# Gravity for a point inside the earth

1. Apr 5, 2009

### loup

Assume the Earth is a perfect sphere,

How do we calculate that a point inside the earth's gravity it experienced?

I know about the formula which is taught in high school. I just cannot understand how it is proved. I know it is related to the integration. Can anybody give some hints to me about how to do the integration?

I am so puzzled.

2. Apr 5, 2009

### Staff: Mentor

What formula are you thinking of?

You might want to look up Newton's Shell theorems, which deal with this.

3. Apr 5, 2009

### arildno

You can do this the hard way, by setting up an integral and derive the answers.

This is a more primitive technique than utilizing Newton's shell theorem directly.

4. Apr 5, 2009

### arildno

Let your point, as measured from the centre of the Earth have position vector $$\vec{r}_{0}=z_{0}\vec{k}$$

The force $\vec{f}_{0}$ per unit mass located at $\vec{r}_{0}$ as acted upon by a uniformly massed ball of radius R and density $\rho$ is given by the integral:
$$\vec{f}_{0}=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\frac{G\rho(\vec{r}-\vec{r}_{0})}{||\vec{r}-\vec{r}_{0}||^{3}}r^{2}\sin\phi{dr}d\phi{d}\theta, \vec{r}=r(\sin\phi\cos\theta\vec{i}+\sin\phi\sin\theta\vec{j}+\cos\phi\vec{k})$$

From this, you should first verify that the force is zero in the i and j-directions, before proceeding further!

5. Apr 5, 2009

### loup

I am sorry. I just don't know what exactly you are talking about. Can you explain more about what you are doing?

6. Apr 5, 2009

### arildno

Okay, let us take this step by step!

Suppose we have some mass object P located at $$\vec{r}_{0}$$, with mass m.

Let an object O fill some region R, so that location $$\vec{r}$$ within R designates a mass point of mass $$dM=\rho{dV}$$, where $\rho$ is the local density.

Thus, we have that the force from O upon P must be given by:
$$\vec{F}_{0}=-\int_{R}\frac{GmdM}{||\vec{r}_{0}-\vec{r}||^{2}}\vec{i}{r},\vec{i}{r}=\frac{\vec{r}_{0}-\vec{r}}{||\vec{r}_{0}-\vec{r}||}$$

Thus, this may be rewritten as:
$$\vec{f}_{0}=\frac{\vec{F}_{0}}{m}=\int_{R}\frac{G\rho(\vec{r}-\vec{r}_{0})}{||\vec{r}_{0}-\vec{r}||^{3}}dV$$

Do you agree to this?

7. Apr 5, 2009

### arildno

Now, let R be a ball, let the origin be the ball's centre, so that in spherical coordinates, the points of the ball constitute the region:
$$0\leq{r}\leq{R},0\leq\theta\leq{2\pi},0\leq\phi\leq\pi$$
Thus, the position vector, expressed by the constant Cartesian unit vectors may be written as:
$$\vec{r}=r(\sin\phi\cos\theta\vec{i}+\sin\phi\sin\theta\vec{j}+\cos\phi\vec{k})$$

No loss of generality is made by aligning the $\vec{k}$-vector along the line from the ball's centre to the mass point at $\vec{r}_{0}$, i.e, we get:
$$\vec{r}_{0}=z_0\vec{k}$$ for some constant z0.

Is that okay?

8. Apr 6, 2009

### loup

But if P is inside the ball. Isn't that not only the things nearer than P to the centre contributes to the gravity, but also the things outside P?

I know the things outside P will cancel each other, but how to prove the things outside P will cancel each other ?

9. Apr 6, 2009

### Phrak

It's very simple. Draw a thin spherical shell. Put some off-center point inside. This is the arbitrary point where we want to know the gravitational force on the point. Imagine two thin cones extending from the point to opposite areas on the shell.

The area of the shell within each cone depends upon how far it is from the point in question. The volume of matter in each patch is proportional to the area. The area is proportional to the distance squared of the patch from the shell. You should be able to convince yourself that this is so by drawing some simple diagrams. The force from each little patch of the shell is complimented be the force from the opposite patch. This is from the inverse square law. The contribuations from all the little patches is zero.

Last edited: Apr 6, 2009
10. Apr 6, 2009

### arildno

You will know by performing the actual calculations!

Now, let us take the easy bits here:

In spherical coordinates, we have $$dV=r^{2}\sin\phi{d\theta}d\phi{dr}$$

Furthermore, we have:
$$||\vec{r}-\vec{r}_{0}||=\sqrt{r^{2}+z_{0}^{2}-2rz_{0}\cos\phi}$$
which you can readily establish, say with reference to the law of cosines, or by laborious calculation.

Thus, let us look at the i-th component of the integral.
$$f_{0,i}=\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}\frac{G\rho{r}^{2}\sin^{2}\phi\cos\theta}{(\sqrt{r^{2}+z_{0}^{2}-2rz_{0}\cos\phi})^{3}}d\theta{d\phi}{dr}$$

Assuming the density to be constant, the full $\theta$-period involved in the integration implies that this becomes 0.
Similarly for the j'th component.

Agreed so far?

11. Apr 7, 2009

### loup

aridno, I agree with what you say. Thank you very much.

12. Apr 7, 2009

### arildno

Okay!

Let us proceed with the component in the k-direction.
This reads, after the $\theta$-integration:
$$f_{0,k}=2\pi\int_{0}^{R}\int_{0}^{\phi}\frac{G\rho{r}^{2}\sin\phi(r\cos\phi-z_{0})}{(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{\frac{3}{2}}}d\phi{d}r$$

Ok?

13. Apr 7, 2009

### arildno

Now, we will perform a partial integration on $\phi$

Note that:
$$\frac{\partial}{\partial\phi}(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{-\frac{1}{2}}=-\frac{rz_{0}\sin\phi}{(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{\frac{3}{2}}}$$

Looks awfully similar, doesn't it?

Thus, we may rewrite our integral as:
$$f_{0,k}=-2\pi\int_{0}^{R}\int_{0}^{\phi}\frac{G\rho{r}(r\cos\phi-z_{0})}{z_{0}}\frac{\partial}{\partial\phi}(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{-\frac{1}{2}}d\phi{d}r$$

Thus, partial integration yields:
$$f_{0,k}=-(2\pi\int_{0}^{R}(\frac{G\rho{r}(r\cos\phi-z_{0})}{z_{0}}(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{-\frac{1}{2}})_{\phi=0}^{\phi=\pi}dr+\int_{0}^{R}\int_{0}^{\phi}\frac{G\rho{r}^{2}\sin\phi}{z_{0}}(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{-\frac{1}{2}}d\phi{d}r)=$$
$$-(2\pi\int_{0}^{R}\frac{G\rho{r}}{z_{0}}(-\frac{r+z_{0}}{|r+z_{0}|}-\frac{r-z_{0}}{|r-z_{0}|}dr+\int_{0}^{R}\int_{0}^{\phi}\frac{G\rho{r}^{2}\sin\phi}{z_{0}}(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{-\frac{1}{2}}d\phi{d}r)$$

Last edited: Apr 7, 2009
14. Apr 7, 2009

### loup

I group what Arildno's for a clearer overview

15. Apr 7, 2009

### arildno

The latter integral can be evaluated as follows:
$$\int_{0}^{\phi}\frac{G\rho{r}}{z_{0}^{2}}rz_{0}\sin\phi(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{-\frac{1}{2}}d\phi=\frac{G\rho{r}}{z_{0}^{2}}(r^{2}+z_{0}^{2}-2rz_{0}\cos\phi)^{\frac{1}{2}}\mid_{\phi=0}^{\phi=\pi}=\frac{G\rho{r}}{z_{0}^{2}}(|r+z_{0}|-|r-z_{0}|)$$

16. Apr 7, 2009

### arildno

Thus, combining, we get:
$$f_{0,k}=-\frac{2\pi{G}\rho}{z_{0}}\int_{0}^{R}(-r(\frac{r+z_{0}}{|r+z_{0}|}+\frac{r-z_{0}}{|r-z_{0}})+\frac{r}{z_{0}}(|r+z_{0}|-|r-z_{0}|)dr$$

17. Apr 7, 2009

### arildno

Note that if z_0 is GREATER than any r between 0 and R, this yields:
$$f_{0,k}=-\frac{2\pi{G\rho}}{z_{0}}\int_{0}^{R}(-r(1-1)+\frac{r}{z_{0}}(r+z_{0}-(z_{0}-r))dr=-\frac{2\pi{G\rho}}{z_{0}}\int_{0}^{R}\frac{2r^{2}}{z_{0}}dr=-\frac{4\pi{G\rho}R^{3}}{3z_{0}^{2}}=-\frac{GM}{z_{0}^{2}}$$
which agrees with what we should have when our point lies outside the sphere.

Last edited: Apr 7, 2009
18. Apr 7, 2009

### arildno

Note that for r-values GREATER than z_{0}, the integrand becomes:
$$(-r(\frac{r+z_{0}}{|r+z_{0}|}+\frac{r-z_{0}}{|r-z_{0}|})+\frac{r}{z_{0}}(|r+z_{0}|-|r-z_{0}|)=-r(1+1)+\frac{r}{z_{0}}(r+z_{0}-r+z_{0})=-2r+2r=0$$

This agrees with the result of the shell method, namely that for a point within a hollow sphere, the net force from that hollow sphere equals zero.

Thus, only the sphere with radius $$0\leq{r}\leq{z_{0}}(\leq{R})$$ counts, and from the previous post, we see we get the result:
$$f_{0,k}=-\frac{4\pi{G}\rho{z}_{0}^{3}}{3z_{0}^{2}}=-\frac{4\pi}{3}G\rho{z}_{0}$$

19. Apr 7, 2009

### loup

It must be the most difficult integration so far I have encountered. But at the same time, it fasinates me.

What is the next step, arildno?

20. Apr 8, 2009

### arildno

As long as you have understood the steps we have been through, we are basically finished.