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Thread: Spin gravity and atmospheric retention

  1. Top | #11
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    Quote Originally Posted by Jokodo View Post
    Quote Originally Posted by Loren Pechtel View Post
    Quote Originally Posted by bilby View Post

    Thanks.

    Presumably, to hold an atmosphere at survivable pressures for the really long term would need an enormous open ring - something on the scale of Niven's Ringworld
    Exactly. Niven handwaved when he needed to but tried to keep the numbers right if he could. 1000 mile walls are what it takes if you want the top open.

    The key to understanding atmosphere is scale height. It's the distance where the pressure drops to 1/e of what it was. For Earth it's about 8.5km at surface temperature and gravity. Thus 100km up the atmosphere has dropped to about 1/(e^12) of what it is at the surface. That's 6 parts in a million of sea level--about half a pascal. Over time that would spill a lot of air. You need to go a lot higher to hold an atmosphere for a long time.
    I'm pretty sure it's 5.5, not 8.5 km where three pressure is half its sea levelvalue. Eg here http://ww2010.atmos.uiuc.edu/(Gh)/gu.../prs/hght.rxml 8.5 sounds like the kind of figure you get by converting 5.5 miles to km, only that it's already in km, and can be avoided bu using metric units exclusively and consistently.

    That's on earth. In a spinning wheel, the drop in "gravity" (acceleration) would be more significant (I.e. half at half the radius) and the drop of in pressure this slower.
    Note that I am talking about 1/e, not 1/2. The dropoff of "gravity" on a Ringworld is much slower than on Earth. I do agree on a spinning wheel the "gravity" drops off much faster, but building a wheel big enough to have vacuum at the hub would be hard!

  2. Top | #12
    Administrator lpetrich's Avatar
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    To settle this question, let us do some hydrodynamics. I will do static and quasi-static hydrodynamics only, to make it easier.

    First, the pressure-equilibrium equation.
     {\vec \nabla} P = {\vec f} = \rho {\vec a} = - \rho {\vec \nabla} V
    where P is the pressure, f is the force density, (rho) is the density, a is the acceleration, and V is the potential.

    For small distances h and acceleration of gravity g,
     V = g h

    For centrifugal acceleration at distance r from a central axis with angular velocity w,
     V = - \frac12 \omega^2 r^2

    Density is related to pressure, temperature T, and molecular weight m by the ideal gas law:
     P = \frac{\rho k T}{m}
    Non-ideality is usually small enough to ignore without loss of much accuracy.

    The pressure-equilibrium equation and the ideal gas law are two equations for three quantities, (rho), P, and T, so we need a third equation.

  3. Top | #13
    Administrator lpetrich's Avatar
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    I will consider two types, a power-law adiabatic,
     P \sim \rho^\gamma
    and isothermal, T = constant.

    Adiabatic first, since that is essentially convective equilibrium, equilibrium over moving up and down. It is easiest to find the temperature, and from the power law, we have
     \rho \sim T^{1/(\gamma-1)} ,\ P \sim T^{\gamma/(\gamma-1)}

    Plugging it into the hydrostatic-equilibrium equation gives us
     \frac{\gamma}{\gamma-1} \frac{k}{m} {\vec \nabla} T = - {\vec \nabla} V
    giving us
     T = T_0 - \frac{\gamma-1}{\gamma} \frac{m}{k} (V - V_0)

    For the gravity case, we get
     T = T_0 - \Gamma (h - h_0) ,\ \Gamma = \frac{\gamma-1}{\gamma} \frac{mg}{k}
    (Gamma) is called the "adiabatic lapse rate".

    The Earth's dry adiabatic lapse rate is 9.8 C / km.

    One can do similar calculations for the centrifugal case, and the center-to-rim temperature difference is half that for constant acceleration = rim acceleration. Thus, if one has 1 g at the rim, then temperature difference between the center and the rim is 4.9 C / km * (radius).

    So there won't be very much difference unless the habitat is very large (radius ~ several kilometers).

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    Administrator lpetrich's Avatar
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    Now the isothermal case. It has
     \frac{kT}{m} {\vec \nabla} \rho = - \rho {\nabla V}
    giving us
     \rho = \rho_0 \exp \left( - \frac{mV}{kT} \right)
    and
     P = P_0 \exp \left( - \frac{mV}{kT} \right)

    Here also, detailed calculation does not show much change in air density unless the habitat is very large.

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    Administrator lpetrich's Avatar
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    Returning to the OP's conundrum, if the hub gets an air leak, the entire station's air will leak out unless the station is divided into sealable compartments. For a large station, I think that such compartments will be a necessity.

  6. Top | #16
    Fair dinkum thinkum bilby's Avatar
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    Quote Originally Posted by lpetrich View Post
    Returning to the OP's conundrum, if the hub gets an air leak, the entire station's air will leak out unless the station is divided into sealable compartments. For a large station, I think that such compartments will be a necessity.
    I suspected that might be the case. A small (ie less than tens of km diameter) space station is essentially at similar risk of decompression regardless of spin gravity - the rotation helps people at the rim to live in earthlike gravity, but does little to protect them from any kind of serious atmosphere leak, even at the hub.

    Which is probably a good thing in the absence of a leak, as it allows workers at the hub to breathe without spacesuits. If an uncontained hull breach at the hub was a minor problem for those at the rim, then working unprotected at the hub would be impossible.

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