I would like to prove that a celestial body's gravitational force is the same as if it would have been a point.

Suppose that the density is the same all over at a certain depth into the body. (It may be less at the surface and more halfway down to the center for example). The density is given by $\displaystyle D(r) \geq 0$, where r is the distance from the center of the body. An object with mass $\displaystyle m$ is laying at the distance d from the center of the celestial body. Suppose it's outside the celestial body, in other words, $\displaystyle r \geq d\ \rightarrow\ D(r) = 0$.

Suppose $\displaystyle F(r)$ is the total force the body is affecting the body with, considering only the mass in the body at a distance from the center within the interval $\displaystyle [0,\ r]$, a mass which still fulfills the density criteria*. Note that $\displaystyle F(0) = 0$, since there is no mass within the interval $\displaystyle [0,\ 0]$, and that $\displaystyle \frac{\partial F(r)}{\partial r} \geq 0$, though I don't want to prove it.

Now, I state that

$\displaystyle

\frac{\partial F(r)}{\partial r}\ =\

\frac{1}{\partial r}\cdot\int_{\alpha\ =\ 0}^{2\pi} \left(

G\cdot m\cdot\overbrace{

D(r)\cdot\underbrace{

sin(\alpha)\cdot 2\pi r

}_\text{

length of latitudinal line

}\cdot\overbrace{

r\cdot\partial\alpha

}^\text{

its "width"

}\cdot\underbrace{

\partial r

}_\text{

its "height"

}

}^\text{mass}

\right.$ $\displaystyle \left.

/\overbrace{\sqrt{(d+cos(\alpha)r)^2 + sin^2(\alpha)r^2}}^

\text{distance from object to line}\ ^2

\right)$

which can be simplified as

$\displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot

\int_0^{2\pi}\left(

\frac{sin(\alpha)}{(d+cos(\alpha)r)^2 + sin^2(\alpha)r^2}

\right)\partial \alpha\ =$

$\displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot

\int_0^{2\pi}\left(

\frac{sin(\alpha)}{d^2 + r^2 + 2cos(\alpha)dr}

\right)\partial \alpha$

But according to a clause (I have only heard of it) that states that the celestial body can be treated as if it where only a point but with the same mass, we would get

$\displaystyle F(r) = \frac{\displaystyle{G\cdot m\cdot\overbrace{\int_0^r\left(

D(r_1)\cdot 4\pi r_1^2

\right)\partial r_1}^\text{mass}}}{d^2}$

so that

$\displaystyle \frac{\partial F(r)}{\partial r} = \frac{G\cdot m\cdot D(r)\cdot 4\pi r^2}{d^2}$

We now has two different formulas for $\displaystyle \frac{\partial F(r)}{\partial r}$:

$\displaystyle G\cdot m\cdot D(r)\cdot 2\pi r^2\cdot

\int_0^{2\pi}\left(

\frac{sin(\alpha)}{d^2 + r^2 + 2cos(\alpha)dr}

\right)\partial \alpha\ \equiv$

$\displaystyle \equiv\ \frac{G\cdot m\cdot D(r)\cdot 4\pi r^2}{d^2}\ \longrightarrow$

$\displaystyle \longrightarrow\

\int_0^{2\pi}\left(

\frac{sin(\alpha)}{d^2 + r^2 + 2cos(\alpha)dr}

\right)\partial \alpha\ \equiv\ \frac{2}{d^2}$

Is this true? Correct me if I'm wrong. Or if I have made any error (probably). Or if I am using the mathematical signs wrong. ;)

* The density critera says that the body shall have the same density all over any given distance from the center. And I just made it up.