# Intro - Saturn's Rings Remember [[Newton's Law of Gravitation - The Intro#Superposition - Gravitation Edition|superposition?]] Well, let's apply this, only this time we'll take into account an actual planet-moon system. The moon is a big object, and there are many factors that affect the gravitational force incident on the moon from our planet, such as the position of the moon in its orbit, the part of the moon relevant to our study... the list goes on and on. As a result, scientists have given up taking the moon as a 'point source', resorting to taking each part of the moon as its own particle. If you're a math guy, you can probably figure out that we're [[Gradient|approximating the moon as a sphere and using an integral]] to get a particle summation of the gravitational force, but I'll try to stay away from that for now. Onto the scenario. Say our moon were to start drifting closer to us. As a *couple of you* may think, yes, it's the lizard people again. Curses! Regardless of why this has happened, we're not going to be having a very good time in a bit. The moon is responsible for our tides Eventually, this object gets ripped apart, with chunks of rock falling towards the planet or settling into a new orbit around the planet. That's the story of Saturn's rings, theorised to be the result of a Mimas-sized planetary object straying too close to the (literal!) planet. A worthy sacrifice, if I'd say! ![[Pasted image 20231101212538.png]] *Saturn and its rings in all their glory! Image Credit: ESA/Hubble* # The Actual Roche Limit - Tidal Forces There are actually two types of Roche limit - the *experimental Roche limit*, which assumes that the entire planet breaks up at once, and the *actual Roche limit*, which is what we see in real life. Let's see **why** a moon would be ripped apart by gravity in the first place. ![[moon-roche.png]] In the diagram above, the point mass $u$ corresponds roughly to a 'point' on the lunar surface that is closest to the planet it orbits. As the moon is about to be 'ripped apart', the forces on $u$ and the centre of mass $p$ (NOT TO SCALE): ![[roche-fb.png]] So if point $u$ is to accelerate outwards towards the planet, the gravitational force of the planet $F_p$ at point $u$ **MINUS** the gravitational force of the planet at the centre of mass will equal to the inwards gravitational force of the planet itself at the Roche limit. This difference in the gravitational force from the planet across the moon is known as the **tidal force**, and is the same thing as the one which causes the tides on Earth! Therefore, our first step will be to find an expression for the tidal force. The tidal force $F_T$ is given as: $F_{T} = F_{r - \Delta r} - F_{r}$ Where $\Delta r$ refers to the radius of the moon, of which one separates the point $u$ and the centre of mass (COM). Note the sign - for this case, since we am taking the closest point to the planet, the force at $u$ must be higher than the force at the COM $r$. Therefore, $F_r$ equals to: $F_{r} = \frac{GMu}{r^2}$ Since the moon is now differentiated into many 'points' - as its radius is no longer negligible to its primary - it is possible to treat $F_r$ also as a point mass that also has mass $u$, the mass we assign to all points in the sphere. Remember - we're only looking at the **instantaneous** gravitational attraction in the centre of the planet Now, let's find $F_{r + \Delta r}$. Writing it out using [[Newton's Law of Gravitation - The Intro|Newton's Law of Gravitation]] gives: $F_{r-\Delta r} = \frac{GMu}{(r-\Delta r)^2}$ Where $M$ is the mass of the planet and $G$ is the universal gravitational constant, roughly $6.67 \times 10^{-11}$. Now, let's remove the $r^2$ term from the fraction: $F_{r-\Delta r} = \frac{GMu}{r^2} \left( \frac{1}{\left( 1-\frac{\Delta r}{r} \right)^2} \right)$ Using the [[Counting Principles - The Binomial Series (Maths)|binomial expansion of a negative coefficient]], we can simplify the rightmost fraction to: $F_{r-\Delta r} = \frac{GMu}{r^2} \left( 1 + \frac{2\Delta r}{r} \right)$Now, notice something. Doesn't $\frac{GMu}{r^2}$ equal to the force experienced by the point at the centre of mass then? This means that we can rearrange our expression to: $F_{r-\Delta r} = F_{r} + \frac{2GMu}{r^3} \Delta r$ When we compare this to our original expression for the tidal force, we get our formula for the tidal force: $F_{T} = \frac{2GMu}{r^3} \Delta r$ Where, once again: - $G$ is the **universal gravitational constant** - $M$ is the **mass of the planet** - $u$ is a **point mass** - $r^3$ is the **distance to the center of mass of the planet** - $\Delta r$ is the **radius of the moon** Now, as we stipulated before, the moon will rip itself apart WHEN this tidal force is greater than its own gravitational force acting on its edges.(feel free to pause and think about **why** if it still doesn't click!) So we can now write the expression: $\frac{2GMu}{r^3} \Delta r = \frac{Gmu}{\Delta r^2}$ Where now $m$ is the **TOTAL mass** of the moon. Our target is $r$, as the distance from the centre of mass of the planet to the moon is the defined 'Roche Limit', so we can solve for it to get: $2M \Delta r ^3 = m r^3 $ $r \approx 1.26 \ \Delta r \left( \frac{M}{m} \right) ^{\frac{1}{3}}$ You've probably found a version that's described as a density. To find it, we can write $m$ as a function of $r_{p}$, which is the radius of the planet: $M= \rho_{p}V = \frac{4\pi r_{p}^3\rho_{{p}}}{3}$ Where $\rho_p$ is the density of the planet. Now, when we substitute this back, we just get: $\frac{\frac{8}{3}r_{p}^3\rho_{p}}{r^3} = \frac{m}{\Delta r^3}$ Now noting that the density of the moon $\rho_{m}$ is equal to $\frac{\frac{3}{4}m}{\Delta r^3}$, we can cancel out terms to give: $r^3 = 2\frac{r_{p}^3\rho_{p}}{\rho_{m}}$ $r \approx 1.26 r_{p}\left( \frac{\rho_{p}}{\rho_{m}} \right)^\frac{1}{3}$ Done! So as long as we know the radius of the original planet, and the densities of the two other orbiting bodies, **we can find the radius to which the moon will be ripped apart!** Was it worth it, Saturn? Sacrificing another one of your moons just for a couple hundred million years of stunning beauty? WAS IT?