will happen, so he uses the super-emergency-only escape device, the Illudium Q-36 thermonuclear time bomb: it doesn’t actually explode, but rather it moves the entire Death Star elsewhere and elsewhen, perhaps to a long time from then in a galaxy far far away. With the only choices being blasted to smithereens or take his chances with time machine, Vader breaks the glass and activates the device, just as Skywalker hits his mark. Poof, the Death Star is gone, Skywalker’s torpedo explodes in empty space, dang! But… the Death Star is gone and Dantooine is saved.
The Death Star isn’t gone exactly, it was just transported to a long time from then, such as… now in a galaxy far far… here, right into the middle of our thought experiment, where we rounded up 2500 earths, evenly spaced in a circular orbit 400,000 km apart. Shoop, here he is, right between two of them, 200,000 km from the one to the right and the one to the left. There is nowhere to go, and no when to go: the super-emergency-only time machine is a one-use device. Vader and his Death Star are stuck, right here.
OK, very well then. Here we are. Now we are. Here and now we stay. Get used to it.
In a fit of Dark Side boredom, Vader uses his last two photon nukes at the two adjacent planets, destroying all life on both. Note to gentle readers: never go over to the Dark Side of the Force. Dark Siders do stuff like this, for no reason other than just pure evil. Why that bad old Vader!
Since the point of this thought experiment is to do a bit of orbit mechanics and math busting, we can go forward, in spite of Vader’s reprehensible behavior.
Before Vader and the Death Star arrived and parked between two of them, the 2500 Earths were in metastable equilibrium. But we, being physicsly-enabled math busters, can see that post-Vader, the system is no longer in equilibrium. So… what happens?
Nothing, one might say, if one is not physicsly-enabled. But if otherwise, then one knows that the Death Star exerts a slight gravitational attraction on the planets on either side. How much attraction is this slight attraction? Well, we go to the specification book in our thought experiment and learn that the Empire defense engineers designed the Death Star with a weight limit of a thousand aircraft carriers. None of the engineers back there and back then actually knew what is an aircraft carrier, those being still long long in the future in a galaxy far far away, but that’s what the spec says: no more than a thousand aircraft carriers. A modern carrier has a mass of about 100,000 tons, so thousand of them are a hundred million tons or a 100 billion kilograms. I do not know how big is a Death Star, but I have been aboard an aircraft carrier, and those things are astonishing beasts. A spacecraft with a mass of a thousand of those is mind-boggling. With that information, we can calculate.
The gravitational force two masses exert on each other is given by the equation
Fg = M1M2G/R^2
The equation tells us that the Death Star exerts a gravitational force on each planet of about 10 million newtons. Each planet pulls back with the same force, but since it is symmetrical in opposite directions, those forces cancel. But the Death Star pulls on the planets. Since the two planets have an Earth mass, about 6E24 kg, then the acceleration toward the Death Star is about 1.67E-16 meters per second squared. Things don’t happen fast, being as it would take 60 trillion of these kilo-carrier Death Stars to make one Earth.
After a year of accelerating toward the Death Star, the two planets have moved about 8 centimeters. OK, this is going to take a while, but this is only a thought experiment, and we can think quickly. Tridactyls live a long time. Proof: Yoda’s comment “When 900 years old you reach, look so good you will not.” After 900 years, the two planets have moved towards each other a distance of about 67 kilometers. At some point, we would need to calculate the gravitational attraction of the two planets on each other (since they are now over 1.3 km closer to each other than they are to their twin on the other side.
To offer a merciful end to our thought experiment, our flight of fancy would end up with the two planets crashing into each other with the Death Star between them, crushing it like a mosquito between two semi trucks speeding toward each other. Semi trucks make a woefully inadequate comparison however, which makes the word picture a gross understatement. It wouldn’t be hyperbole, it would be the opposite: hypobole?
A fully-loaded NASA Crawler, with a service structure and a Saturn V rocket has a mass of about 10 million kg. The internet tells me a mosquito has a mass of about 5 milligrams. Since the internet never lies, let’s assume it true.
The ratio of mass of our 1000-carrier Death Star to planet earth is analogous to a mosquito vs 27 fully-loaded NASA crawlers (must include service structure and Saturn V rocket.) My friends, space is big. We are not.
The two planets will eventually crush the Death Star like two caravans of 27 loaded NASA crawlers crushing a mosquito. It takes a long time of course, and every life form aboard that Death Star would have long since died of boredom. So being crushed between two dead planets would come as a merciful end to any remaining life.
Well OK then. That was dramatic: a two-planet collision. But wait, there’s more. A blogster always enjoys every opportunity to use the phrase “but wait, there’s more.”
It isn’t really as clear as a string of 2498 earths spaced evenly at 400,000 km except for a single 2 earth mass glob of molten rock with the remains of a Death Star somewhere within, spaced 600,000 km from its neighbors. The two-earth (plus Death Star) mass is about 1.5 times farther away from its neighbors, so even with (very slightly more than) twice the mass, the squared distance is 2.25 times as much. The distance squared goes in the denominator. So 2.0000…0001 times the mass divided by 2.25 means the double earth tugs on its neighbors slightly less than their neighbors. So… they system is not in equilibrium. The earths adjacent to the two-earth mass accelerate gradually away from the two-earth mass until they too crash into their own neighbors. When that double-event happens, we now have three double earth mass planets and 2494 earth mass planets. Then the process repeats again and again, until we eventually end up with 1250 double-earth mass planets. And once again… the whole system is in equilibrium, with those 1250 double planets spaced at 800,000 km rather than 400,000. All is fully at equilibrium once agin.
Well, almost. Extremely almost, but not extremely exactly. One of those double-mass planets has an extra Death Star mass within it. So that one attracts its two neighbors to itself until they crash together to form a 6 Earth Planet mass. Then the previously-described process creates a 622 four-earth masses plus a 6 earth mass on the opposite side of the sun.
But that system still isn’t in gravitational equilibrium. Once the Death Star arrives, there is no new equilibrium. That additional 60 trillionth of an earth mass destabilizes the whole system.
Oh please Math Geezer, can we stop this brutal thought experiment now? It is far too depressing! Sure we can. I do hope this flight of fancy has been educational. If nothing else, it illustrates why one should not go over to the Dark Side: it wrecks other people’s perfectly orderly thought experiments. Why that bad old Vader.