# Dependence of gravitational potential

1. May 14, 2015

### henrybrent

it mentions that a uniform sphere has 'dependence' on its gravitational potential, in terms of spherical harmonics.

[ite

I don't understand the term dependence? What does this mean exactly?

Cheers

Last edited: May 14, 2015
2. May 14, 2015

### Staff: Mentor

Can you quote the exact text and its context?

3. May 14, 2015

### henrybrent

Just trying to find the exact passage again, it was a quote in a powerpoint that had referenced a single textbook, so it was rather misleading to say I was reading a textbook...

"Within a uniform body of spherical nature, of radius R and density rho, the gravitational force g inside the sphere varies linearly with distance r from the centre, giving
$V(r)= \frac{2}{3}\pi G \rho(r^2-3R^2)$ "

4. May 14, 2015

### Staff: Mentor

This has nothing to do with spherical harmonics.

5. May 14, 2015

### nasu

And no mention of a "dependence".:)

6. May 15, 2015

### henrybrent

Apologies, typo. Spherical Co ordinates.

7. May 15, 2015

### Staff: Mentor

The gravitational potential depends on the radius. That should not be surprising: you need energy to climb up a ladder.

8. May 16, 2015

### vanhees71

Well, let's calculate the gravitational potential for a homogeneous sphere. There are two ways. One is to use the superposition principle, and "add" Newton's potential for a point mass up, which leads to
$$g(\vec{x})=-G \int_{V} \mathrm{d}^3 \vec{x}' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
A bit more mathematically spoken, that's just folding the mass-density distribution $\rho$ with the Green's function of the Laplace operator. This means that you can reformulate Newton's gravitation law in terms of a local field equation, which is very clever compared to write down the above integral form:

\label{1}
\Delta g=4 \pi G \rho.

Now you have a (partial) differential equation instead of an integral.

Since we deal with a highly symmetric situation, it can be simplified to an ordinary differential equation. Just introduce spherical coordinates $(r,\vartheta,\varphi)$. Now due to spherical symmetry, the potential and $\rho$ only depend on $r$. Now you use the Laplacian in spherical coordinates to get
$$\frac{1}{r} \frac{\mathrm{d}^2}{\mathrm{d} r^2} [r g(r)]=4 \pi G \rho.$$
Now for $r<R$ (where $R$ is the radius of the homogeneous sphere) you have $\rho=\rho_0=\text{const}$. You can immidiately integrate the equation:
$$\frac{\mathrm{d}^2}{\mathrm{d} r^2} [r g(r)]=4 \pi G r \rho_0.$$
The first integration gives
$$[r g(r)]'=2 \pi G \rho_0 r^2+C_1,$$
where $C_1$ is an integration constant, and the second integration leads to
$$g(r)=\frac{2 \pi G \rho_0}{3} r^2 + C_1 + \frac{C_2}{r}, \quad r<R.$$
For $r>R$ you just need to set $\rho_0=0$ in this equation, which gives
$$g(r)=C_1'+ \frac{C_2'}{r}, \quad r \geq R.$$
We can choose the constant arbitrarily since it doesn't provide anything to the gravitational force to a test mass, $m$, which is given by $\vec{F}=-m \vec{\nabla} g$. Usually one likes the potential to vanish at infinity. So we set $C_1'=0$. For $r<R$, we shouldn't have a singularity at $r=0$ since we deal with a smooth mass distribution there, which leads to $C_2=0$. So far we thus have

\label{2}
g(r)=\begin{cases}
\frac{2 \pi G \rho_0}{3} r^2 + C_1 & \text{for} \quad r<R,\\
\end{cases}

Now we somehow have to determine the remaining constants $C_1$ and $C_2'$. To that end we go back to Eq. (\ref{1}). Now we can use Gauss's integral law to integrate this equation over an arbitrary volume. We choose a sphere $K_a$ around the center with radius $a>R$. Then Gauss's Law tells us
$$\int_{\partial K_a} \mathrm{d}^2 \vec{F} \cdot \vec{\nabla} g=4 \pi G \int_{K_a} \mathrm{d}^3 \vec{\rho}.$$
Now, on the right-hand side the volume integral integrates over the entire mass distribution, i.e., it's simply the total mass of the spherically symmetric body! The left-hand side is easily calculated. Since the surface $\partial K_a$ is at $r=a>R$, we have according to Eq. (\ref{2})
$$\vec{\nabla} g=g'(r) \vec{e}_r=-\frac{C_2}{r^2} \vec{e}_r.$$
Since further $\mathrm{d}^2 \vec{F}=\mathrm{d} \varphi \mathrm{d} \vartheta a^2 \sin \vartheta \vec{e}_r$, we find
$$-4 \pi C_2=4 \pi G M \; \Rightarrow \; C_2=-G M,$$
i.e., outside the mass distribution we have
$$g(r)=-\frac{G M}{r}=-\frac{4 \pi G R^3 \rho_0}{3r}, \quad r>R.$$
This is Newton's Gravitational Law for a point particle with the total mass of the sphere (it's easy to prove that this is true for any radially symmetric mass distribution extending over only a finite region around the center).

Finally the constant $C_1$ is found by making the gravitational potential continuous at $r=R$, leading to
$$\frac{2 \pi G \rho_0 R^2}{3}+C_1=-\frac{4 \pi G R^2 \rho_0}{3} \; \Rightarrow \; C_1=-2 \pi G \rho_0 R^2.$$
Put this in (\ref{2}), we finally have
$$g(r)=\begin{cases} \frac{2 \pi G \rho_0}{3}(r^2-3R^2) &\text{for} \quad r \leq R,\\ -\frac{4 \pi G \rho_0 R^3}{3R} & \text{for} \quad r >R. \end{cases}$$
This shows the correctness of Posting #3.

Suppose you drill a small hole along a diameter through the sphere. Then a body of mass $m$ inside this hole feels the gravitational force
$$\vec{F}=-m \vec{\nabla} g=-\frac{4 \pi m G \rho_0}{3} r \vec{e}_r=-\frac{G M}{R^3} r \vec{e}_r.$$
This is the equation of motion of a harmonic oscillator, i.e., a mass started at rest at the surface will oscillate back and forth forever (assuming no friction).

Of course, that's unrealistic for the earth, which is not homogeneous. There's a great paper by a graduate student (!!!) in AJP about this issue. It's really a great read:

Alexander R. Klotz, The gravity tunnel in a non-uniform Earth, Am. J. Phys. 83, 231 (2015)
http://dx.doi.org/10.1119/1.4898780
http://arxiv.org/abs/1308.1342

Of course, it's still a bit academic, because so far nobody has drilled a whole through the entire earth through the center, and we cannot check, how long it takes for a body to fall through the earth from one end to the other ;-).

9. May 24, 2015

### StellarCore

vanhees71,

Excellent derivation and very insightful !
How would gravitational potential in a star with non-uniform density modify your derivation?

10. May 25, 2015

### vanhees71

The derivation would be very similar if the density is still radially symmetric since then the potential is also radially symmetric, and outside the stellar matter would still look like a 1/r potential (even in General Relativity you get uniquely one static vacuum solution for a radially symmetric star, the Schwarzschild solution). For non-symmetric mass distributions you'd have to use more advanced methods like trying to fold the mass distribution over the Green's function or the multipole expansion for the field outside of the star etc. You find a lot about this in textbooks on electrodynamics. Electrostatics in the gauge, where you describe the electric field by a scalar potential only, is exactly the same math as Newton's gravitational law.

11. May 25, 2015

### StellarCore

Thank you kindly for your robust and helpful explanation. In particular I will revisit JD Jackson's and Marcus Zahn's texts in EM/ED. Your response affirms my expectations. Reviewing texts such as Hansen & Kawaler; Kaurganoff; Kippenhahn; Zeldovich; et.al. it seems to me that a polytropic model (assuming spherical stars) may be appropriate for such tasks in which ρ(r) is a complex function . In this situation ρ could be a power function of r, for example, ρ(r^[1/n]). I believe there is also an influence (perhaps minimal) of the initial type of star and the evolution of its stellar interior (obviously, this could span many millions/billions of years though for low to medium+ mass stars) but for the purposes of conceptual understanding of mechanism and structure this latter point does not seem relevant. If one considers the reality of stellar rotation and oscillation then things get very complex but it is fascinating to at least attempt to grasp the nature of things.