I’m holding on to the tether for one of these really fast tether cars. Mine’s so fast, it goes the speed of light (ignore the physical problems with this; maybe my car’s composed of a single photon). How long must the tether be so that my revolution only takes, say, 2 seconds? I’m holding my end of the tether at arm’s length, about three feet.

You want the circumference of your circle to be two light-seconds. Dividing by 2pi, we find that the radius is over 95 million meters. I think we can safely ignore the length of your arm.

Smokin’

The strength of your arm, however, might be of some importance.

My first thought was that it was that easy (the radius of a 2 light-second circle) but for some reason I got hung up on the length of my arm, and ended up asking here.

So the tether’d have to be about 59,300 miles long. Thanks!

Except, the energy required to rotate the tether goes up as the speed increases. Plus, the apparent mass gets heavier. Plus, at relativistic speeds there’s no such thing as a straight rope.

Basically, the factor is IIRC 1/SQRT(1-(v^2)/(c^2)) so as V increases, it becomes 1 over a smaller and smaller number, until at C (which you’ll never reach) energy required is like 1/0 (work it out - 1, 2, 3, 4, …1000, …1000000, etc.

The closer V gets to C, the huger the factor by which apparent weight goes up, the huger the energy needed to accelerate a bit more, etc.

Plain old Newton a=vt and F=ma stop working at those speeds, they’re approximations that work well with speed dencetly below c.

Making the photon change direction to swing in a circle is going to take energy, so it is probably going to start spitting out gamma radiation (which you are going to supply the energy for via the tether). Better wear lead lined underpants.

So lets be somewhat more practical - outside of the event horizon of a black hole is a photon sphere, a zone that allows suitable photons to orbit the black hole forever (or at least until the black hole evaporates). So how overweight do you have to be to allow photons to orbit you at 53,900 miles?

And lest you think I am joking, they have created artificial black hole analogues using refractive materials to illustrate the concept

Wild. Tether cars and black holes. I love the Dope.

j

So much force must our hammer-thrower exert to keep the photon in spin at the end of the tether?

It depends on his/the black hole’s spin relative to the photon’s right? If he had 32,000 Solar Masses and were spinning at near *c*, it would be require less force to whip it, right?

High school physics problem formulation required ASAP, please…

That’s way more complicated than it needs to be. Just make a big hoop with a mirrored inner surface, and let the photon bounce around inside that. No gamma rays, no gamma factor.

I’m not going near the relativistic implications of a rapidly spinning black hole. The OP specified a 30rpm (2 light second diameter) spin rate, is all.

To be honest, my math is pretty much a ballpark guess - there is a bunch of relativity (frame dragging and space-time distortion) to account for, and I didn’t do any of that. And concepts like *force* have no real meaning to a photon, because it has no mass. Which is why we end up talking about curved space-time, cause it avoids all those nasty divide-by-zero/NaN/infinity issues.

The amount of force required depends on the wavelength

The concept of the gravitational force on a photon AFAIC see has zero practical meaning. However the spin of the black hole does affect where the circular orbits of photons exist (it’s worth pointing out though that I believe all orbits of photons around a black hole are unstable and certainly the circular ones).

For a (non-rotating) Schwarzschild black hole there exists circular orbits in all planes at 1.5 times the Schwarzschild radius, but for a (rotating) Kerr black hole there exists two circular orbits for photons, both in the equatorial plane: one between 0.5-1.5 times (depending on the BH’s angular momentum parameter) the Schwarzschild radius for photons orbiting in the same direction as the black hole’s rotation and one at 1.5-2 times (ditto) the Schwarzschild radius for photons orbiting in the opposite direction to the black hole’s rotation. A near-extremal Kerr black hole (one which is spinning so fast it is almost a naked singularity) will have the orbits at just above 0.5 times the Schwarzschild radius and just below 2 times the Schwarzschild radius.

on the off chance that it’s not clear, the car in the video is tethered to the central pole, and not the man.