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Density of a hydrogen cloud

  1. Nov 3, 2014 #1

    I wanted to ask, what does the density of a hydrogen cloud in space depend upon?

    That might be a silly question given the definition of density, but here's the context;

    Considering a particle at rest within a molecular cloud at radius [itex]r[/itex] from the centre, I have shown that the acceleration this particle feels is approximately [itex]a \approx \frac{Gm}{r^{2}} \approx \frac{4 \pi G \rho r}{3}[/itex] (from mass being density x volume).

    Using the equations of motion for constant acceleration I have determined that the 'free fall' time of this particle is independent of [itex]r[/itex] and can be approximated by [itex]t \approx \frac{1}{\sqrt{G \rho}}[/itex].

    Two questions;

    Why is this time independent of the particles distance from the center?

    How do you determine the density of hydrogen when the external pressure is presumably close to zero?

  2. jcsd
  3. Nov 8, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Nov 11, 2014 #3
    The free-fall time is only constant for constant density. You can get this from Kepler's 3rd law t2~r3/m.(the square of the orbital period is proportional to the cube of the radius divided by the enclosed mass). Since for a constant density m~r3, this means that t=const.

    A cloud with constant density would be gravitationally unstable though, In hydrodynamic equilibrium, the cloud density decreases like 1/r2. In that case, m~r (not ~r3), so that t2~r3/m ~ r2 i.e. t~r
  5. Nov 24, 2014 #4

    Ken G

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    Gold Member

    Because at larger r, there is more mass that is pulling the particle inward, so the force is greater in a way that ends up being proportional to r-- just like a simple harmonic oscillator (for which the time is also independent of r).
    You need to use the external pressure-- the pressure in the interstellar medium is not zero.
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