Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Diameter of critical object

  1. Oct 13, 2007 #1
    Diameter of "critical" object

    1. The problem statement, all variables and given/known data

    A pressure of [itex] \textrm{2} \cdot \textrm{10}^{9} [/itex] Pa on a time scale of [itex] \textrm{10}^{9} [/itex] yr, holds half way to the center of the "critical" objects.
    Estimate the diameter using two densities: [itex] \rho = 3 \textrm{g}/ \textrm{cm} ^{3} [/itex] and [itex] \rho = 1 \textrm{g}/ \textrm{cm} ^{3} [/itex].

    2. Relevant equations

    I was told to use the equation for hydrostatic equilibrium

    [tex] \frac{dp_}{dr} = -\frac{GM_{r}\rho}{{r}^{2}} [/tex]

    where

    [tex] M_{r} = \frac{4\pi {r}^{3}\rho}{3} [/tex]

    3. The attempt at a solution

    I chose 2r as the radius for the entire object, then r is the radius half way.
    I then inserted [itex]M_{r}[/itex] into the equation for hydrostatic equilibrium and integrated the new equation

    [tex]p = \int_{2r}^r -\frac{{4\pi G} \rho^{2}}{3}r^\prime\, dr^\prime = \frac{{4\pi G} \rho^{2}}{3} ((2r)^{2}-r^{2}) = 2 \pi G \rho^{2} r^{2}[/tex]

    Then I just inserted all the constants and solved the equation for r.


    [tex]\rho = 3 \textrm{g}/ \textrm{cm} ^{3} \Rightarrow r \approx 728,183 km \Rightarrow d=4r \approx 3000km [/tex]

    and

    [tex]\rho = 1 \textrm{g}/ \textrm{cm} ^{3} \Rightarrow r \approx 2184,550 km \Rightarrow d=4r \approx 8740 km [/tex]

    I was told that I should get a diameter between 500-600 km. Obviously I'm no where near that, so I really need help with this because I have to present a solution on October 18th.
     
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted