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Homework Help: Comparing the the magnitude of grav. field at top of a plateau vs. a nearby city

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    My instructor wants me to calculate the magnitude of the gravitational field at the top of a plateau and then do the same for a nearby city.



    2. Relevant equations

    [tex]\vec{g}_{density} = \int_{v}\frac{\rho(\vec{r})\hat{r}}{r^{2}}d\tau[/tex]



    3. The attempt at a solution

    Our instructor first tells us to assume the plateau is an infinite layer with thickness h = 300m and mass density [tex]\rho = 2000 kg/m^{3}.[/tex] He says that this assumption is correct as long as we stay away from the edges. He then asks if we compress this volume into a thin sheet with mass density per unit area [tex]\sigma[/tex] what is the value of [tex]\sigma[/tex].

    [tex]\sigma = \frac{dM}{dA} , \rho = \frac{dM}{dV} \Rightarrow dM = \rho dV

    \Rightarrow \sigma = \frac{\rho dV}{dA} , dV = dA(h) \Rightarrow \sigma = \rho h [/tex]

    So [tex]\sigma = 600,000 kg/m^{2}[/tex] Is this right?

    He then says that if we call [tex]\vec{r}'[/tex] (r prime) our source vector at point (x',y',0) and [tex]\vec{r}[/tex] the point we are solving the field for at (0,0,h) the equation I wrote under relevant equations becomes

    [tex]\vec{g}_{density} = \int_{v}\frac{\sigma(\vec{r}-\vec{r}^{'})}{|\vec{r}-\vec{r}'|^{2} |\vec{r}-\vec{r}'|}dx'dy'[/tex] = [tex]-G\sigma h\int_{v} \frac{dx'dy'}{(x'^{2} + y'^{2} + h^{2})^{3/2}} \hat{z}[/tex] (This is given to us)


    Now using cylindrical coordinates we can show that this equals:

    [tex](g_{z})_{density} = - 2\pi G \sigma h \int_{0}^{\infty} \frac{r dr}{(r^{2} + h^{2})^{3/2}}[/tex]

    evaluating this integral I get:

    [tex](g_{z})_{density} = -2\pi G \sigma \approx -2.51 * 10^{-4} m/s^{2}[/tex]

    So that is the value of g at the top of the plateau. Now for the city he wants us to compute it a different way using the formula:

    [tex] g_{z}^{h} = -\frac{GM}{(R + h)^{2}} + \frac{GM}{R^{2}}[/tex] (This describes the change in gravity caused by a change in distance from the center of the earth)

    where [tex]\frac{GM}{R^{2}} = g = 9.81 m/s^{2}.[/tex]

    With little bit of algebra it can be shown that this can be rewritten as:

    [tex] g_{z}^{h} = g(1 - \frac{1}{(1 + \frac{h}{R})^{2}})[/tex]

    Now using this formula we can compute g at the city using [tex] R \approx 6 * 10^{-6}m[/tex]. I get [tex] g_{z}^{h} = 9.81 * 10^{-4} m/s^{2}[/tex]

    Now did I do something wrong? I don't understand how I get a negative answer for my first g and and a positive for my second.


    Thanks in advance,

    KEØM

    (Any help will be appreciated)

    EDIT: Sorry for the long-winded post but I guess if I were to boil it down to one question it would be: Is it wrong that my two values of g are opposite in sign?
     
    Last edited: Feb 9, 2010
  2. jcsd
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