# Gauss' Theorem for gravitational force

Hello,
I wonder that the gauss' theorem for gravitational force area.

$$\int\int_S \vec{g}\hat{n}dS=-4\pi GM=\int\int\int_V \vec{\nabla}\stackrel{\rightarrow}{g}dV$$

$$\vec{g}=-G\frac{M}{r^2}\hat{r}\Rightarrow\hat{r}=\frac{\vec{r}}{r}\Rightarrow\vec{g}=-G\frac{M}{r^3}\vec{r}$$

for $$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$$ and $$r=\sqrt{x^2+y^2+z^2}$$

$$\vec{\nabla}\vec{g}=-\frac{\partial}{\partial x}G\frac{M}{r^3}x-\frac{\partial }{\partial y}G\frac{M}{r^3}y-\frac{\partial }{\partial z}G\frac{M}{r^3}z=0$$

The divergence of g has 0 so $$\int\int_S\vec{g}\hat{n}dS=0$$

Uh, I'm not sure I understand all of your equations there. By $$\vec{\nabla}\vec{g}$$, did you mean, $$\vec{\nabla}\cdot\vec{g}$$?

If so, then you should know that $$\vec{\nabla}\cdot\vec{g}$$ is not zero. The correct expression is,

$$\vec{\nabla}\cdot\vec{g} = -4\pi G\sum_{i=0}^n m_i \delta^3(\vec{r} - \vec{r_i})$$

When dealing witih point masses, the divergence of the gravitational field is a sum of Dirac delta functions. That way when you take the surface integral of the gravitational field, the volume integral that you have to take on the right hand side will give you $$4\pi G$$ times the sum of the point masses inside the surface of integration. This is actually a very common error, and Griffiths' E&M book discusses it in the first chapter on vector calculus.

Hope that helps!

Last edited:
nrqed
Homework Helper
Gold Member
Hello,
I wonder that the gauss' theorem for gravitational force area.

$$\int\int_S \vec{g}\hat{n}dS=-4\pi GM=\int\int\int_V \vec{\nabla}\stackrel{\rightarrow}{g}dV$$

$$\vec{g}=-G\frac{M}{r^2}\hat{r}\Rightarrow\hat{r}=\frac{\vec{r}}{r}\Rightarrow\vec{g}=-G\frac{M}{r^3}\vec{r}$$

for $$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$$ and $$r=\sqrt{x^2+y^2+z^2}$$

$$\vec{\nabla}\vec{g}=-\frac{\partial}{\partial x}G\frac{M}{r^3}x-\frac{\partial }{\partial y}G\frac{M}{r^3}y-\frac{\partial }{\partial z}G\frac{M}{r^3}z=0$$

The divergence of g has 0 so $$\int\int_S\vec{g}\hat{n}dS=0$$

You seem to use
$$\frac{\partial}{\partial x} \frac{1}{r^3} = 0$$
and similarly for the derivatives with respect to y and z. That's not the case!

you pointed out in your derivation what r was equal to but did not use it when you were taking the partial I think.

Uh, I'm not sure I understand all of your equations there. By $$\vec{\nabla}\vec{g}$$, did you mean, $$\vec{\nabla}\cdot\vec{g}$$?

If so, then you should know that $$\vec{\nabla}\cdot\vec{g}$$ is not zero. The correct expression is,

$$\vec{\nabla}\cdot\vec{g} = -4\pi G\sum_{i=0}^n m_i \delta^3(\vec{r} - \vec{r_i})$$

Hope that helps!
Yes,$$\vec{\nabla}\cdot\vec{g}$$ I mean. Thanks for your helps but I found that $$\vec{\nabla}\cdot\vec{g}$$ is zero.Let's I show it,

$$\vec{\nabla}\cdot\vec{g}=-GM(\frac{\partial}{\partial x}\frac{x}{(x^2+y^2+z^2)^{3/2}}+\frac{\partial}{\partial y}\frac{y}{(x^2+y^2+z^2)^{3/2}}+\frac{\partial}{\partial z}\frac{z}{(x^2+y^2+z^2)^{3/2}})$$

Now I calculate first partial derivative after generalize the other derivatives.

$$-GM\frac{\partial}{\partial x}\frac{x}{(x^2+y^2+z^2)^{3/2}}=-GM\frac{(x^2+y^2+z^2)^{3/2}-3x^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3}$$
Then
$$\vec{\nabla}\cdot\vec{g}=-GM(\frac{(x^2+y^2+z^2)^{3/2}-3x^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3}+\frac{(x^2+y^2+z^2)^{3/2}-3y^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3}+\frac{(x^2+y^2+z^2)^{3/2}-3z^2(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3})$$

$$\vec{\nabla}\cdot\vec{g}=-GM(\frac{3(x^2+y^2+z^2)^{3/2}-3(x^2+y^2+z^2)(x^2+y^2+z^2)^{1/2}}{(x^2+y^2+z^2)^3})=0$$

Where did I make wrong?I wonder it.Thanks.

jtbell
Mentor
What happens when x = y = z = 0?

Physically speaking, the divergence of g should depend upon mass density. There is a monopole source of gravity...mass!

Also, as a suggestion...work in spherical coordinates.