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Homework Help: Aggregation of debris

  1. Jul 6, 2014 #1
    1. The problem statement, all variables and given/known data

    We know the terrestrial planets formed by aggregation of debris from the solar nebula. We want
    to calculate the maximum size of object that can form by aggregation before self-gravity causes it to pull itself into a round shape.

    Our analysis is assisted by considering Earth’s mountains. It is clear from Earth’s geology that of its many mountains, none are taller than 10 km high. This indicates that the strength of rock at the center of the base of a mountain cannot support a structure higher than 10 km. The pressure at the middle of the base of a mountain is equal to the weight of a column of rock (with unit cross sectional area) as high as the mountain. If a mountain were taller than 10 km, it would slump or begin to be overtaken by self-gravity. We want to use this fact to determine the maximum size of an object that can form without being overtaken by self-gravity.

    Assume you have two cubes of rock (density of 3000 kg/m3) of equal mass (M) and size (length d) that are face to face. They are so large that the only force holding them together is pressure caused by gravitational attraction, but not so large that they flow into a spherical shape due to self-gravity.

    Determine the maximum size of rectangular body that could be made from the two cubes.

    2. Relevant equations

    Force between cubes is F = GM2/d2

    3. The attempt at a solution

    V = d3
    M = ρV = ρd3
    F = GM2/d2 = Gρ2d4

    I think I have the formulas okay, but I don't know how to get a "maximum" size out of the equation.
  2. jcsd
  3. Jul 6, 2014 #2


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    Gold Member

    You can determine the maximum pressure that can be exerted by the rock from the information in the second paragraph. From there you can determine the maximum force, etc...
  4. Jul 6, 2014 #3
    I'm not sure if I follow.

    So a column of height h and radius r would have a volume of πr2h.

    multiply that by the density of the rock (3000 kg/m3) and you get 3000πr2h, which is the mass of the column.

    Assuming that's the mass, I don't see where the pressure comes in.
  5. Jul 6, 2014 #4


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    Gold Member

    The rock at the very bottom of the column has to support the weight of the column. The pressure is then the force that the rock exerts divided by the area over which it acts (the cross sectional area of the cylinder).
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