Lower limit to the diameter

  1. I have a homework question that I am having troubles with.

    Q: By equating the pressure at the centre of an icy planetesimal to the maximum pressure that cold ice can sustain without deforming, about 40 MPa, find a lower limit to the diameter of an icy minor planet.

    The part I dont understand is the "lower limit to the diameter"
    Do I use: Pcentral >[itex]\frac{GM^{2}}{8πr^{4}}dm[/itex]

    I just dont know how to get the diameter....
     
  2. jcsd
  3. mfb

    Staff: Mentor

    I don't know where your formula comes from, but you need some relation between size (like the radius r in your formula?) and pressure in the center, and then let the pressure in the center be 40MPa.
    This looks more like an upper limit, however.
     
  4. I did:
    [itex]\frac{dP}{dr}[/itex]=[itex]ρg[/itex]
    and [itex]g[/itex]=[itex]\frac{GM}{r^2}[/itex]
    so [itex]\frac{dP}{dr}=\frac{-GMρ}{r^2}[/itex] (Hydrostatic equilibrium equation)
    and [itex]\frac{dM}{dr}[/itex]=[itex]4πr^{2}ρ[/itex] (equation of mass conservation)

    by dividing the two equations: [itex]\frac{dP/dr}{dM/dr}[/itex]=[itex]\frac{dP}{dM}[/itex]=[itex]\frac{-GM}{4πr^4}[/itex]

    integration: [itex]P_{c}-P_{s}[/itex]=-[itex]\int^{M_{c}}_{M_{s}}[/itex]([itex]\frac{GM}{4πr^4}[/itex])[itex]dM[/itex]
    [itex]P_{c}[/itex] and [itex]P_{s}[/itex] are pressure at centre and surface of the planet
    and by setting [itex]M_{c}=0[/itex] and by switching the intergral:

    [itex]P_{c}-P_{s}[/itex]=[itex]\int^{M_{s}}_{0}[/itex]([itex]\frac{GM}{4πr^4}[/itex])[itex]dM[/itex]

    and

    [itex]\int^{M_{s}}_{0}[/itex]([itex]\frac{GM}{4πr^4}[/itex])[itex]dM[/itex] > [itex]\int^{M_{s}}_{0}[/itex]([itex]\frac{GM}{4πr^{4}_{s}}[/itex])[itex]dM[/itex] = [itex]\frac{GM^{2}_{s}}{8πr^{4}_{s}}[/itex]

    hence

    [itex]P_{c}-P_{s}[/itex]>[itex]\frac{GM^{2}_{s}}{8πr^{4}_{s}}[/itex]

    and by approximating that the pressure at the surface to be zero ([itex]P_{s}=0[/itex])
    we get:

    [itex]P_{c}[/itex]>[itex]\frac{GM^{2}_{s}}{8πr^{4}_{s}}[/itex]


    So are you saying I should do:

    [itex]40MPa[/itex]>[itex]\frac{GM^{2}_{s}}{8πr^{4}_{s}}[/itex]?

    I figure I have to solve for r and then obtain the diameter from there but Im stuck because I am not given the mass... (or do I use a mass of an icy minor planet like Ceres?)
     
    Last edited: Nov 3, 2013
  5. mfb

    Staff: Mentor

    Your formula differs from the one in the first post now.

    You know the density of ice, this gives the relation radius<->mass,
     
    1 person likes this.
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