# Comparing the the magnitude of grav. field at top of a plateau vs. a nearby city

1. Feb 9, 2010

### KEØM

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

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

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 $$\rho = 2000 kg/m^{3}.$$ 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 $$\sigma$$ what is the value of $$\sigma$$.

$$\sigma = \frac{dM}{dA} , \rho = \frac{dM}{dV} \Rightarrow dM = \rho dV \Rightarrow \sigma = \frac{\rho dV}{dA} , dV = dA(h) \Rightarrow \sigma = \rho h$$

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

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

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

Now using cylindrical coordinates we can show that this equals:

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

evaluating this integral I get:

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

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:

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

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

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

$$g_{z}^{h} = g(1 - \frac{1}{(1 + \frac{h}{R})^{2}})$$

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

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.