Escape velocity via rotational motion

Cycle

Escape velocity via rotational motion
« on April 28th, 2017, 10:30 PM »Last edited on April 29th, 2017, 11:11 PM by Cycle
So we all know that in order to slip the surly bonds of earth and dance the skies on laughter-silvered wings, we first need to attain an escape velocity of 11.186 km/s. If we go faster than that, we rise from earth even faster.

And we also know that the linear motion equations are fungible with their angular motion counterparts, thus kinematic linear motion principles hold true for kinematic rotational motion.

Linear Motion Variables:
--------------------
displacement = x1 -  x0
velocity = deltax / deltat
acceleration = deltav / deltat
--------------------

Rotational Motion Variables:
--------------------
angular displacement = o1 -  o0
angular velocity = deltao / deltat
angular acceleration = deltaw / deltat
--------------------

This means that we can design a rotational machine which exhibits "anti-gravity" effects.

How fast would this machine need to rotate in order for it to rise against Earth's gravity? Well, we need to make an assumption first. Let's assume the machine has a rotational radius of 1 meter.

Circumference = 2 * pi * r, so its circumference would be 6.2832 meters. For each revolution of this machine, its rotational component travels 6.2832 meters.

11.186 kilometers is 11,186 meters, so it would need to rotate at 1780.303 revolutions per second, or 106,818.18 RPM to attain escape velocity of 11.186 km/s.

Now, there's no way any metal known to man could withstand the centripetal force of 106,818.18 RPM at a radius of 1 meter without ripping apart, and in order to lift more than just the rotational part of the machine, it would need to spin even faster than that. So a solid metal disc is definitely out.

But...

We all know about mercury vapor lamps, they're a gas-discharge lamp which ionizes mercury via an electric arc in a quartz arc tube, then that plasma emits light at the spectral lines of mercury. They generally operate at close to atmospheric pressure, so there's no high pressure risk of explosion.

Now, what if one were to make a quartz arc tube and a circular 1 meter tube of thick high-strength glass, ringed with coils. Those coils would be powered in a round-robin fashion at 106,818.18 RPM or higher to cause that mercury plasma to rotate in the tube at the desired speed. If one mirrored the inside of the tube, more of the energy would remain in the tube (rather than being emitted as light) to keep the mercury plasmized, reducing energy consumption. Thermally insulating the tube would reduce energy consumption even more, although the extra weight would require driving the coils at a higher rotational frequency.

The laws of physics dictate that the machine would rise.

Let's look at something larger, perhaps 10 meters in radius, giving us a circumference of 62.8318 meters, and a minimum required rotational speed of 178.0307 revolutions per second, or 10,681.8431 RPM. That's easily doable, and pushing the rotational frequency higher gives more lifting capacity.

Cycle

Re: Escape velocity via rotational motion
« Reply #1, on April 28th, 2017, 11:26 PM »Last edited on April 29th, 2017, 07:36 AM by Cycle
To recoup the cost to experimenters of purchasing the mercury, here's an idea... electrolyze water such that you're separately capturing the resultant hydrogen and oxygen. Ionize the hydrogen to strip the electrons, leaving you with protons.

Then fire high-energy electrons at a tungsten target to get bremsstrahlung radiation (circularly polarized MeV gamma rays). Some of the gamma rays will scatter from the tungsten target to produce electron-positron pairs, which can be separated via a magnetic field.

Run your protons from the first step through the positrons you're generating in the second step. Some of the protons will convert to neutrons.

Fire those neutrons at your mercury, and you've got yourself gold. You've transmuted the 0.15% of 196Hg in your mercury sample into 197Hg. It is radioactive, and will decay via electron capture to 197Au, gold's only stable isotope.

Spin the mercury in a centrifuge as you do so, so the gold settles out away from the neutron beam and isn't transmuted any further. Sell the gold to recoup the cost of the mercury. :D

As for the rest of the mercury, the remaining 99.85%... you'd have to hit it with electrons from a high voltage source of at least 30,000 volts in hopes of forcing an electron into the nucleus. Do that and one of the protons converts to a neutron, making an unstable nucleus which will decay to gold... but it'd be a radioactive isotope, and will eventually decay to another element.