Section 12

V. Hunter Adams
In [1]:
from IPython.display import Latex
import matplotlib.pyplot as plt
%matplotlib inline
import numpy
from matplotlib import animation, rc
from IPython.core.display import HTML 
from IPython.display import HTML
from ipywidgets import *
from scipy.integrate import odeint
from IPython.display import Image

Rings and the Roche Limit

The Roche Limit

We learned previously that tidal forces are caused by differential gravity felt across a body. The tides on the Earth, for example, are a consequence of the fact that the Moon is pulling harder on one side of the Earth than on the other, causing the oceans to bulge. We've discussed how this effect becomes more pronounced the closer that one gets to a gravitational body.

Eventually, if something like a moon gets too close to a very graviational body, these differential forces can rip that moon apart. The distance from the gravitational body at which this happens is called the Roche Limit, and it's given by the following equation:

\begin{align} d &= r\left(2\frac{M}{m}\right)^{\frac{1}{3}} \end{align}

where $r$ is the radius of the satellite, $m$ is the mass of the satellite, and $M$ is the mass of the parent body. Let's consider this for a second. As $r$ goes up, what happens to $d$? It goes up, because the larger body is more susceptible to differential forces. What happens as $M$ goes up? $d$ again goes up, since the tidal forces from a more massive parent body will be stronger. How about if $m$ goes up? In this case, $d$ goes down because the more massive satellite is better able to maintain its structure with its own self gravity.

Roche Limit in terms of Density

Using some known relationships, we could re-write the above expression in terms of other characteristics of the two bodies, such as the densities. Recall that the density of a planet is given by its mass divided by its volume. That is:

\begin{align} \rho &= \frac{M}{V} \end{align}

If we assume that each body is a sphere, then the volume is given by:

\begin{align} V &= \frac{4}{3}\pi r^3 \end{align}

where $r$ is the radius of the body. We can then substitute this into the expression for the density to get:

\begin{align} \rho &= \frac{M}{\frac{4}{3}\pi r^3}\\ &= \frac{3M}{4\pi r^3} \end{align}

Solving this expression for the mass (so that we can substitute it into the Roche Limit equation), we have:

\begin{align} M &= \frac{4}{3}\pi r^3 \rho \end{align}

Substituting into the Roche Equation:

\begin{align} d &= r\left(2 \frac{\frac{4}{3}\pi R^3\rho_{parent}}{\frac{4}{3}\pi r^3 \rho_{satellite}}\right)^{\frac{1}{3}} \end{align}

Simplifying:

\begin{align} d &= r\left(2 \frac{R^3\rho_{parent}}{r^3 \rho_{satellite}}\right)^{\frac{1}{3}}\\ &= \frac{r}{r}R\left(2 \frac{\rho_{parent}}{\rho_{satellite}}\right)^{\frac{1}{3}}\\ &= R\left(2 \frac{\rho_{parent}}{\rho_{satellite}}\right)^{\frac{1}{3}} \end{align}

Is Enceladus Safe?

Enceladus has a density of 1.61 g/cc. Given that Saturn has a density of 0.687 g/cc and a radius of 58,000 km, what is the Roche Limit for moons of this density in the Saturn system?

Plug and chug!

\begin{align} d &= R\left(2 \frac{\rho_{parent}}{\rho_{satellite}}\right)^{\frac{1}{3}}\\ &= 58000\left(2\frac{0.687}{1.61}\right)^{\frac{1}{3}}\\ &= 55,015 \text{ km} \end{align}

Enceladus orbits 238,000 km from the center of Saturn. What is the ratio of its orbital radius to the Roche Limit?

\begin{align} \text{ratio} &= \frac{238000}{55015}\\ &= 4.33 \end{align}

So...yes, Enceladus is safely outside the Roche Limit. In fact, it's over four times as far as it must be to remain coherent.

But what if....

Let us imagine that Saturn's rings all came from one moon that got ripped apart by tidal forces. Let's pretend that this moon had the density of Enceladus, and that it orbited at approximately the same distance as Enceladus. How large would this moon needed to have been, given that the total mass of Saturn's rings is $3 \times 10^{19}$ kg?

Consider the Roche Limit equation in its original form:

\begin{align} d &= r\left(2\frac{M}{m}\right)^{\frac{1}{3}} \end{align}

Solve for $r$:

\begin{align} r &= \frac{d}{\left(2\frac{M}{m}\right)^{\frac{1}{3}}} \end{align}

Use Saturn's density and radius to solve for its mass:

\begin{align} \rho_{Saturn} &= \frac{0.687\text{ g}}{1\text{ cc}}\cdot\frac{1kg}{1000g}\cdot \left(\frac{100cm}{1m}\right)^3 \cdot\left(\frac{1000m}{km}\right)^3= 6.87\times 10^{11}\text{ kg/}km^3\\ R_{Saturn} &= 58,000km \end{align}

Solve for the mass:

\begin{align} M_{Saturn} &= \rho V\\ &= \rho\left(\frac{4}{3}\pi R_{Saturn}^3\right)\\ &= 6.87\times 10^{11}\left(\frac{4}{3}\pi 58000^3\right)\\ &= 5.61\times 10^{26} kg \end{align}

Substitute into the equation for $r$:

\begin{align} r &= \frac{d}{\left(2\frac{M}{m}\right)^{\frac{1}{3}}}\\ &= \frac{238,000}{\left(2\frac{5.61\times 10^{26}}{3 \times 10^{19}}\right)^{\frac{1}{3}}}\\ &= 711\text{ km} = 711,000\text{ m} \end{align}

RADAR Depth Sounding

RADAR is capable of penetrating below a top surface. The depth of penetration is given by:

\begin{align} L &= \frac{\lambda}{2\pi \sqrt{\epsilon}\tan{\delta}} \end{align}

Treating a quick example, let's consider some imaginary medium with $\epsilon=2.8$ and $\tan{\delta} = 10^{-3}$. What wavelength would be required to probe to a depth of 3km in this medium?

Let's solve for $\lambda$:

\begin{align} \lambda &= 2\pi L\sqrt{\epsilon}\tan{\delta}\\ &= 2 \pi \left(3000\right)\sqrt{2.8}\cdot10^{-3}\\ &= 31.54m \end{align}

Jupiter

Click on this: https://www.nasa.gov/centers/goddard/images/content/388625main_Jupiter_Approach.gif. This gif is made from photographs taken over 60 Jupiter days by Voyager 2 as it approached Jupiter. In this video, you can see many of the things that we'll be discussing in today's section.

Atmosphere

Jupiter's atmosphere is very dynamic. The clouds are banded, with high-altitude white zones of ammonia clouds and lower altitude red belts. Oval-shaped structures in the atmosphere are storms, with the most famous being the Great Red Spot (a high pressure zone).

The composition of Jupiter is very close to the composition of the Sun (86 percent hydrogen by volume, 13 percent helium, and some methane, ammonia, and other stuff). If it were about 80 times more massive, Jupiter could have become a star.

Interior

As previously mentioned, Jupiter is composed largely of hydrogen. As one plunges deeper into Jupiter's interior, the pressure (and thus the temperature) increases. For this reason, we see molecular hydrogen in the outer layers of Jupiter (gaseous and liquid), with metallic hydrogen deeper inside. In the deep interior there is a layer of ice surrounding a rock and metallic core.

In [3]:
PATH = "/Users/hunteradams/Documents/PhD Semester 5/Sections/"
Image(filename = PATH + "jovian.png", width=600, height=600)
Out[3]: