<p>This sounds like a homework problem so I'll give you most of it but you'll have to work out the details yourself.</p>
<p><a href="https://en.wikipedia.org/wiki/Shell_theorem" rel="nofollow noreferrer">Newton's shell theorem</a> tells us that for any spherical mass distribution, the potential at a distance <span class="math-container">$r_0$</span> from the center is the same as if all the mass within the shell of radius <span class="math-container">$r_0$</span> were at <span class="math-container">$r=0$</span>, and all the mass outside of the shell didn't exist at all (it cancels):</p>
<p><span class="math-container">$$m(r_0) = \int_0^{r_0}4 \pi r^2\rho(r)dr$$</span></p>
<p>and if we let <span class="math-container">$\rho(r)$</span> to be constant (the sphere is of uniform density):</p>
<p><span class="math-container">$$m(r_0) =  4 \pi \rho \int_0^{r_0}r^2 dr = \frac{4}{3} \pi \rho r_0^3$$</span></p>
<p>The gravitational potential for a point mass is</p>
<p><span class="math-container">$$V(r) = \frac{Gm}{r}$$</span></p>
<p>Put them together and you have</p>
<p><span class="math-container">$$V(r) = \frac{4 \pi G \rho r^3}{3r} = \frac{4}{3} \pi \rho G r^2$$</span></p>
<p>and a parabolic <span class="math-container">$~r^2$</span> potential is called a <a href="https://en.wikipedia.org/wiki/Harmonic_oscillator" rel="nofollow noreferrer">harmonic oscillator potential</a>. <a href="https://en.wikipedia.org/wiki/Bertrand%27s_theorem#Harmonic_oscillator" rel="nofollow noreferrer">Motion in this potential</a> (one or many dimensions) is <a href="https://en.wikipedia.org/wiki/Simple_harmonic_motion" rel="nofollow noreferrer">simple harmonic motion</a>. That's because the potential is proportional to <span class="math-container">$r^2 = x^2 + y^2 + z^2$</span> and the restoring force is proportional to <span class="math-container">$\nabla r^2 / 2 = x \mathbf{\hat{x}} + y \mathbf{\hat{y}} + z \mathbf{\hat{z}}$</span>. Oscillations have the same restoring force in all directions, so the periods <span class="math-container">$\omega_x, \omega_y, \omega_z$</span> will be the same.</p>
<p>Yes, these will be ellipses and since <span class="math-container">$r = \sqrt{x^2 + y^2 + z^2}$</span> the value of <span class="math-container">$r$</span> will reach a maximum twice during each orbit.</p>
<p>If for example <span class="math-container">$\mathbf{r}(t) = \cos(\omega t)\mathbf{\hat{x}} + \frac{1}{2}\sin(\omega t)\mathbf{\hat{z}}$</span> then you will see that <span class="math-container">$r(t)$</span> has two maxima and two minima per cycle.</p>
<hr />
<p><a href="https://i.sstatic.net/evSmY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/evSmYm.png" alt="plot" /></a></p>
<pre><code>import numpy as np
import matplotlib.pyplot as plt

wt = np.linspace(0, 2*np.pi, 1001)

x, z = np.cos(wt), 0.5*np.sin(wt)
r = np.sqrt(x**2 + z**2)

plt.figure()
plt.plot(wt, x)
plt.plot(wt, z)
plt.plot(wt, r)
plt.show()
</code></pre>

CC BY-SA 4.0
2021-05-06T14:10:03.677Z
Thank you. From your answer I understand that I can say that, since $U \propto r^2$ and motion is a simple harmonic motion, if I have, for example, an elliptic trajectory in two dimensions, I'll have oscillations along $x$ and $y$ around a common center (the center of the ellipse, and center of gravity) and that's why, if I have an ellipse of semimajor axis $a$ and semiminor axis $b$, pericenter and apocenter distances matches $a$ and $b$.
Martrin
CC BY-SA 4.0
2021-05-06T14:10:07.490Z
While in the case of keplerian potential, which provide a force that goes as $1/r^2$, Kepler's first law says that the center of gravity is in one of the ellipse foci, so the smaller and longer distance from center of gravity are both on the same axis.
Martrin
CC BY-SA 4.0
2021-05-06T14:13:35.107Z
@Martein yes you can say that I think. For consistency only you should say Keplerian orbits come from $1/r$ *potentials* (rather than saying $1/r^2$ forces).
uhoh
CC BY-SA 4.0
2021-05-06T12:54:08.350Z
2021-05-06T13:06:43.070Z
2021-05-06T13:06:43.070Z
uhoh
<p>For the first Kepler's law the orbit of a planet is an ellipse, with the Sun in one of the foci, but <a href="http://astro.utoronto.ca/%7Ebovy/AST1420/notes/notebooks/04.-Orbits-in-Spherical-Potentials.html" rel="nofollow noreferrer">I've read that</a> in the case of an homogeneous sphere distribution, orbits can be elliptic, but the azimutal period is twice the radial period, meaning that the gravity source is at the center of the ellipse and pericenter and apocenter distances match the semiminor and semimajor axes respectively. How can the latter be found?</p>

CC BY-SA 4.0
2021-05-06T12:59:52.977Z
Link added to the question.
Martrin
<orbital-mechanics><newtonian-gravity>
Non-keplerian orbit features of orbits in uniform mass distribution
Martrin
# Non-keplerian orbit features of orbits in uniform mass distribution

(Asked by: username_0 on 2021-05-06)

For the first Kepler's law the orbit of a planet is an ellipse, with the Sun in one of the foci, but [I've read that](http://astro.utoronto.ca/%7Ebovy/AST1420/notes/notebooks/04.-Orbits-in-Spherical-Potentials.html) in the case of an homogeneous sphere distribution, orbits can be elliptic, but the azimutal period is twice the radial period, meaning that the gravity source is at the center of the ellipse and pericenter and apocenter distances match the semiminor and semimajor axes respectively. How can the latter be found?

**Tags:** orbital-mechanics, newtonian-gravity

> **Comments:**
> **username_0**: Link added to the question.  

---

## 1 answers

### Answer by username_1 on 2021-05-06. Score: 1 | Accepted answer

This sounds like a homework problem so I'll give you most of it but you'll have to work out the details yourself.

[Newton's shell theorem](https://en.wikipedia.org/wiki/Shell_theorem) tells us that for any spherical mass distribution, the potential at a distance $r\_0$ from the center is the same as if all the mass within the shell of radius $r\_0$ were at $r=0$, and all the mass outside of the shell didn't exist at all (it cancels):

$$m(r\_0) = \int\_0^{r\_0}4 \pi r^2\rho(r)dr$$

and if we let $\rho(r)$ to be constant (the sphere is of uniform density):

$$m(r\_0) = 4 \pi \rho \int\_0^{r\_0}r^2 dr = \frac{4}{3} \pi \rho r\_0^3$$

The gravitational potential for a point mass is

$$V(r) = \frac{Gm}{r}$$

Put them together and you have

$$V(r) = \frac{4 \pi G \rho r^3}{3r} = \frac{4}{3} \pi \rho G r^2$$

and a parabolic $~r^2$ potential is called a [harmonic oscillator potential](https://en.wikipedia.org/wiki/Harmonic_oscillator). [Motion in this potential](https://en.wikipedia.org/wiki/Bertrand%27s_theorem#Harmonic_oscillator) (one or many dimensions) is [simple harmonic motion](https://en.wikipedia.org/wiki/Simple_harmonic_motion). That's because the potential is proportional to $r^2 = x^2 + y^2 + z^2$ and the restoring force is proportional to $\nabla r^2 / 2 = x \mathbf{\hat{x}} + y \mathbf{\hat{y}} + z \mathbf{\hat{z}}$. Oscillations have the same restoring force in all directions, so the periods $\omega\_x, \omega\_y, \omega\_z$ will be the same.

Yes, these will be ellipses and since $r = \sqrt{x^2 + y^2 + z^2}$ the value of $r$ will reach a maximum twice during each orbit.

If for example $\mathbf{r}(t) = \cos(\omega t)\mathbf{\hat{x}} + \frac{1}{2}\sin(\omega t)\mathbf{\hat{z}}$ then you will see that $r(t)$ has two maxima and two minima per cycle.

---

[![plot](https://i.sstatic.net/evSmYm.png)](https://i.sstatic.net/evSmY.png)

```
import numpy as np
import matplotlib.pyplot as plt

wt = np.linspace(0, 2*np.pi, 1001)

x, z = np.cos(wt), 0.5*np.sin(wt)
r = np.sqrt(x**2 + z**2)

plt.figure()
plt.plot(wt, x)
plt.plot(wt, z)
plt.plot(wt, r)
plt.show()

```

> **Comments:**
> **username_0**: Thank you. From your answer I understand that I can say that, since $U \propto r^2$ and motion is a simple harmonic motion, if I have, for example, an elliptic trajectory in two dimensions, I'll have oscillations along $x$ and $y$ around a common center (the center of the ellipse, and center of gravity) and that's why, if I have an ellipse of semimajor axis $a$ and semiminor axis $b$, pericenter and apocenter distances matches $a$ and $b$.  
> **username_0**: While in the case of keplerian potential, which provide a force that goes as $1/r^2$, Kepler's first law says that the center of gravity is in one of the ellipse foci, so the smaller and longer distance from center of gravity are both on the same axis.  
> **username_1**: @Martein yes you can say that I think. For consistency only you should say Keplerian orbits come from $1/r$ *potentials* (rather than saying $1/r^2$ forces).  

---

