<p>The short answer which may or may not be an "Aha!" answer is that what is plotted is what's left over after a much larger, uniform force is subtracted. </p>

<p>The uniform force is the Force from the Moon evaluated at the center of the Earth, and the arrows show the <em>deviation</em> of the actual force from that average.</p>

<p>Why do we do it that way? When looking at the tidal forces on the oceans we treat the Earth as a rigid body with spherical symmetry. With that, we can use a variation of Newton's <a href="https://en.wikipedia.org/wiki/Shell_theorem" rel="nofollow noreferrer">Shell theorem</a> to say that the extended Earth will move the same way as if it were a point mass at its center.</p>

<p>Now the oceans are fluid (the opposite of rigid) and each bit responds to the Moon's force locally.</p>

<p>That force is (skipping constants unnecessary to make the cartoon plot) </p>

<p><span class="math-container">$$F = -\frac{\mathbf{\hat{r}}}{|r|^2}  = -\frac{\mathbf{r}}{|r|^3}$$</span> </p>

<p>where the vector <span class="math-container">$\mathbf{r}$</span> is drawn from the Moon to some point on the Earth and <span class="math-container">$\mathbf{\hat{r}}$</span> is its unit vector. If the Earth's center is at <span class="math-container">$\mathbf{\hat{x}} R$</span> (<span class="math-container">$R$</span> is the Moon-Earth distance) and you subtract <span class="math-container">$-\mathbf{\hat{x}}/R^2$</span> you'll get that image.</p>

<p>In the plot below I've chosen the Earth-Moon distance to be only 10 Earth radii to highlight the slight left-right asymmetry. The tidal force is stronger on the side closer to the Moon.</p>

<p><a href="https://i.sstatic.net/MqXP0.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MqXP0.png" alt="Tidal forces with Moon at 10 Earth radii"></a></p>

<pre><code>import numpy as np
import matplotlib.pyplot as plt

R = 10.0
r_moon = np.array([R, 0], dtype=float)[:, None]
earth  = np.zeros(2)[:, None]
theta = np.linspace(0, 2*np.pi, 49)
positions = earth + np.array([f(theta) for f in (np.cos, np.sin)])

r = positions-r_moon
F = -r * ((r**2).sum(axis=0))**-1.5

r = earth-r_moon
Fmean = -r * ((r**2).sum(axis=0))**-1.5

Ftide = F - Fmean

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    (x, y), (Fx, Fy) = positions, 50.*Ftide
    plt.quiver(x, y, Fx, Fy, width=0.005)
    plt.plot(x, y, '-b')
    plt.xlim(-2, 2)
    plt.ylim(-1.5, 1.5)
    plt.gca().set_aspect('equal')
    plt.subplot(4, 1, 3)
    for thing in F:
        plt.plot(thing)
    plt.subplot(4, 1, 4)
    for thing in Ftide:
        plt.plot(thing)
    plt.show()
</code></pre>

CC BY-SA 4.0
2020-03-25T19:08:07.367Z
Thanks. Very useful. If I understand correctly, the r vector is from any point (relevantly on the surface of the earth) to the center of the moon (which can be considered a single point of mass). Right? So your first "x" should be "r", right? Then I am not sure what is ˆx
Jean-Denis Muys
CC BY-SA 4.0
2020-03-25T19:09:07.553Z
and R being the moon-earth distance, is more precisely the distance between their respective centers? Right?
Jean-Denis Muys
CC BY-SA 4.0
2020-03-25T19:09:54.793Z
I am not sure what is the meaning of the "^" on top of r and x. Is it just to say these are vectors?
Jean-Denis Muys
CC BY-SA 4.0
2020-03-25T22:00:33.130Z
@Jean-DenisMuys This is a pretty standard notation. I'll leave a comment now and modify the text when I can. Bold face font denotes vectors, so $\mathbf{F}$ is a force vector,$ \mathbf{r}$, is a vector from the lunar center to some point, $\mathbf{\hat{x}}$ is a *unit vector* in the x direction (pointing in the Moon to Earth direction). It's pronounced "x hat".
uhoh
CC BY-SA 4.0
2020-03-25T22:17:24.063Z
Sorry for having been ignorant, and thank you for making me slightly less so. :-)
Jean-Denis Muys
CC BY-SA 4.0
2020-03-25T22:29:57.883Z
@Jean-DenisMuys omg nothing to be sorry about; this is Stack Exchange and explaining stuff is what it's all about! Let me get some coffee (it's morning here) and I'll make some nice edits.
uhoh
CC BY-SA 4.0
2020-03-22T03:34:39.343Z
2020-03-26T02:37:16.100Z
2020-03-26T02:37:16.100Z
uhoh
<p>I am seeking to understand why the tidal force is pushing towards the center of the earth (centripetal), at a point that is making with the center of the earth a 90º angle to the moon-earth axis.</p>

<p>This is very well seen in the common picture (from the Wikipedia "<a href="https://en.wikipedia.org/wiki/Tidal_force" rel="noreferrer">Tidal force</a>" article):</p>

<p><a href="https://i.sstatic.net/CKYb7.png" rel="noreferrer"><img src="https://i.sstatic.net/CKYb7.png" alt="from Wikipedia &quot;Tidal force&quot; article"></a></p>

<p>I remember a vectorial construction of that from the French edition of Scientific American ("Pour la Science"), in their August 2001 issue. But I lost it, and it is now behind a pay wall:</p>

<p><a href="https://www.pourlascience.fr/sd/geosciences/les-caprices-des-marees-4441.php" rel="noreferrer">https://www.pourlascience.fr/sd/geosciences/les-caprices-des-marees-4441.php</a></p>

<p>I used to have the physics/mathematical background to understand that, and perhaps I could dig it out of the decade-old sediments of my memory. The vectorial construction looked interesting to me as slightly easier to explain to people around than the differential equations.</p>

CC BY-SA 4.0
2020-03-22T12:24:34.370Z
I'd recommend viewing this video on the [PBS Spacetime channel](https://www.youtube.com/watch?v=pwChk4S99i4) on YouTube which explains the tricky idea behind tides.
StephenG - Help Ukraine
CC BY-SA 4.0
2020-03-24T17:12:31.963Z
Your diagram, while correct, is misleading.  That's what uhoh's answer points out
Carl Witthoft
CC BY-SA 4.0
2020-03-25T19:13:31.010Z
misleading how? BTW, this is not my diagram, but Wikipedia's. But I think it reflects rather well the local experience.
Jean-Denis Muys
<tidal-forces>
Vectorial construction of tidal forces - or why is it centripetal at low tide
Jean-Denis Muys
# Vectorial construction of tidal forces - or why is it centripetal at low tide

(Asked by: username_0 on 2020-03-22)

I am seeking to understand why the tidal force is pushing towards the center of the earth (centripetal), at a point that is making with the center of the earth a 90º angle to the moon-earth axis.

This is very well seen in the common picture (from the Wikipedia "[Tidal force](https://en.wikipedia.org/wiki/Tidal_force)" article):

[![from Wikipedia "Tidal force" article](https://i.sstatic.net/CKYb7.png)](https://i.sstatic.net/CKYb7.png)

I remember a vectorial construction of that from the French edition of Scientific American ("Pour la Science"), in their August 2001 issue. But I lost it, and it is now behind a pay wall:

<https://www.pourlascience.fr/sd/geosciences/les-caprices-des-marees-4441.php>

I used to have the physics/mathematical background to understand that, and perhaps I could dig it out of the decade-old sediments of my memory. The vectorial construction looked interesting to me as slightly easier to explain to people around than the differential equations.

**Tags:** tidal-forces

> **Comments:**
> **username_1**: I'd recommend viewing this video on the [PBS Spacetime channel](https://www.youtube.com/watch?v=pwChk4S99i4) on YouTube which explains the tricky idea behind tides.  
> **username_2**: Your diagram, while correct, is misleading.  That's what username_3's answer points out  
> **username_0**: misleading how? BTW, this is not my diagram, but Wikipedia's. But I think it reflects rather well the local experience.  

---

## 1 answers

### Answer by username_3 on 2020-03-22. Score: 4

The short answer which may or may not be an "Aha!" answer is that what is plotted is what's left over after a much larger, uniform force is subtracted.

The uniform force is the Force from the Moon evaluated at the center of the Earth, and the arrows show the *deviation* of the actual force from that average.

Why do we do it that way? When looking at the tidal forces on the oceans we treat the Earth as a rigid body with spherical symmetry. With that, we can use a variation of Newton's [Shell theorem](https://en.wikipedia.org/wiki/Shell_theorem) to say that the extended Earth will move the same way as if it were a point mass at its center.

Now the oceans are fluid (the opposite of rigid) and each bit responds to the Moon's force locally.

That force is (skipping constants unnecessary to make the cartoon plot)

$$F = -\frac{\mathbf{\hat{r}}}{|r|^2} = -\frac{\mathbf{r}}{|r|^3}$$

where the vector $\mathbf{r}$ is drawn from the Moon to some point on the Earth and $\mathbf{\hat{r}}$ is its unit vector. If the Earth's center is at $\mathbf{\hat{x}} R$ ($R$ is the Moon-Earth distance) and you subtract $-\mathbf{\hat{x}}/R^2$ you'll get that image.

In the plot below I've chosen the Earth-Moon distance to be only 10 Earth radii to highlight the slight left-right asymmetry. The tidal force is stronger on the side closer to the Moon.

[![Tidal forces with Moon at 10 Earth radii](https://i.sstatic.net/MqXP0.png)](https://i.sstatic.net/MqXP0.png)

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

R = 10.0
r_moon = np.array([R, 0], dtype=float)[:, None]
earth  = np.zeros(2)[:, None]
theta = np.linspace(0, 2*np.pi, 49)
positions = earth + np.array([f(theta) for f in (np.cos, np.sin)])

r = positions-r_moon
F = -r * ((r**2).sum(axis=0))**-1.5

r = earth-r_moon
Fmean = -r * ((r**2).sum(axis=0))**-1.5

Ftide = F - Fmean

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    (x, y), (Fx, Fy) = positions, 50.*Ftide
    plt.quiver(x, y, Fx, Fy, width=0.005)
    plt.plot(x, y, '-b')
    plt.xlim(-2, 2)
    plt.ylim(-1.5, 1.5)
    plt.gca().set_aspect('equal')
    plt.subplot(4, 1, 3)
    for thing in F:
        plt.plot(thing)
    plt.subplot(4, 1, 4)
    for thing in Ftide:
        plt.plot(thing)
    plt.show()

```

> **Comments:**
> **username_0**: Thanks. Very useful. If I understand correctly, the r vector is from any point (relevantly on the surface of the earth) to the center of the moon (which can be considered a single point of mass). Right? So your first "x" should be "r", right? Then I am not sure what is ˆx  
> **username_0**: and R being the moon-earth distance, is more precisely the distance between their respective centers? Right?  
> **username_0**: I am not sure what is the meaning of the "^" on top of r and x. Is it just to say these are vectors?  
> **username_3**: @username_0 This is a pretty standard notation. I'll leave a comment now and modify the text when I can. Bold face font denotes vectors, so $\mathbf{F}$ is a force vector,$ \mathbf{r}$, is a vector from the lunar center to some point, $\mathbf{\hat{x}}$ is a *unit vector* in the x direction (pointing in the Moon to Earth direction). It's pronounced "x hat".  
> **username_0**: Sorry for having been ignorant, and thank you for making me slightly less so. :-)  
> **username_3**: @username_0 omg nothing to be sorry about; this is Stack Exchange and explaining stuff is what it's all about! Let me get some coffee (it's morning here) and I'll make some nice edits.  

---

