So I’ve started out telling you that momentum-exchange tethers are great. Why? you might ask. Well, because a spinning momentum-exchange tether is capable of transferring a fair fraction of the orbital energy and angular momentum in its orbit to a payload in just a few minutes. That’s a pretty impressive trick. It gets even better when you learn that you can put that energy and angular momentum back just using electricity over a period of weeks to months, but that’s another lesson.
So we might want to try to figure out what kind of size of tether it would take to do these nifty tricks. The first place to start is to ask how big the tether needs to be–is it huge or tiny relative to our payload? To answer that question we need to know two things: what the velocity of the spinning tether is at its tip, and what it’s made out of.
There is a simple relationship between the mass of an untapered tether and the payload it is meant to catch and throw. Actually, it’s a ratio, much like a mass ratio in the rocket equation:
MR is the mass ratio, or the mass of the tether divided by the mass at the tip.
VR is the velocity ratio, which is the tip velocity divided by the characteristic velocity of the material.
The characteristic velocity of the material is the square root of two times the tenacity divided by the product of the safety factor and the material density.
An example might make this more clear:
Let’s say we wanted to throw a payload from LEO to a geosynchronous transfer orbit (GTO). That’s about a 2400 m/s delta-V beyond LEO, applied impulsively at LEO altitude. Assuming that the tether is in an orbit intermediate between GTO and LEO, and assuming that the tether will give half of the delta-V at catch and the other half at throw, let’s run the numbers:
Let’s assume we’re using a material with a characteristic velocity of 1600 m/s. To get the 2400 m/s of DV we need a tip velocity of 1200 m/s. This gives us a velocity ratio of:
Plugging VR = 0.75 into the equation gives:
So the untapered tether will have a mass 2.57 times that of its payload, in this scenario. This very simple analysis also assumes that the tether is connected to a counterweight that is, for all intents and purposes, infinite, so as to save us the step of computing the mass of the tether on the other side of the center-of-mass. So assuming we had a payload of, say, 1000 kg, the mass of the tether would be 2570 kg.
The equation I’ve just described shows that the characteristic velocity of a material has a physical meaning. If you look at the denominator of the equation, you can see that when the velocity ratio goes to one, the denominator goes to zero and the mass ratio goes to infinity. So the characteristic velocity is the maximum tip velocity of an untapered tether (of any length) with a safety factor of 1, at which speed it will break under its own tension.
In my next post, this simplified approach to tether modeling will get more complicated by describing TAPERED tethers whose velocity ratio can be greater than one. But the mathematics will be more complicated.
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