Since Angular speed is simply a mathematical concept to measure the rate of angular change...say RPM it is not a true speed measuring distance per time. To start by saying the angular speed is half the speed of light is in error. If the object is travelling half the speed of light. This will remain constant and the RPM will double.

The physics will hold true.

The object will traverse the same distance in the same time. The diameter of the circle has halved, therefore, the circumference of the circle has halved. This means the distance around the orbit has halved.

In the same time it took 1 revolution before it was reeled in, now it will take 2 revolutions. The distance and speed the objects travel remains constant.

Hi techno. "To start by saying the angular speed is half the speed of light is in error. If the object is travelling half the speed of light. This will remain constant and the RPM will double."

Check again! The challenge talks tangent velocity (V_t), not angular velocity (ω or RPM). BTW, the RPM will quadruple (not double) when the radius becomes half. This is the only way angular momentum can be conserved (L = mr²ω, because V_t=rω).

Jorrie

I think the problem that I have is with the string.

It is not the same as gravity.

I agree, to maintain a half radius orbit with gravity the same mass has to quadruple the speed due to the acceleration effect of gravity. To balance the forces of gravity and momentum at a lower orbit requires more speed for a constant mass.

If an object has a specific speed without gravity and only its momentum and direction constrained by a string, how does the object increase speed? In a gravity environment I understand it is from the gravity (sling shot effect).

Hi techno, interesting question you asked: "If an object has a specific speed without gravity and only its momentum and direction constrained by a string, how does the object increase speed?"

We all know from experience and by the principle of conservation of angular momentum that the angular velocity and tangential velocity must go up for smaller radii, but what is the physical mechanism behind it?

Hmm... Will have to sleep on that a bit. Getting late in my valley, especially over a glass or so of wine!

Anyone out there with a ready answer?

Jorrie

It's 'cause you can't see it . . . The physical mechanism behind the increasing speed is the increasing tension in the string. Momentum is conserved.

Hi Bill: you wrote: "The physical mechanism behind the increasing speed is the increasing tension in the string."

Must be close to the best explanation. You also wrote: "Momentum is conserved."

Nor quite - angular momentum is conserved, but instantaneous linear momentum and energy are not; they both increase. The increased energy comes from the work done by the cable and mechanism that reels the masses in.

In a non-circular orbit there is no net work done because the kinetic energy change is balanced by an opposite potential energy change. In the case of the spinning masses of the challenge, there is only work done when the radius is changing. If the radius decreases, the work is done by the reel-in mechanism and transferred to the masses by the increased tension in the cable.

Techno, does this answer your question?

Jorrie

That helps.

So the sharper the curve of the orbit, the more force is being applied to deflect the normally straight trajectory.

The force in the forward direction (momentum) + force towards the center results in a increase in the total vector length. Since mass is fairly constant, The result is an increase in speed.

" . . . angular momentum is conserved."

Good point! I wasn't thinking about the work done by the winding mechanism.

Now that we've got the mechanism sorted, how about the angular or tangent speed at half radius?

It is obviously a relativistic problem and one may expect that for the same rotation rate, relativistic angular momentum (which is conserved) will be higher than Newtonian angular momentum .

The trick lies in JD's reply #3 and my reply #4. Any further ideas?

Jorrie

Jorrie, your equation 6.8 of Relativity 4 Engineers gives the angular momentum parameter as a function of r, v_r, v_t, g_tt and g_rr. I presume that in a gravity-free place, g_rr and g_tt are both equal to 1, right?

That leaves below the line: √[1-v_r²/c² - v_t²/c²], meaning that we need to know both the radial and the tangent velocities to calculate the angular momentum. You have not given a radial velocity, so I cannot see an analytical way to solve your puzzle.

Hi Guest, well spotted.

You wrote: "... equation 6.8 of Relativity 4 Engineers gives the angular momentum parameter as a function of r, v_r, v_t, g_tt and g_rr. I presume that in a gravity-free place, g_rr and g_tt are both equal to 1, right?"

Right. You are also right about the radial velocity, but at v_t > 0.5c, the slow reel-in specified must give a negligible v_r. If you want to be (un-engineering-like) purist, assume a small radial velocity and add it vector-wise to the original 0.5c and then calculate the angular momentum. You'll then have more difficulty extracting the required v_t from the equation at r = 0.5r_o, but it can be done.

Jorrie

Given that the rate of change of "r" is very small WRT v_t, it nonetheless gives a resultant velocity, "v_R." What does this mean?

Hi Bill. You wrote: "Given that the rate of change of "r" is very small WRT v_t, it nonetheless gives a resultant velocity, "v_R." "

The v_r here means radial velocity, the dr/dt in your drawing, which is very small. The resultant velocity remains v_t for all practical purposes.

Jorrie

I guess I should have said, "Does this have any significance with respect to your original question?"

"Assuming that the conservation of angular momentum holds (which is true) and that the cable does not snap (which is doubtful), what will happen when the cable is reeled in to half its original length (or less)?"

I do realize that the resultant "remains v_t for all practical purposes."

Even though I was joking in my original post when I stated, "So that's how black holes are formed!," what happens when two massive objects rotating about a center of gravity, such as a binary star system, finally reach the limit of rotation speed and tangential velocity? Gravity is trying to pull them together, and tangential velocity is trying to force them apart.

Mathematical challenge, for sure . . .

Hi Bill, you asked: "... what happens when two massive objects rotating about a center of gravity, such as a binary star system, finally reach the limit of rotation speed and tangential velocity?"

Due to gravitational wave radiation, things like binary neutron stars do lose angular momentum and slowly spiral into close orbits. They do reach a limit in tangential speed of about 0.5c in their common frame,^[1] where the orbits become unstable and they rapidly spiral into each other, forming a single neutron star. If the combined mass is high enough, they may form a black hole.

Jorrie

[1] The chapter "How Orbits Work" on my website gives a good idea of the orbital dynamics in highly relativistic scenarios.

PS: Also look at the solution that I will post shortly.

The solution has been added to the opening post above.

Jorrie

Hello, are you asking about microstellar ojects re:gravity (so called gravitrons), or about the practical apps. using this force, on earth's enviornment? - Louis.

I am trying to find a used wheelchair, that is beyond repair, and ready for discarding,, and build a gravity wheelchair with my wheels. - Again - Louis.

As you know I'm not well up on this subject, but I remember reading somewhere that as an object increases in speed so does its mass? So hypothetical as the masses move towards the centre their mass will increases because of the increase in velocity, so to keep the formula in balance the speed increase will not be proportional to the distance, and the increase of the masses will ensure that they never reach the speed of light? Just a guess.

Regards JD.

Hi JD, not bad intuition!

The issue of mass that increases with velocity is no longer considered valid - mass means the rest mass and that's invariant. Some refer to relativistic mass, but that's also confusing. The best is to consider that the total energy of a mass in the frame of reference changes under a velocity change. The trick is how and how does it influence the angular momentum?

Jorrie

Hardly at all until you find a cable that can hold it.

I suspect the speed of light can be exceeded but our yardstick to measure it by will become meaningless.

Just intuition

Brad

Hi Jorrie,

Does the answer lie in terms of relative velocity? The relative velocity of mass1 to mass2 would have been the speed of light, at the start of the question. Is that velocity now 2c?

I am dedicated to making every conceivable mistake until I learn enough to become perfected. If you haul in the line, that adds to the velocity of the object so when the velocity of the object in orbit and the velocity of its being hauled into the center add up to C, you will be unable to haul it any farther toward the center. You will reach a stasis wherein you can haul it closer to the center at a decreasing velocity until the velocity of the object reaches C and then you will be unable to haul it any closer to the center because to do so would violate a fundamental law of physics.

You could apply any amount of energy to the drive mechanism you please, but it would stall regardless. You would have created an immovable object which would necessarily win out over the so-called irresistable force.

One measure of mass is how much energy it takes to move it, so if an infinite amount of energy will not move the object, I think you've made a case for its mass being infinite.

What do you say to that, jorrie?

Hi rbixby. You said: "If you haul in the line, that adds to the velocity of the object so when the velocity of the object in orbit and the velocity of its being hauled into the center add up to C, ..."

You're partially right, but remember, the one operative word about the reeling-in was 'slowly'. At more than half the speed of light tangentially, the radial effect is negligible. I take it that you reckon the objects will reach the speed of light just before the cable length halves?

Jorrie

I reckon that it will reach C at some point and before it reaches C, the sum of the orb's velocity in one direction and its velocity as it approaches the center of the string will add up to C, even if it is being reeled in at a speed of a beard-second per century (a beard-second is a very small distance someone suggested in the Journal of Irreproducible Results in the 60s to compete with light-years for very large distances; it is the distance a beard grows in the period of one second). anything plus x will eventually equal C, as long as x is always increasing, regardless of the rate of increase in "x" or the tiny-ness of "anything," as long as both are greater than zero. Henceforth this shall be known among all learned peoples as Gump's Law.

Now let us consider that when C is achieved, a person standing on one orb will see a seeming paradox in the sky--another orb of equal mass at the end of a string apparently unmoving on the sky. Would it look like that? Would a person be able to climb the string toward the other object? Could the orb be massive enough that a person could stand on the far side of it, while centrifugal force would seem to be throwing them away from the center at the speed of light? Wouldn't these tidal forces cause both orbs to take on the shape of a one-dimensional line of great length somehow tied to the string at one end. What part of this infinite line would actually be traveling at the speed of light? It would have to be the farthest tip of the line, perhaps several billion light years away, wouldn't it? Because if it were any other part of the line, any part farther from the center would be traveling faster than the speed of light, correct?

If two objects were spinning at either end of a shortening string their farthest ends traveling at close to the speed of light relative to the center of the string, how fast would the near ends be moving? They would appear not to move at all, I think, unless you watched them for many millions of years, after which they might appear to have rotated some very tiny bit of an arc second.

Hi rbixby, you said: "I reckon that it will reach C at some point and before it reaches C,..."

At what point (radius) do you feel it will reach C?

If someone could ride on one of the masses (which is possible if the cable is long enough) and if C can be reached (which is not possible), that person would have seen a weird world, but let's hold this discussion over a bit. There is still the original challenge to be solved. What happens when the radius is halved? What tangential speed, etc...?

Jorrie

So that's how black holes are formed! I didn't know until now . . .

Hi Bill, well, if you keep reeling in those masses, who knows?

Just a thought, are we talking about velocity or RPM, shouldn't time be included, as the string gets shorter the RPM increases but the velocity remains the same, masses travelling a shorter distance appear to be relatively moving faster referenced from the centre, energy is conserved by the increase in RPM, not velocity?

JD.

Hi JD, you said: "... as the string gets shorter the RPM increases but the velocity remains the same, ... "

Not quite. Both will increase. In Newton's scheme of things, at half radius the velocity would have doubled and the RPM would have quadrupled. Check reply #2 above. It is slightly different in Einstein's scheme, but both will still increase.

"... energy is conserved by the increase in RPM, not velocity?"

No, neither is conserved - both will increase. It is only angular momentum (L= m r v_t = m r² ω) that is conserved.

Jorrie