How fast can an accelerator fire an object?

Saying it operated in the vacuum of space. is there a speed limit to
how fast an electro-magnetic repulsion accelerator can launch an
object? Naturally, there is the speed of light barrier, but I don't
think that's possible ... or is it?
And whatever speed limit such an accelerator had, could it be reached
if the accelerator was 60,000 miles long? If so, what would the rate
of acceleration from a standing start to achieving that speed at the
end of the 60,000 miles? If 60,000 miles wouldn't be enough to
accelerate an object to that speed limit, how fast could it get at the
end of a 60,000-mile run?
Thanks in advance!
Scott
Reply to
STJensen
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Dear STJensen:
A function of the amount of energy you can deliver, how you plan to dissipate excess heat, and how much mass you want to boost.
No.
The longer it is, the more trouble it is. The longer it is the lower acceleration you can have, and still get a good speed.
42.
No limit. No secret number. The more engineering you put into it, the more you get out.
David A. Smith
Reply to
N:dlzc D:aol T:com (dlzc)
There have been accelerators that reach speeds of a high fraction of c for about fifty years. They use a circular track to allow compact electromagnets. The particles involved are usually very light - protons etc.
Sychrotrons, cyclotrons etc.
Brian Whatcott Altus OK
Reply to
Brian Whatcott
The absolute speed limit is c (the speed of light). This has been experimentally verified in all high energy accelerators.
You could add all the energy you wish to the object in your accelerator, it still would not reach c. Get near to it but not reach it.
You can use the special relativity equations to verify it if you know how, or if not, I can give you an equivalent equation very easy to use that will show you.
Andr=E9 Michaud
Reply to
srp2inc
| Saying it operated in the vacuum of space. is there a speed limit to | how fast an electro-magnetic repulsion accelerator can launch an | object?
Let's put it this way. How fast can a gun fire a bullet? Let's assume the bullet leaves the gun at 1000 m/s. The gun will recoil. If the gun's mass is 1 kg and the bullet's mass is 1gram, how fast does the gun go backwards? Momentum is conserved, mV + Mv = 0 So, (1 * 1000) + (1000 * -1) = 0, the gun goes backwards at 1 m/s.
BUT! If the gun is going backwards at 1m/s then the bullet can only be going forwards at 999 m/s. Why is that?
And what if the gun was already moving backwards at 999 m/s when it fired? Or moving forwards at 1000 m/s? In the vacuum of space what do these numbers mean anyway?
| Naturally, there is the speed of light barrier, but I don't | think that's possible ... or is it?
There is no "speed of light barrier". What we DO have is a limit to how fast the electro-magnetic field that is propelling the projectile can leave the gun. If the projectile leaves the gun at 300,000,000 m/s and the gun is moving (relative to what?) at 1000 m/s then the projectile is travelling at 300,001,000 m/s. ALL motion is relative. Do not let cranks tell you the speed of light is... without say"But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v" Ref:
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| And whatever speed limit such an accelerator had, could it be reached | if the accelerator was 60,000 miles long? If so, what would the rate | of acceleration from a standing start to achieving that speed at the | end of the 60,000 miles? If 60,000 miles wouldn't be enough to | accelerate an object to that speed limit, how fast could it get at the | end of a 60,000-mile run? | | Thanks in advance!
A baseball cannot exceed the speed of the arm that threw it. Even if the pitcher runs forward (as a bowler in cricket does), the speed of the ball leaving the arm is still the speed of the arm. The speed of the ball arriving at the bat changes, all speed is relative. Therefore in the vacuum of space, move the accelerator, making it longer will not help.
Reply to
Androcles
Androcles:
Yeah, certainly.
Really?
Maybe a Newtonian twin paradox (?), or, more probably, because simple mechanics is too hard for you.
/mel
Reply to
mL
You don't seem to have gotten a completely straight answer, so I'll throw my $.02 in.
No. But you could get so close to c that it would be hard to measure the difference. The thing about the relativistic speed limit is that it takes more and more energy as you approach c. It makes more sense to talk about your energy, or your relativistic "gamma" factor, than it does to talk about speed for objects very close to c.
Thus, there are accelerators with gamma factors of 100, and others with gamma = 1000. In both cases the velocity is just a tiny bit different from c, but the energy of a particle with gamma = 1000 is 10 times the energy of the same particle with gamma = 100, and it has 10 times the momentum also.
You can never reach v = c (gamma->oo as v->c), but there's no limit to high gamma can be. An incredibly high-energy particle detected a few years ago, nicknamed the "Oh-my- god particle", had a gamma somewhere in the neighborhood of 10^20 as I recall. But still a speed v < c.
You could use work = Force * distance to figure out how much energy you could get on such an accelerator.
Let's say you can accelerate a mass of 1 kg at 1000 g's, i.e. you can exert a force of 9800 N on that mass. Then over 60000 miles (96500 km) you would give the mass 9800 N *
96,500,000 m = 9.46*10^11 J.
But actually, this isn't a relativistic speed. Let's check it with the Newtonian formula, which is OK at low speeds: KE = 0.5*mv^2
v = sqrt(2*9.46*10^11/m) = 1.38 *10^6 m/sec, or about 0.0046 c.
That gives a gamma of 1/sqrt(1 - v^2/c^2) = 1.00001
Not a sensible question since you can't achieve c. How about "what acceleration would be required to reach a gamma of 1000 at the end of the 60000 miles?"
So we want an energy of 1000*mc^2 = 9*10^19 J over 96500 km, which means the force should be 9.326*10^11 N, or the (initial) acceleration should be 9.326*10^11 m/sec^2 (95 billion g's).
At that acceleration, it would reach half the speed of light in 0.16 milliseconds, at which point it has already traveled 121 km.
- Randy
Reply to
Randy Poe
| > Saying it operated in the vacuum of space. is there a speed limit to | > how fast an electro-magnetic repulsion accelerator can launch an | > object? | | You don't seem to have gotten a completely straight answer, | so I'll throw my $.02 in. | | > Naturally, there is the speed of light barrier, but I don't | > think that's possible ... or is it? | | No. But you could get so close to c
Relative to what?
that it would be hard | to measure the difference. The thing about the relativistic speed | limit is that it takes more and more energy as you approach | c.
Relative to what?
It makes more sense to talk about your energy, or your | relativistic "gamma" factor, than it does to talk about speed | for objects very close to c.
Relative to what?
| | Thus, there are accelerators with gamma factors of 100, and | others with gamma = 1000. In both cases the velocity is just | a tiny bit different from c,
Relative to what?
| but the energy of a particle with | gamma = 1000 is 10 times the energy of the same particle | with gamma = 100, and it has 10 times the momentum | also. | | You can never reach v = c Relative to what?
(gamma->oo as v->c Relative to what?
), but there's | no limit to high gamma can be. An incredibly high-energy | particle detected a few years ago, nicknamed the "Oh-my- | god particle", had a gamma somewhere in the neighborhood | of 10^20 as I recall. But still a speed v < c. Relative to what?
| | > And whatever speed limit such an accelerator had, could it be reached | > if the accelerator was 60,000 miles long? | | You could use work = Force * distance to figure out how | much energy you could get on such an accelerator. | | Let's say you can accelerate a mass of 1 kg at 1000 g's, | i.e. you can exert a force of 9800 N on that mass. Then | over 60000 miles (96500 km) you would give the mass | 9800 N * 96,500,000 m = 9.46*10^11 J. | | But actually, this isn't a relativistic speed. Let's check it | with the Newtonian formula, which is OK at low speeds: | KE = 0.5*mv^2 | | v = sqrt(2*9.46*10^11/m) = 1.38 *10^6 m/sec, or about | 0.0046 c.
Relative to what?
| | That gives a gamma of 1/sqrt(1 - v^2/c^2) = 1.00001
Relative to what?
| > If so, what would the rate | > of acceleration from a standing start to achieving that speed at the | > end of the 60,000 miles? | | Not a sensible question since you can't achieve c.
Relative to what?
Not a sensible answer, "But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v". - Einstein.
Reply to
Androcles
Note that a struck baseball does move faster than the swung bat.
Reply to
PD
v> Even if the pitcher runs forward (as a bowler in cricket does), the
Has been explained before. He went berserk.
Dirk Vdm
Reply to
Dirk Van de moortel
Cosmic rays (and the Greisen-Zatsepin-Kuzmin, GZK, limit). Protons with 6x10^19 eV enegy are routinely observed, and that is the energy of a well-thrown baseball. If you have the budget you can impart an arbitrarily large energy to a body. Special Relativity then tells you how fast it will appear to travel.
A .357 magnum handgun pulls 50,000 gravities as a bullet rides down its barrel going from zero to 1100+ feet/second in 4.5 inches. If you extend that barrel to a lightyear length footnotes will intrude.
Reply to
Uncle Al
A lot sooner than that - (in Newtonian mechanics) at 50,000 g (500 km/s^2) it takes 600 seconds to reach the speed of light, implying a barrel 90 million km long - less than the length of a Golden Ship, iirc.
-- Peter Fairbrother
Reply to
Peter Fairbrother
Just a note. Cyclotrons were abandoned early in the 1940's precisely because they couldn't get particles anywhere near c, due to their dependence on mass remaining constant.
They then switched for circular track accelerators to the betatron and then to the various flavors of synchrotron.
Andr=E9 Michaud
Reply to
srp2inc
Now the Pentagon is gonna bother us for specs, estimates, and a PERT chart. First, we're gonna substitute an alloy .223... "8^>)
Reply to
Uncle Al
Of course you never rad my proof by wave mekanics thas a mote can reach celerity with finite energy. Hmm, the term I was a'looking for was coherential length, which becomes positive at .5c and above.
Reply to
Autymn D. C.

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