Re: Rail Guns don't recoil
- From: "Neil B." <neil_delver@xxxxxxxxxxxxxxx>
- Date: Sat, 6 Dec 2008 19:12:50 +0000 (UTC)
"Tom Roberts" <tjroberts137@xxxxxxxxxxxxx> wrote in message
news:mt2.1-11417-1228409365@xxxxxxxxxxxxxxxxxxxxxxxxxx
Canup, Robert E. (JSC-EV)[ESCG] wrote:Tom, I'm glad someone finally brought up the issue of momentum being
[absence of measured recoil on the rails of a railgun]
The attached photo shows a shot of the Naval prototype rail gun
firing.
Note the thin pedestal mounts used to hold the rail gun apparatus.
http://en.wikipedia.org/wiki/Image:Railgun_usnavy_2008.jpg
I'm not going to read that 93-page thesis. But I note that if one
instead fired a rocket along the rails of a railgun, one would not
expect to measure any significant reaction force on the rails. That
picture looks indistinguishable from a rocket doing just that.
So one must ask: how much momentum is contained in all that recoiling
fire and gas? How much is in the high-power electromagnetic radiation
involved? And where does all that fire and smoke come from?
carried in the "exhaust" analog of the rail gun: electromagnetic waves.
After all, that's the point of Feynman's charged disk paradox (from
_Lectures_ Vol. II). But I don't think there's enough in EM waves to
carry off the many kg*m/s shot out by a rail gun (Anymore than in the
case of me pressing two magnets together and then letting them go, etc.)
It takes a lot of EM power to have significant (linear) momentum. Note
that we need about 300 MW of power from a photon rocket to get a mere
one newton of thrust!
Also of course, statements about momentum being directed out to the
sides don't solve anything - the conservation law is a vector identity!
It seems to me the rail magnets, in near proximity to the projectile,
should have nearly complete reaction forces right against them instead
of needing to appeal to equipment below. So this empirical result
(correct I suppose?) is still perplexing.
Actually, I am skeptical of the claim of negligible reaction momentum
since I remember seeing proposals to use the reaction from rail guns to
apply thrust to asteroids! (Mine ferrous slugs for projectiles and maybe
nuclear plants for the power, make the asteroid into a "spaceship" etc.)
OTOH, that was likely a theoretical calculation, perhaps simplistically
assuming action-reaction: what if actual results are contradictory?
There has been a paradox going around about classical derivation of spin
from forces between neighboring charges, it really is weird. See for
example about the curious work and claims of Anders O. Wistrom and V. M.
Khachatourian, at the link
http://www.aip.org/tip/INPHFA/vol-9/iss-4/p8.html from _The Industrial
Physicist_, August/September 2003 (Volume 9, Issue 4):
"Spin and energy?free?
Most physicists would not expect startling new theoretical conclusions
to emerge from electrostatics, whose basic mathematical structure was
completed 150 years ago. Yet two researchers at the University of
California, Riverside, arrived at conclusions that, if true, would be
revolutionary. In a forthcoming paper ( J. Physics A: Math. Gen. 2003,
36, 6495), Anders O. Wistrom and V. M. Khachatourian say they have
proven mathematically that electrostatic forces among three charged,
perfectly conducting spheres will cause them to start spinning. This
conclusion, which the authors have derived from Coulomb?s law,
contradicts long-held assumptions about how electrical fields behave.
Equally striking, it implies that, in theory, an arrangement of three
such spheres could transfer unlimited amounts of energy into the
spinning spheres?a violation of conservation of energy."
See also commentary and responses at
http://www.aip.org/tip/INPHFA/vol-9/iss-6/p4.html.
I'm surprised not to hear more about this work anymore - anyone have
scoop?
(Note about rocket thrust: Take f = v(dm/dt) where dm/dt is the rate of
mass exiting the rocket. That's how we can derive the thrust from a
photon rocket, since we have f = c(dm/dt) equivalent = (dE/dt)/c.
However, there is a pseudo-paradox with rockets, since technically the
momentum of the whole rocket itself seemingly should have a term dp/dt =
v(dm/dt) where the same term refers to the changing "mass of the
rocket." However this is spurious because it is a changing redefinition
of the mass being considered; for any one particle the usual f = ma =
dp/dt still holds. (Yet many students and even text-book writers still
get confused.)
.
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