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BitWiz
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jbriggs444 said:There is no point invoking relativistic complications until this much is understood.
Thanks, jbriggs, I'll work on it.
Chris
jbriggs444 said:There is no point invoking relativistic complications until this much is understood.
That's not how rockets work, jbriggs.jbriggs444 said:For propulsion of a rocket ship at non-relativistic speeds, if it takes mass ratio 1/x to get from 0 m/s to 1 m/s then it takes mass ratio 1/x2 to get from 0 m/s to 2 m/s
BitWiz said:That's why I'm having trouble with rocket KE measured from Earth. A rocket is an independent object in its own frame as soon as its acceleration overcomes gravity, imo, and its KE is undefined. It's KE would be exhibited in a collision with Earth, but then it would no longer be independent.
yeah. But once the exhaust velocity is a significant fraction of the speed of light, the classical equation no longer works. The relativistic equation (when the exhaust velocity approaches the speed of light) is:BitWiz said:With classic Tsioilkovsky, delta-v is based in part on the difference in mass before and after the burn. In the example where I accelerated a single proton as the propellant, these masses are effectively the same such that
ln Mbefore/Mafter
is close to zero, which requires the specific impulse (Isp) to be huge. That's fine, but Isp, is based on exhaust velocity, and in relativistic Tsiolkovsky, I think Isp = effective velocity = actual exhaust velocity in a vacuum.
Since effective velocity is capped < c, we can never get to this Isp unless we allow the mass of the propellant to grow. I don't see this term in relativistic Tsiolkovsky. I presume I'm missing something very important, I just don't know what it is.
BruceW said:yeah. But once the exhaust velocity is a significant fraction of the speed of light, the classical equation no longer works. The relativistic equation (when the exhaust velocity approaches the speed of light) is:
[tex]\Delta v = c \frac{m_0^2 - m_1^2}{m_0^2+m_1^2}[/tex]
where ##m_0## is total rest mass of the rocket (payload and propellant) at the start of the journey, and ##m_1## is the final rest mass of the rocket. And since we don't care about a return journey, we can just say ##m_1## is equal to the rest mass of the payload.
D H said:That's not how rockets work, jbriggs.
yeah, true. So for the first stage of the journey, we could accelerate the rocket to something like 10% speed of light, then coast for most of the journey, then near the end of the journey, we'd need to accelerate in the opposite direction, to get back to zero velocity relative to the sun. The mass of fuel required for the deceleration is roughly 10% of the payload. And so (with a quick calculation), the mass of fuel required for the acceleration would need to be 11/10 of the mass for the deceleration. So the total mass of fuel needs to be 21% of the payload.sophiecentaur said:But, presumably, we want to slow up at the other end of the journey and stop of at the star Shangrila.
BruceW said:yeah. But once the exhaust velocity is a significant fraction of the speed of light,
sophiecentaur said:I know the answer to this (my following) comment will be "Technology will take care of it" but . . .
It will not just be a matter of traveling to another particular star - chosen as a result of Earthbound (or at least, Solar bound) observations. This imagined trip would need to be totally open ended. It is highly unlikely that our Scientists will have successfully selected one ideal planet, remotely. The 'trip' will be more of a 'tour', which would include visits to a number of candidate solar systems, separated by 'inter-stellar' distances (i.e. probably thousands of light years) before a serious candidate could be found for 'colonisation' and terraforming. So you need to multiply all the previous resource calculations several times. The 'ship' will need to be incomprehensibly massive and a nice place to live in.
I don't think this is so much of a technological problem as a sociological one. Any ship that's designed to suit human passengers on a permanent basis (i.e. over several generations) is actually going to be a very suitable environment to live in anyway. If a suitable ship design were ever to be completed then you would already have your ideal replacement for Earth. It would need to be; after all, it would consume a large chunk of terrestrial (or even solar system) resources. Humans would have, in fact, come up with their own artificial planet environment so why would they want to get off that onto a hostile, unknown world?
If the reason for leaving Earth were some impending disaster then, by leaving the Solar System, they would have achieved that. All they would need is a 'nearby' star to orbit (no Goldilocks planet needed) with some nearby asteroids that could be used as a source of materials.
As for the need for the human population to expand (a commonly held belief), in the next few generations, humans will have intellectually grown out of the Darwinian urge to breed and breed and be going for quality of life rather than quantity of people. Birth rate and standard of living are already strongly connected - despite the influence of the existing primitive religious beliefs.
sophiecentaur said:"Obligation"? To whom?
Ryan_m_b said:It's probably for the best if we all try to steer the discussion back to the original question on the technical/economic feasibility of propulsion good enough to make interstellar crossings in a reasonable time frame. Discussions of why this should be done (if possible) are interesting but tangential.
phyzguy said:...and that you can keep a 1 GWatt laser aimed at it over interstellar distances
That could be an excellent idea for some scenarios but, eventually, the inverse square law comes into play and there will be some distance where you just can't focus your 'motive beam' effectively and most of your energy gets lost. I don't know the optics of this but I reckon you would be limited to launching small vehicles into big solar orbits and no more.phyzguy said:I'll shut up about the motivations. One comment on the feasibility that hasn't come up in these discussions. This is the idea that you "leave your rocket at home". A large stationary laser or mass driver can fire a beam into space which the space vessel intercepts to provide propulsion. This way the fuel does not have to be carried by the vessel, and it greatly improves the trade-offs. For example, suppose I want to send a small probe to Alpha Centauri. I build a large stationary laser, and aim it at the probe, which has a large reflector to reflect the laser light, thus continuously gaining momentum. Since for light, E = pc, the probe will accelerate with an acceleration a = 2P/(mc), where P is the power of the laser, and m is the mass of the probe. A 100 kg probe and a 1 Gigawatt laser will give you a proper acceleartion of about 0.07 m/s^2, and get you to Alpha Centauri in about 40 years, at which point you are traveling at about 0.2c. Of course, this assumes that a 100 kg probe is big enough to actually be useful, that you don't want to decelerate when you get there, and that you can keep a 1 GWatt laser aimed at it over interstellar distances, but you get the idea.
Nugatory said:When we say "the frame of <something>" or "its frame" where "it" is some thing, that's just a convenient and somewhat sloppy shorthand for the more precise "a frame in which <something> has velocity zero".
Thus, the kinetic energy of the rocket is perfectly well defined in both the Earth's frame and the rocket's frame, just as the kinetic energy of the Earth is defined in both frames. All objects can always be described in any and all frames, and the dynamical quantities such as momentum and kinetic energy are defined for all objects no matter which frame you choose to do the arithmetic in. The numerical values of some of these (momentum and kinetic energy, for example) may be frame dependent, but they are not undefined.
It's not undefined, it's zero - this follows from the fact that the comet has zero velocity relative to you. You don't need to open the window to know that the comet has zero velocity relative to you.BitWiz said:T
A hollow comet is closing in on Earth. I am perched inside the comet, and to my way of thinking, I'm weightless and at rest. There is no measurable velocity or acceleration. To me, my comet has undefined KE.
Its the same physics whether you're looking out the window or not. The comet's KE was zero relative to you before you looked out the window and it's still zero after you've looked out the window.I drift over to my comet window and look out. Holy cow, a planet is coming at me. Look at all that KE! And it's accelerating. What incredible power is required to make a planet accelerate that fast?
Is something similar going on when using Earth-bound KE measurements to determine the fuel requirements of an interstellar rocket? I am inside my rocket. When not accelerating, I am virtually at rest. Whenever I DO accelerate, I will always experience the same amount of acceleration per fuel unit (disregarding mass loss of the propellant), yet Earth-bounders will see huge jumps in KE. Where is the disjunct?
very true. I'm only trying to counter what BitWiz seems to think - that it is not even possible in principle. I agree that sending people on an interstellar voyage is definitely not possible with technology in the near future. It would be the very far future. And by that time, it is possible that other technology would have arisen, which we can't even predict at the moment. So maybe rockets are not the kind of technology that would get people there anyway.Ryan_m_b said:An exhaust velocity that high would require significant advances (but then again every suggestion in this thread would). The highest estimate I could find is for a beamed-core antimatter rocket that could have an effective exhaust velocity of ~.7c
Beamed[/PLAIN] Core Antimatter Propulsion: Engine Design and Optimization
Ronan Keane, Wei-Ming Zhang
But that would require significant amounts of antimatter. The second order consequences of a world in which antimatter can be mass produced are far more daunting than sending something interstellar.
Nugatory said:It's not undefined, it's zero - this follows from the fact that the comet has zero velocity relative to you. You don't need to open the window to know that the comet has zero velocity relative to you.
Its the same physics whether you're looking out the window or not. The comet's KE was zero relative to you before you looked out the window and it's still zero after you've looked out the window.
BruceW said:yeah. But once the exhaust velocity is a significant fraction of the speed of light, the classical equation no longer works. The relativistic equation (when the exhaust velocity approaches the speed of light) is:
[tex]\Delta v = c \frac{m_0^2 - m_1^2}{m_0^2+m_1^2}[/tex]
where ##m_0## is total rest mass of the rocket (payload and propellant) at the start of the journey, and ##m_1## is the final rest mass of the rocket. And since we don't care about a return journey, we can just say ##m_1## is equal to the rest mass of the payload.
BruceW said:very true. I'm only trying to counter what BitWiz seems to think - that it is not even possible in principle. I agree that sending people on an interstellar voyage is definitely not possible with technology in the near future. It would be the very far future. And by that time, it is possible that other technology would have arisen, which we can't even predict at the moment. So maybe rockets are not the kind of technology that would get people there anyway.
BitWiz said:So I come back to my original question: are Earth-observer-based KE calculations fair when determining fuel requirements for a rocket?
I didn't explain myself very well. I'm the observer. I'm trying to determine the KE of the comet to something else. There is no something else, therefore I think my KE is undefined.
haha, awesome. Well, they were definitely talking about the near future. No-one can say anything for certain about technology in the far future. I think the bad news is that neither of us will be getting a starship anytime in the near future.BitWiz said:Hi, Bruce,
No, I'm not advocating the pessimistic position of the Joint Propulsion people, and in fact, I'm pretty upset about it. I really want my own starship, and how am I going to get one if the propulsion people have already given up?!
I admit to trying to "hide" my feelings since what I really want is unbiased information, but perhaps I went too far advocating the propulsion expert's position, which -- if the posts I've seen so far both here on PF and elsewhere represent the trend -- is a majority opinion. However, "Poppycock!" is not a reasoned counterargument, so I need bullets. Pass the ammo, please.
Chris
Nugatory said:...
Yes, because no matter which observer you use to base the kinetic energy calculations on, you will get the same answer for the fuel burn and amount of energy that has to be generated by the rocket's propulsion system to send the rocket on its journey. ...
The kinetic energy in the Earth's frame is well defined and makes sense but the real question is does that accurately represent the energy actually expended from the ships point of view. I argued this with a friend who says yes, it does. I'm still struggling with the concept.Nugatory said:When we say "the frame of <something>" or "its frame" where "it" is some thing, that's just a convenient and somewhat sloppy shorthand for the more precise "a frame in which <something> has velocity zero".
Thus, the kinetic energy of the rocket is perfectly well defined in both the Earth's frame and the rocket's frame, just as the kinetic energy of the Earth is defined in both frames. All objects can always be described in any and all frames, and the dynamical quantities such as momentum and kinetic energy are defined for all objects no matter which frame you choose to do the arithmetic in. The numerical values of some of these (momentum and kinetic energy, for example) may be frame dependent, but they are not undefined.
What about the recoil of the gun? If you choose to use a frame where the gun is moving, the energy subtracted from it by recoil is non-negligible and accounts for the discrepancy you see.bob012345 said:So the question is how did the bullet end end up with 4K of kinetic energy wrt the ground frame when only K of additional chemical energy was added to its initial energy of K wrt the ground? I'm stumped.
The recoil would make the relative velocity wrt the ground frame less, not more. I could devise an equivalent frame such as another planet moving past us such that the recoil effects would be the same as in the ground frame. Basically, I think we can ignore the recoils. Treat it as an idealized problem please.jbriggs444 said:What about the recoil of the gun? If you choose to use a frame where the gun is moving, the energy subtracted from it by recoil is non-negligible and accounts for the discrepancy you see.
This is incorrect. Recoil is the source of the discrepancy. The recoil of the gun makes the gun's energy less, not more. That's what balances the books with the extra energy that shows up in the bullet.bob012345 said:The recoil would make the relative velocity wrt the ground frame less, not more. I could devise an equivalent frame such as another planet moving past us such that the recoil effects would be the same as in the ground frame. Basically, I think we can ignore the recoils. Treat it as an idealized problem please.
I don't get that especially if the recoil is not going to change the velocity of the plane much since it has an almost infinite mass by comparison. Also, I showed a scenario where the recoil is the same in both frames.jbriggs444 said:This is incorrect. Recoil is the source of the discrepancy. The recoil of the gun makes the gun's energy less, not more. That's what balances the books with the extra energy that shows up in the bullet.
Infinite mass times an infinitesimal change in velocity gives a finite change in energy. The answer I have given twice now is still correct. Do the math.bob012345 said:I don't get that especially if the recoil is not going to change the velocity of the plane much since it has an almost infinite mass by comparison.
I think you are saying that the whole plane slowed down due to recoil just enough to 'balance' the energy so the total amount of energy of the system of plane, gun and bullet is only increased by the the amount of energy the gun powder released. The bullet is going about 2v and has about 4K but borrowed some of it from the planes vast reserves.jbriggs444 said:Infinite mass times an infinitesimal change in velocity gives a finite change in energy. The answer I have given twice now is still correct. Do the math.
Yes, that is correct.bob012345 said:I think you are saying that the whole plane slowed down due to recoil just enough to 'balance' the energy so the total amount of energy of the system of plane, gun and bullet is only increased by the the amount of energy the gun powder released. The bullet is going about 2v and has about 4K but borrowed some of it from the planes vast reserves.