On Thu, Jul 10, 2003 at 11:29:26PM -0400, David Hobby wrote:
> Erik Reuter wrote:
> > but since R_e = 6370km, and h = 1km, (1 + h / R_e) = 1 is an
> > excellent approximation so the formula becomes
> > P/P0 = exp[ -( h / R_e )( R_e m g / k / T ) ]
> > = exp[ -739 ( h / R_e )]
> I can't find the post where you derived the potential energy at
> a height h above the rim of a habitat of radius R. So here's mine,
> assuming artificial gravity on the rim of 1 g.
Ah, now we're getting somewhere (I like equations)! You were right all
along, but it wasn't being communicated to me without equations. Your
physical intuition is superb, but I'm afraid it doesn't make for very
good communication.
> The radius from the axis is R-h, and centrifical force goes as radius,
> so the force must be (mg/R)*(R-h). We choose the zero of potential
> energy to be when h = 0, just as in your formula for the Earth. We
> get this potential U by integrating the force, so we have:
> U = Integral(0,h) of (mg/R)*(R-t) dt
>
> = (mg/R)*[Rt - t^2/2] Evaluate(0,h)
>
> = (mg/R)*[Rh - h^2/2]
>
> = mgh*[1 - (h/2R)]
Thank you! In my previous post, I only had the second term in the
potential. I'm not sure how I dropped the first one, but your equation
is correct as far as I can tell. I guess I should type up my math steps
next time something like this comes up. That way we won't get caught in
a loop with me making math mistakes and you saying the result doesn't
make sense to you but you don't want to work out the math.
By the way, how did you know, before working out the equation, that
the potential function of the spinning habitat had to be approximately
the same as that due to gravity? Did you intuit that from a principle
of the equivalence of a gravitational field with that of a centrifugal
force field in a rotating frame? That seems a key insight here. If I had
realized that, I would have recognized my mistake a lot sooner.
Anyway, since the internal chemical potential of an ideal gas is
u_i = k T ln[ n / nq ]
then the total chemical potential, which must be independent of height
in equilibrium, is the sum of u_i and the potential energy
u = k T ln[ n[h] / nq ] + m g h ( 1 - 0.5 h / R ) = k T ln[ n[0] / nq ]
where the right hand side is obtained by substituting h=0. Then
k T ln[ n[h] / n[0] ] = - m g h ( 1 - 0.5 h / R )
n[h] / n[0] = Exp[ - m g h ( 1 - 0.5 h / R ) / k / T ]
So, assuming I don't make another math mistake, the formula for pressure
ratio is the same as that for n[h] / n[0] which can be written
P/P0 = exp[ - ( h / R ) ( 1 - 0.5 h / R ) ( m g R / k / T ) ]
= exp[ - h m g / ( k T ) ] exp[ + 0.5 h^2 m g / ( k T R ) ]
If R=5km, m=4.85e-26, g=9.8, k=1.381e-23, T=300, and we note that h must
be in km, then
P/P0 = exp[ -0.115 h ] exp[ +1.15e-2 h^2 ] , h in km, R=5km, h <= R
For Rama, with R=8km,
P/P0 = exp[ -0.115 h ] exp[ +7.17e-3 h^2 ] , h in km, R=8km, h <= R
For the earth, the equivalent formula is quite similar (as David
predicted)
P/P0 = exp[ -0.115 h ] exp[ +1.81e-5 h^2] , h in km, h/6370 is small
As a check, one of my textbooks gives a curve fit of actual Earth
pressure gradient data to exp[ -h / hc ] and they find hc=8.5km. Since
1/.115 = 8.7km, it seems this formula is reasonable. At h=1km, the
pressures in the 5km habitat, Rama, and Earth are within 1% of each
other.
However, note that for the R=5km habitat, the second exponential factor
is equal to 1.33 at the habitat center where h=5km. In contrast,
on the earth when h=5km, the second exponential factor is only
1.0005. So the smaller radius of the habitat does come into play as one
approaches the center of the habitat. This is what I was referring to
previously. Apparently it holds true despite the mistaken formula that
first suggested it to me. Here is a table of the second exponential
factor
-------------------------------------------
Non-Linear Correction Factor
h(km) 5km habitat 8km Rama 6370km Earth
===========================================
1 1.0116 1.0072 1.0000
2 1.0471 1.0291 1.0001
3 1.1090 1.0667 1.0002
4 1.2020 1.1216 1.0003
5 1.3331 1.1963 1.0005
-------------------------------------------
Basically, since gravity or centrifugal force gets weaker as h
increases, the pressure needs a correction factor as shown above. This
factor becomes significant when h/R grows, but is insignificant in
Earth's atmosphere since the atmosphere is gone by the time h/R becomes
significant. So, the only difference between the three formulas is the
"non-linear-potential correction factor" which is negligible for small
h/R, but becomes significant when h/R increases, which happens for a
small habitat but is insignificant for the Earth.
Apparently, my willingness to work through the math, and David's
physical reasoning to catch my mistakes seem to make a good team! Sorry
for the mistake, R.C., I guess you'll have to do your program yet again
(is that emacs Lisp?). I came up with 0.5 atmospheres as the surface
pressure for Rama (pre-melting of the sea), which seems a little nicer,
huh? I guess Clarke got it right after all. This was an entertaining
problem.
--
"Erik Reuter" <[EMAIL PROTECTED]> http://www.erikreuter.net/
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