Artistic Skills Put to the Test: Tides Near and Far

Recall the reading assignment about tides. Draw a circle representing the Earth (mass $M_E$) with 8 equally spaced point masses, $m$, placed around the circumference. Also draw the Moon with mass $M_m$ to the side of the Earth. In the following, do each item pictorally, with vectors showing the relative strengths of various forces at each point. Don't worry about the exact geometry, trig and algebra. I just want you to think about and draw force vectors qualitatively, at least initially.

a) What is the gravitational force due to the Moon, $\vec{F}_{m, cen}$, on a point at the center of the Earth?

This is a pretty straightforward application of Newton's law of universal gravitation, $\frac{GMm}{r^2}$. However, we're not worrying about the math of this (at least not right yet), so we can pretty intuitively draw a vector from the center of the Earth in the direction of the Moon--this is the effect that the Moon's gravity is having on the Earth as a whole.

Not to scale.

b) What is the force vector on each point mass, $\vec{F}_m$, due to the Moon? Draw these vectors at each point.

Again, we aren't too worried about the exact magnitude and direction of each force vector. We could calculate them if we felt a burning desire to--the only tricky part lies in figuring out the exact distances and directions from each point mass to the center of the Moon, which at worst is a slightly ugly geometry problem. But those details aren't really that important here. What we're concerned with is the relative strength of each force vector, which is just a function of distance. As intuition would suggest, the closer point masses are going to feel a stronger tug from the Moon, while the more distant ones on the far side of the planet are going to feel less.

Still not to scale.

It's worth noting that, while it seems like the top and bottom point masses should have a fairly extreme angle, that's due to the difficulty in scaling a drawing like this. As you'll no doubt recall, the Moon is far enough a way that the force vectors are almost (though not quite) parallel.


c) What is the force difference, $\Delta\vec{F}_m$, between each point and Earth's center? This is the tidal force.

What we're looking at here is how hard each of our point masses is being pulled relative to the Earth as a whole. As we just showed, the point masses closer to the Moon are getting pulled harder (since the force vectors have a greater magnitude), and those farther away are feeling a weaker attraction.

It's a bit unclear on that drawing, since it's hard to tell what's the Moon's gravitational force (the original $F$ vectors) and what's the tidal force. But if you take out the extraneous stuff, you're left with something that looks like this:

(Source)

The result is that, when we do some basic vector subtraction, the point masses closer to the Moon are getting pulled with magnitude $\Delta\vec{F}_m$ toward the Moon, while those farther away are (somewhat counterintuitively) getting pulled with magnitude $\Delta\vec{F}_m$ away from the Moon. This seems bizarre at first, but remember that we're looking at the difference in force between each point and the center of the Earth. Since the far side of the Earth is getting pulled less strongly than the planet as a whole, it's actually "stretching" out away from the planet's center. One way to imagine this is by thinking about the Earth as being like a rubber band held at each end. Pulling on one side of the rubber band stretches both ends out away from the center, even if the other hand is holding still.*

Meanwhile, 90 degrees away from the Moon, the force vectors are actually pointing inward. Since the magnitude of the force vectors on these two point masses is pretty close to being the same as that affecting the center of the Earth, $\Delta\vec{F}_m$ is very weak at each one. However, that tiny vertical component of each $\vec{F}_m$ remains, along with an even tinier horizontal component. Because these two points are the slightest bit farther away from the Moon than the center of the Earth, this horizontal component actually points away from the Moon. You can use the rubber band analogy for this part as well--at least for the vertical component of $vec{F}_m$--imagine how, as you stretch out a rubber band, it gets narrower as well as longer. While different processes are at work here, it definitely helps as a way of visualizing what's happening.


d) What will this do to the ocean located at each point?

No tricks here--since the ocean is (surprise!) fluid, it will flow in the direction of the force vectors. As a result, you get high tides on the side of the Earth directly in line with the Moon (on both sides of the planet!), and low tides 90 degrees away.


e) How many tides are experienced each day at a given location located along the Moon's orbital plane?

Again, this follows intuition. Since the Moon orbits us slowly enough that it doesn't affect the number of times the configuration we drew will occur each day, each Earthly point in the Moon's orbital plane will experience two high tides for every rotation of the Earth--one when the Moon is directly overhead, and one when it is at the nadir.


f) Okay, now we will use some math. For the two points located at the nearest and farthest points from the Moon, which are separated by a distance $\Delta r$ compared to the Earth-Moon distance $r$, show that the force difference is given by \[\Delta\vec{F}_m = \frac{2GmM_m}{r^3}\Delta r\]
Alas, our happy time of rough pictoral and geometric approximations has come to an end. Fortunately, though, the worst thing we'll be up against here is a not-terribly-bad derivative. If we define a function $f(r)$ as the universal law of gravitation, this actually works out quite nicely. The force felt by each point mass can be defined as $f(r_0)$ and $f(r_0+2R_E)$. Or, defined slightly differently, $f(r_0)$ and $f(r_0+\Delta r)$, where in this case $\Delta r = 2R_E$. The question also provides a helpful hint in reminding us that \[\lim_{x \to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x} = \frac{d}{dx}f(x) \approx \frac{\Delta f(x)}{\Delta x}\] We have an $f(x)$ and an $f(x+\Delta x)$. Thus, if we take the derivative of $f(r)$, it should point us in the right direction. As with most constant-filled equations, $\frac{df}{dr}$ looks worse than it is, and a simple application of the power rule tells us that \[\frac{df}{dr} = \frac{-2GmM_m}{r^3} \approx \frac{\Delta f(r)}{\Delta r}\] We can ignore the negative sign since what we have here is a comparison--it'll be positive or negative depending on whether we're going from the closer point to the farther one or vise-versa--and if we multiply the $\Delta r$ up to the other side (and remember that $\Delta f(r) = \Delta \vec{F}$), we see that, as we expected \[\Delta \vec{F} = \frac{2GmM_m}{r^3}\Delta r \]


g) Compare the magnitude of the tidal force $\Delta \vec{F}_m$ caused by the Moon to $\Delta \vec{F}_s $ caused by the Sun. Which is stronger and by how much? What happens when the Moon and the Sun are on the same side of the Earth?

For this one, we're really just taking our solution to f), plugging in two sets of numbers, and dividing them. The nice part is that a lot cancels--${2Gm}$ is constant in both, and we're left with $\Delta \vec{F}=\frac{M_s \times r^3}{M_m \times a^3}$ where $a$ is the distance from the Earth to the Sun, 1 AU. It comes out to $0.425$, telling us that the Moon, despite being vastly smaller than the Sun, has almost two and a half times as much tidal affect on the Earth. When the two line up, however, you get a spring tide, where their cumulative affect leads to the highest possible tide. Meanwhile, when the two are 90 degrees apart , the two tides are acting against each other, leading to a neap tide. Thus, there are two configurations that lead to a spring tide--when the moon is directly between the Earth and the Sun, and when the Earth is directly between the Moon and Sun--and two configurations that lead to a neap tide--at the first and last quarter (confusingly, when the Moon is half full either waxing or waning).

This image actually only shows one of the two neap tides, but you can imagine the other--just put the Moon below the Earth instead of above it (Source)

h) How does the magnitude of $\Delta \vec{F}_m$ caused by the Moon compare to the tidal force caused by Jupiter during its closest approach to the Earth ($r \approx $ 4 AU$)?

Mighty Jupiter is going to look a bit weak here, thanks to the fact that $\Delta\vec{F}$ is decreasing as an inverse cube of the distance between the two objects. We'll set up the same way we did in part g), giving us $\Delta \vec{F}=\frac{M_j \times r^3}{M_m \times d_j^3}$. When we once again plug in the quantities, we get a rather disappointing result of $6.68 /times 10^{-6}$--even at its closest approach, Jupiter's effect is six orders of magnitude weaker than the Moon's. Along the Bay of Fundy, which experiences the highest tidal variation in the world, this would mean Jupiter was raising the water level a staggering 0.11 millimeters--about the width of a human hair.

The Bay of Fundy. Jupiter probably isn't making much difference here (Source)

As usual, I'm reserving the end for a shoutout to my collaborators--this time, Anne Madoff and Scott Zhuge.

*I'm not sure who originally came up with this analogy--it wasn't me, though I'm not sure where I first heard it.