Re: Physic help

[kgstar@most.fw.hac.com (Kelly Starks x7066 MS 10-39)]

At 4:28 PM 5/5/96, DotarSojat@aol.com wrote:
>On 5/3/96 at 9:07 am EDT, Kelly Starks wrote-
>
>>I have a table where I list various fusion fuel cycles.  It
>>lists the resulting energy in Mev.  For those of us who arn't
>>familure with translating Mev into anything, could someone
>>tell me what the speed of the resulting particals is?
>
>Yes, but I don't know what you can do with the answer.
>
>First of all, you have to coax two charged particles to penetrate
>each other's Coulomb barriers with some amount of accelerator
>(bombardment) energy, a few MeV, before there can be a nuclear
>reaction (or use very high temperature and pressure to cause a
>thermonuclear reaction).  This bombardment energy conveys a
>motion to the center of mass of the reacting particles which must
>be added vectorially to the velocity (which is in a random direc-
>tion) of each resulting particle with respect to their center of
>mass after the reaction.  We ignore this center-of-mass motion in
>the following analysis.
>
>Let's define the following quantities-
>     m = mass of lighter reaction product, in atomic mass units
>         (e.g., mass of proton = 1.00813 amu)
>     M = mass of heavier reaction product, in amu
>     E = "resulting energy" in MeV of reaction products (this is
>          with respect to their center of mass)
>        = [(m1 + M1) - (m2 + M2)] 931 MeV
>          (where 1 designates particles before the reaction and
>           2 designates particles after the reaction; for a mass
>           of 1.00000 amu, mc^2 = 931 MeV)
>     v = velocity of the lighter reaction product with respect to
>         the center of mass
>     V = velocity of the heavier reaction product with respect to
>         the center of mass
>
>After the reaction, the momentums of the two particles are equal
>and opposite (and in a random direction), and the sum of the kin-
>etic energies of the particles is equal to "the resulting energy,"
>i.e.,
>    m v = M V
>  0.5 m v^2 + 0.5 M V^2 = E
>  0.5 m v^2 + 0.5 M (m v/M)^2 = E
>  0.5 m v^2 [1 + (m/M)] = E
>   (v/c)^2 = 2[E/(931 MeV)]M/[m(M + m)]
>
>So, the velocity of the lighter particle, in units of c, is
>    v/c = sqrt(2[E/(931 MeV)]M/[m(M + m)])
>
>and the velocity of the heavier particle, in units of c, is
>    V/c = (m/M)v/c
>        = sqrt(2[E/(931 MeV)]m/[M(M + m)])
>
>I hope this answers your question.
>
>(The velocity calculations can be made simpler without great loss
>of accuracy by considering only whole numbers of amu, e.g., 1 in-
>stead of 1.00813 for a proton.)
>
>Regards, Rex


???

Hum, I'll have to try going through that step by step with some real data.

Unfortunately given the numbers from Tim's responce, and assuming no
losses, it looks like we can't get even close to the specific impluse I was
assuming.  (I feel like an air head.)  I had thought the numbers seemed
higher than the articals I was reading seemed to imply.  But I was assured
by someone (one of the physics students) in the group that I could assume
the higher vecocities/spec impluse.  Obviously with the lower numbers, the
ship isn't usable even for a flight to Alpha Centauri.

I am bumbed!  I feel like whiping the explorer class files of the Daves
proto-LIT server.

Kelly


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Kelly Starks                       Internet: kgstar@most.fw.hac.com
Sr. Systems Engineer
Magnavox Electronic Systems Company
(Magnavox URL: http://www.fw.hac.com/external.html)

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