Worksheet 14.2, Problem 1: Tides

Draw a circle representing the Earth (mass \(M_\oplus\)), with 8 equally-spaced point masses, m, placed around the circumference. Also draw the Moon with mass \(M_{moon}\) to the side of the Earth. In the following, do each item pictorially, 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.

The Earth has mass \(M_\oplus\) and the Moon has mass \(M_{moon}\). The red dots along the circumference of the
Earth represent point-masses, m.

a) What is the gravitational force due to the Moon, \(\vec{F}_{moon, cen}\), on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head).

The Earth experiences a gravitational force from the Moon. The red vector
represents the force of the Moon on a point at the center of the Earth.

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

Each red vector arrow represents the force of the Moon on a given point on the Earth. Different points on Earth will experience different forces from the Moon due to their varying distances away from the Moon.

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

The black vectors represent the tidal forces, which are the differences between the force vectors at each point along the circumference of the Earth and the force vector at the center of the Earth.

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

The light blue around the Earth depicts the effect of the tidal forces on the ocean.

Since the point on Earth closest to the Moon experiences a stronger gravitational pull from the Moon than the point at the center of the Earth, it accelerates more than the center of the Earth, and thus causes the ocean on this side to 'stretch' towards the Moon. Similarly, the point on Earth furthest from the Moon has a weaker gravitational pull from the Moon than the center of the Earth, causing it to accelerate less than the center. Thus here, the two points stretch apart from each other, giving the appearance that the ocean is stretching away from the Earth. These are what we know as the tides.

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

There are two tides on the Earth at any given moment of the day: on the side facing the moon and the side opposite the moon. Since the Earth completes one full rotation per day, any given location will experience a tide twice a day.

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 

\(\Delta F=\dfrac{2GmM_{moon}}{r^3}\Delta r\)

(Hint: recall that 

\(\displaystyle\lim_{\Delta x \to 0} \dfrac{f(x+\Delta x)-f(x)}{\Delta x}=\dfrac{d}{dx}f(x)\approx \dfrac{\Delta f(x)}{\Delta x}\)

And \(\Delta r\ll r\).)

The distance between the Moon and closest point on the Earth is r, and the diameter of the Earth (the distance between the closest and furthest points on the Earth) is \(\Delta r\). Let's call the force on the point on the Earth closest to the Moon \(F_1\) and the force on the point furthest from the moon \(F_2\). The difference in the gravitational forces on these two points is equal to:

\(\Delta F=F_1-F_2=\dfrac{GmM_{moon}}{\left(r-\frac{\Delta r}{2}\right)^2}-\dfrac{GmM_{moon}}{\left(r+\frac{\Delta r}{2}\right)^2}\)

           \(=GmM_{moon}\left[\dfrac{1}{\left(r-\frac{\Delta r}{2}\right)^2}-\dfrac{1}{\left(r+\frac{\Delta r}{2}\right)^2}\right]\)

           \(=GmM_{moon}\left[\dfrac{1}{r^2\left(1-\frac{\Delta r}{2r}\right)^2}-\dfrac{1}{r^2\left(1+\frac{\Delta r}{2r}\right)^2}\right]\)

We can Taylor expand this using the Taylor expansions \(\frac{1}{(1-\epsilon )^2}\approx 1+2\epsilon\)   and   \(\frac{1}{(1+\epsilon )^2}\approx 1-2\epsilon\).

 \(\Delta F=GmM_{moon}\left[\dfrac{1}{r^2}\left(1+\frac{\Delta r}{r}\right)-\dfrac{1}{r^2}\left(1-\frac{\Delta r}{r}\right)\right]\)

We simplify this to get the final answer:

\(\Delta F=\dfrac{2GmM_{moon}}{r^3}\Delta r\)

g) Compare the magnitude of the tidal force \(\Delta F_{moon}\) caused by the Moon to \(\Delta F_\odot\) 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?

We can use the expression we found from part (f) to compare the tidal forces from the Moon and from the Sun:

\(\dfrac{\Delta F_{moon}}{\Delta F_\odot}=\dfrac{\frac{2GmM_{moon}}{r_{moon}^3}\Delta r}{\frac{2GmM_\odot}{r_\odot^3}\Delta r}=\dfrac{M_{moon}}{M_\odot}\left(\dfrac{r_\odot}{r_{moon}}\right)^3\)

We know that \(M_{moon}=7.35\times 10^{25}\) g, \(M_\odot =1.99\times 10^{33}\) g, \(r_{moon}=3.63\times 10^8\) m, and \(r_\odot =1.50\times 10^{11}\) m. Plugging these values in, we get:

\(\dfrac{\Delta F_{moon}}{\Delta F_\odot}=\dfrac{7.35\times 10^{25}\text{ g}}{1.99\times 10^{33}\text{ g}}\left(\dfrac{1.50\times 10^{11}\text{ m}}{3.36\times 10^8\text{ m}}\right)^3\approx 2.6\)

The magnitude of the tidal force caused by the Moon is about 3 times stronger than the tidal force caused by the Sun. While the Sun is significantly more massive, it is much further away compared to the Moon, which is why it has a smaller tidal force.

Since both the Moon and the Sun cause a tidal force on the Earth, both forces are interacting with the Earth at the same time in varying ways. When the Sun and the Moon are on the same side of the Earth, their tidal forces will add to create larger tides on Earth. These tides are called Spring Tides. Conversely, when the Sun and the Moon are on opposite sides of the Earth, the tidal forces will subtract from each other to make smaller tides on Earth, called Neap Tides.

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

Again, we can use the expression from part (f) to create a ratio:

\(\dfrac{\Delta F_{moon}}{\Delta F_J}=\dfrac{M_{moon}}{M_J}\left(\dfrac{r_J}{r_{moon}}\right)^3\)

The mass of Jupiter is \(M_J=1.90\times 10^{30}\) g and its distance from the Earth on its closest approach is \(r_J\approx 4\text{ AU}\approx 7.50\times 10^{11}\) m. Plugging these in, we get:

\(\dfrac{\Delta F_{moon}}{\Delta F_J}=\dfrac{7.35\times 10^{25}\text{ g}}{1.90 \times 10^{30}\text{ g}}\left(\dfrac{7.50\times 10^{11}\text{ m}}{3.36\times 10^8\text{ m}}\right)^3\approx 3.4\times 10^5\)

We have found that the tidal force from the Moon is about 400,000 times stronger than the tidal force from Jupiter. So the tidal force from Jupiter can be considered negligible, and we can see that the Moon is the most dominant cause for tides on the Earth.

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their collaboration in solving this problem.