# Gauss' Theorem for gravitational force

• coki2000
In summary: That's where the symmetry is.In summary, the conversation discusses the Gauss' theorem for gravitational force area and the use of symmetry in choosing coordinates. It also touches upon the incorrect use of the divergence of the gravitational field and the correct expression for it, as well as the suggestion to work in spherical coordinates for better symmetry.
coki2000
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:
coki2000 said:
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.

arunma said:
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.

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.

## What is Gauss' Theorem for gravitational force?

Gauss' Theorem for gravitational force, also known as Gauss' Law of Gravitation, is a fundamental law of physics that describes the force of gravity between two objects. It states that the strength of the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

## Who discovered Gauss' Theorem for gravitational force?

Gauss' Theorem for gravitational force was discovered by German mathematician and physicist Carl Friedrich Gauss in the late 18th century. He developed the theorem based on his extensive research and experiments on the laws of gravitation.

## What is the significance of Gauss' Theorem for gravitational force?

Gauss' Theorem for gravitational force is significant because it explains the universal nature of gravity and allows for accurate predictions and calculations of the force between any two objects. It also forms the basis for understanding the motion of planets and other celestial bodies in our solar system.

## How is Gauss' Theorem for gravitational force related to Newton's Law of Universal Gravitation?

Gauss' Theorem for gravitational force is mathematically equivalent to Newton's Law of Universal Gravitation, but it provides a more general and elegant formulation of the law. While Newton's law only applies to point masses, Gauss' theorem can be applied to objects of any size or shape, making it a more versatile and useful tool for calculations.

## Can Gauss' Theorem for gravitational force be applied to non-spherical objects?

Yes, Gauss' Theorem for gravitational force can be applied to objects of any shape or size. This is because the theorem takes into account the distribution of mass within the objects, rather than just their overall mass. This allows for accurate calculations of the gravitational force between any two objects, regardless of their shape.

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