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Homework Help: Gravitation vertical mine shaft question

  1. Jul 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that, at the bottom of a vertical mine shaft dug to depth D, the measured value of [tex]g[/tex] will be

    [tex]g = g_{s}\left(1-\frac{D}{R}\right)[/tex]


    [tex]g_{s}[/tex] being the surface value. Assume that the Earth is a uniform sphere of radius [tex]R[/tex].

    2. Relevant equations

    [tex]F = \frac{GMm}{r^{2}}[/tex]

    [tex]V_{S} = \frac{4}{3}\pi r^{3}[/tex]

    3. The attempt at a solution

    I thought you could just plug in [tex](R-D)[/tex] in the force equation but when I looked at the solution they did something with a ratio of masses that looked like this:

    [tex]\frac{M(r)}{\frac{4}{3}\pi r^{3}} = \frac{M}{\frac{4}{3}\pi R^{3}}[/tex]

    where [tex]M[/tex] is the total mass.

    Then the solution went on to this:

    [tex]F = \frac{GM_{E}m}{r^{2}}\left(\frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi R^{3}}\right)[/tex]

    I don't really understand these last two steps. Can someone please explain what is happening here, and why are they doing a ratio of masses (related by density) and then multiplying by this ratio?
     
  2. jcsd
  3. Jul 13, 2008 #2
    Right, they are figuring out what part of the earth contributes to the gravity because as you can imagine, if you are a distance D in a mine shaft, the mass of the ground above you is not going to attract you towards the center of the earth while the ground below you will. So they used a volume ratio to represent the mass of the earth still "below" you aka between you and the center
     
  4. Jul 13, 2008 #3

    dynamicsolo

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    Homework Helper

    Unfortunately, it's not quite that simple, since that equation expresses the force outside a sphere of mass M and radius R.

    What the solution is doing, which crytoguy is saying in another way, is to compare the gravitational acceleration, g_s, at the Earth's surface, to a faked-up planet which has the same average density of Earth, but is smaller in radius by an amount D.

    The value for Earth's surface gravity is [tex]g_s = \frac{GM}{R^2} [/tex]; you can use a similar equation for the alternative planet's surface gravity, [tex]g = \frac{GM(r)}{(R-D)^2} [/tex], with r = R - D (I am referring to the solution's manual's notation in part of this).

    Putting these pieces together will lead to the desired equation.
     
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