<p><a href="https://en.wikipedia.org/wiki/Tidal_force" rel="noreferrer">Tidal forces</a> generated are proportional to <span class="math-container">$m/r^3$</span>, where <span class="math-container">$m$</span> is the mass of the Earth satellite and <span class="math-container">$r$</span> is the semi-major axis (we assume circular orbits for simplicity).  The derivation of this relationship is performed nicely by <a href="http://butikov.faculty.ifmo.ru/TidesOD.pdf" rel="noreferrer">Butikov</a>.</p>
<p>So <span class="math-container">$m_m/r_m^3 = m_c/r_c^3$</span>, where</p>
<p><span class="math-container">$m_m \approx 7.3 \times 10^{22} \rm\, kg$</span> is the mass of the Moon,</p>
<p><span class="math-container">$m_c \approx 9.1 \times 10^{20} \rm\, kg$</span> is the mass of Ceres,</p>
<p>and <span class="math-container">$r_m \approx 385,000 \rm\, km$</span> is the semi-major axis of the Moon's orbit around the Earth</p>
<p><strong>Solve for the Ceres orbit distance <span class="math-container">$r_c$</span> to get about <span class="math-container">$89,000\rm\, km$</span> to exert the same tidal effects as the Moon.</strong></p>
<p><strong>Note:</strong> Tidal forces are due to a <em>difference</em> or <em>gradient</em> of gravitational forces across a body instead of simply the magnitude of the gravitational force on the body.  That is why the lunar tides on Earth are more powerful than the solar tides, even though the Sun exerts 177 times more gravitational pull on the Earth than the Moon exerts on the Earth. <a href="https://oceanservice.noaa.gov/education/tutorial_tides/tides02_cause.html" rel="noreferrer">NOAA has an excellent description of this</a> with the following illuminating graphic:
<a href="https://i.sstatic.net/dadOX.gif" rel="noreferrer"><img src="https://i.sstatic.net/dadOX.gif" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2021-06-21T22:23:38.990Z
The Ceres moon would have a pretty short month (~3.5 days?), so we'd get spring & neap tides much more frequently.
PM 2Ring
CC BY-SA 4.0
2021-06-22T02:21:20.037Z
@DavidHammen:  I don't think the gold badge cares if the answer is correct, it just cares if people vote for it (teasing for the second "correct" in your sentence").  I agree with the sentiment and have contributed my vote.
Ross Millikan
CC BY-SA 4.0
2021-06-22T07:11:58.827Z
@DavidHammen I share your sympathy but that's not how [the badge](https://astronomy.stackexchange.com/help/badges/39/populist) works: the accepted answer needs to have a score of at least 11.
Glorfindel
CC BY-SA 4.0
2021-06-22T11:28:30.707Z
I was wondering if that distance was within the Roche limit, but for the Moon that's about 9500 km. So Ceres would likely also survive at 90,000 km.
CJ Dennis
CC BY-SA 4.0
2021-06-22T11:28:53.350Z
@CJDennis Yes, I included that below. Roche limit for Ceres on Earth is around 11175 km.
User123
CC BY-SA 4.0
2021-06-23T15:22:15.047Z
@DavidHammen Here you have it, he got the golden Populist badge.
User123
CC BY-SA 4.0
2021-06-21T16:58:24.037Z
2021-06-21T20:13:04.250Z
2021-06-21T20:13:04.250Z
Connor Garcia
<h2>It can't cause tides of the same strength</h2>
<p>Though you could have Ceres exert the same force on the ocean it has to be at a much smaller orbit. A smaller orbit means a shorter orbital period. This means that the Ceres tides can't be same as the current ones.</p>
<p>To calculate the wanted equations, we need to use <a href="https://www.britannica.com/science/sidereal-period" rel="noreferrer">sidereal periods</a>: The Earth rotates <a href="https://en.wikipedia.org/wiki/Sidereal_time#Comparison_to_solar_time" rel="noreferrer">once every 23.935 hours</a> and the Moon orbits the Earth <a href="https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html" rel="noreferrer">every 655 hours</a>. Ceres would need to be orbiting at <span class="math-container">$\approx 90,000\rm\,km$</span> to exert the same pull (as shown by <a href="https://astronomy.stackexchange.com/a/44460">Connor's answer</a>) – this would a give it an orbital period of <a href="https://keisan.casio.com/exec/system/1224665242" rel="noreferrer">73 hours</a>.</p>
<p>The comparitive periods mean that over the period of rotation Ceres has moved much further than the Moon would have done and so the net angle of pull over a day differs.</p>
<p>Under an assumption that Ceres would rotate in the same direction as Moon (counter-clockwise), we can set an example: After 12 hours, the Earth has completed half a rotation and the ocean on this side of the Earth is now experiencing a pull in the opposite direction. The Moon has moved ~2% of its rotation whereas Ceres has moved ~16% of its orbit. The Moon pulls in the exact opposite direction on the ocean after ~12.5 hours compared to Ceres' 17.9 hours.</p>
<p>To get the above difference in time periods I just did:
<span class="math-container">$$
T_{\text{Earth-Moon}}\omega_{\text{Earth}}-T_{\text{Earth-Moon}}\omega_{\text{Moon}} = \pi
$$</span>
and
<span class="math-container">$$
T_{\text{Earth-Ceres}}\omega_{\text{Earth}}-T_{\text{Earth-Ceres}}\omega_{\text{Ceres}} = \pi
$$</span>
under an assumption that Ceres rotates counter-clockwise: in the same direction as the Moon.</p>
<p><span class="math-container">$$
T_{\text{Earth-Moon}} = \frac{\pi}{\omega_{\text{Earth}}-\omega_{\text{Moon}}}=\frac{1}{2(\frac{1}{23.935}-\frac{1}{655})}\rm\, h=12.42\rm\, h
$$</span></p>
<p><span class="math-container">$$
T_{\text{Earth-Ceres}} = \frac{\pi}{\omega_{\text{Earth}}-\omega_{\text{Ceres}}}=\frac{1}{2(\frac{1}{23.935}-\frac{1}{73})}\rm\, h=17.81\rm\,h
$$</span>
On the image:
<a href="https://i.sstatic.net/28NoZ.png" rel="noreferrer"><img src="https://i.sstatic.net/28NoZ.png" alt="Image" /></a></p>
<hr />
<p>Other effects:</p>
<ul>
<li>The Moon would be between the Earth and Sun more often, meaning the pull of Moon and Sun together creates a greater pull and larger tides (larger difference between <a href="https://oceanservice.noaa.gov/facts/springtide.html" rel="noreferrer">neap and spring tide</a>)</li>
<li>The Moon would more often be at right angles to the Sun-Earth line – a point at which the pull on the oceans is smaller and the tides are smaller (smaller difference between neap and spring tide)</li>
</ul>

CC BY-SA 4.0
2021-06-22T15:57:00.617Z
Frequency doesn't matter for the answer, just how high the tides are (the distance Ceres needs to be so that they're about equally high).
John
CC BY-SA 4.0
2021-06-22T17:05:55.273Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/126772/discussion-between-lio-elbammalf-and-user123).
Lio Elbammalf
CC BY-SA 4.0
2021-06-22T17:38:21.810Z
@John Frequency **does** matter. The Moon is currently receding from the Earth at significantly more than the average recession rate over the last several hundred million years. One key reason is the current shape of the North Atlantic. The shape is just right to enable resonances and hence greater losses with the M2 tide (frequency = one cycle per 12.42 hours).  Replacing the Moon with Ceres at ~89000 km from the Earth's center most likely would not result in the huge tides that are currently seen in the Bay of Fundy, or similar seen high swings elsewhere in the North Atlantic.
David Hammen
CC BY-SA 4.0
2021-06-23T18:06:51.360Z
This answer should probably mention that resonances are important for tide height.  Ignoring that, a longer interval between low tides give more time for the water to "slosh" further to the bulges. So in parts of the world that don't resonate significantly, you could maybe match the tide height with somewhat smaller tidal forces.  The finer details like spring/neap would of course be different, along with the interval itself not being close to syncing with 24 hours, of course.
Peter Cordes
CC BY-SA 4.0
2021-06-22T15:17:10.020Z
2021-06-22T18:52:16.643Z
2021-06-22T18:52:16.643Z
Lio Elbammalf
<p>As is nicely put on the Wikipedia page about tidal forces, the tidal force is given by
<span class="math-container">$$T=Gm\frac{2r}{d^3}$$</span>
where <span class="math-container">$T$</span> is the tidal force (see below), <span class="math-container">$G=6.67\cdot 10^{-11}\rm\,\frac{m^3}{kg s^2}$</span> is the gravitational constant, <span class="math-container">$r$</span> is the radius of the Earth, and <span class="math-container">$d$</span> is the distance between the centers of the two objects. This is not force in correct sense (in Newtons), but is given by <span class="math-container">$\frac{m}{s^2}$</span>, like acceleration. You said, that the tidal forces must be the same, so:
<span class="math-container">$$T_M=T_C$$</span>
<span class="math-container">$$GM_M\frac{2r_E}{d_M^3}=GM_C\frac{2r_E}{d_C^3}$$</span>
<span class="math-container">$$\frac{M_M}{d_M^3}=\frac{M_C}{d_C^3}$$</span>
<span class="math-container">$$d_C=d_M\left(\frac{M_C}{M_M}\right)^\frac{1}{3}\approx 89000\rm\, km$$</span>
The height of the center above the surface would thus need to be around 82500 km.</p>
<p>But note that the tidal forces of Earth on Ceres would be large. Would Earth destroy Ceres? To answer this, we need an equation for <a href="https://en.wikipedia.org/wiki/Roche_limit#Rigid-satellite_calculation" rel="noreferrer">Roche limit</a>
<span class="math-container">$$d=R_m\left(2\frac{M_M}{M_m}\right )^\frac{1}{3}$$</span>
where <span class="math-container">$R_m$</span> is the radius of secondary, <span class="math-container">$M_M$</span> is the mass of primary, and <span class="math-container">$M_m$</span> is the mass of secondary. With inserting the data for Earth and Ceres, we get
<span class="math-container">$$d=11175\rm\, km$$</span>
and <span class="math-container">$11175\rm\, km &lt;89000\rm\, km$</span>. Thus, Ceres wouldn't be destroyed.</p>
<hr />
<p>But, as @JamesK said, how would the tides look like? By the equation:
<span class="math-container">$$a_c=a_g$$</span>
<span class="math-container">$$\frac{v^2}{r}=\frac{GM}{r^2}$$</span>
and
<span class="math-container">$$v=\frac{2\pi r}{t_0^2}$$</span>
we get
<span class="math-container">$$t_0=\sqrt{\frac{4\pi^2r^3}{GM}}=\sqrt{\frac{4\pi^2 (8.9\cdot 10^7)^3}{6.67\cdot 10^{-11}\cdot 6\cdot 10^{24}}}\rm\, s=3.052\rm\, d$$</span>
So, Ceres would make one <strong>sidereal rotation period</strong> in 3.052 days. If Ceres rotates clockwise, then the one synodic period around the Earth is <span class="math-container">$((1\rm\, d)^{-1}+(3.052\rm\, d)^{-1})^{-1}=0.7532\rm\, d$</span>. But if Ceres rotates counter-clockwise, then the one synodic period around the Earth is <span class="math-container">$((1\rm\, d)^{-1}-(3.052\rm\, d)^{-1})^{-1}=1.4873\rm\, d$</span>. The two graphs for total tidal forces are drawn below (with their Python code):
<a href="https://i.sstatic.net/lcJUm.png" rel="noreferrer"><img src="https://i.sstatic.net/lcJUm.png" alt="counter-clockwise Ceres" /></a>
Python code:</p>
<pre><code>import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yCeres = []
ySun = []
y = []
for i in x:
    yCeres.append(np.sin(i * 4 * PI / 1.487) * 1.098e-6)
    ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
    y.append(yCeres[-1] + ySun[-1])

plt.title('Total tidal force: Sun and Ceres (counter-clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show()
</code></pre>
<hr />
<p><a href="https://i.sstatic.net/zfhhV.png" rel="noreferrer"><img src="https://i.sstatic.net/zfhhV.png" alt="clockwise Ceres" /></a>
Python code:</p>
<pre><code>import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yCeres = []
ySun = []
y = []
for i in x:
    yCeres.append(np.sin(i * 4 * PI / 0.7532) * 1.098e-6)
    ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
    y.append(yCeres[-1] + ySun[-1])

plt.title('Total tidal force: Sun and Ceres (clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show()
</code></pre>
<hr />
<p>Compare it to the real current tides:
<a href="https://i.sstatic.net/a8HwH.png" rel="noreferrer"><img src="https://i.sstatic.net/a8HwH.png" alt="with Moon" /></a>
Python code:</p>
<pre><code>import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yMoon = []
ySun = []
y = []
for i in x:
    yMoon.append(np.sin(i * 4 * PI / 1.035087719) * 1.098e-6)
    ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
    y.append(yMoon[-1] + ySun[-1])

plt.title('Total tidal force: Sun and Moon (counter-clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show()
</code></pre>
<p>Which is pretty natural because the tides are at their highest on full and new moon.</p>

CC BY-SA 4.0
2021-06-22T11:19:10.057Z
@DavidHammen Ok, corrected and added Python graphs.
User123
CC BY-SA 4.0
2021-06-22T13:53:28.680Z
Thanks for the correction. I've deleted my negative comments and have retracted my downvote.
David Hammen
CC BY-SA 4.0
2021-06-22T14:00:17.207Z
As a side note, the Roche limit pertains to objects consisting of a fluid or particles that are held together by gravity and nothing else. Multiple nations and multiple corporations have placed lots and lots of objects in low Earth orbit, which is well below the Roche limit. Those objects don't break up because they are held together by chemical bonds that are multiple orders of magnitude stronger than gravitation. This would apply to Ceres as well if it was in low Earth orbit.
David Hammen
CC BY-SA 4.0
2021-06-22T14:17:05.100Z
Your initial answer was quite wrong. I myself have provided incorrect answers to questions, so that your initial answer was wrong didn't bug me that much. What did bug me, at the fingernails dragged across a chalkboard level, was that people upvoted your answer. What bugged me even more was that the OP accepted your incorrect answer. To me, that was the equivalent to everyone in the room coming to the chalkboard to simultaneously drag their fingernails across it. Thanks again for correcting your answer.
David Hammen
CC BY-SA 4.0
2021-06-22T14:48:50.713Z
@DavidHammen If he asked the question, he couldn't know the correct answer, so he thought that this is the one. That's why he accepted it (and no other answer was given at the moment.) And people trusted his checkmark, so they upvoted (people love such simple hypothetical questions). But the answer was wrong, so this was just an unfortunate sequence of events. I often wait 1 or 2 days before I accept an answer because often many great answers emerge.
User123
CC BY-SA 4.0
2021-06-22T14:55:11.843Z
@DavidHammen About your first comment: thus the Roche limit for Ceres would be a bit lower, but in no case higher => Ceres would be safe.
User123
CC BY-SA 4.0
2021-06-22T15:50:59.587Z
@DavidHammen By the time I accepted User123's answer, it already had upvotes and there was no other. Their answer looked reasonable to me. When I discovered Garcia's one (that was the last time I checked) his had none while the above had 4 upvotes, so I trusted User123. The upvotes were quite understandable too, as their answer looked reasonable. I corrected which answer I'd accept eventually.
John
CC BY-SA 4.0
2021-06-22T22:04:58.057Z
@John which is exactly why you wait and don't immediately accept the first and only answer (and why people shouldn't upvote "reasonable" looking answers if they have no actual clue if they are correct).
eps
CC BY-SA 4.0
2021-06-23T06:09:30.083Z
@eps On meta someone asked the question why their answers remain un-accepted. But better not to click that button at all. Sometimes there are multiple good answers. Upvoting suffices.
John
CC BY-SA 4.0
2021-06-24T18:42:15.220Z
Maybe tidal forces wouldn't destroy Ceres, but how tight could the orbit get before atmospheric drag dropped it out of the sky?
Mark Ransom
CC BY-SA 4.0
2021-06-21T15:12:01.573Z
2021-06-22T11:42:32.683Z
2021-06-22T11:42:32.683Z
User123
<p>How close to Earth would Ceres have to be to cause tides of the same strength as by the Moon, above the region it orbits? Ceres has 1.3% the Moon's mass, but that doesn't mean it must be at 1.3% the Moon's distance, right? Is it even possible for Ceres or is Ceres too low-mass to cause similar tides?</p>

CC BY-SA 4.0
2021-06-21T20:25:17.643Z
Note that, even with the same tidal force, the actual tides may be different. Actual tides are a complex harmonic and resonance effect, driven by the sun and moon, but acting on the flows of water around the coastline.  With different driving frequency from a smaller, closer moon, the resonances would be different.
James K
CC BY-SA 4.0
2022-12-24T08:44:52.107Z
I am amazed that the phrase “cube root” appears nowhere on this page!  Because the strength of tide is inversely proportional to distance cubed, the number you want is the cube root of 0.013, which is 0.235.
Anton Sherwood
<tidal-forces><ceres>
If we replaced the Moon with Ceres, how close would Ceres have to orbit to cause the same tides?
John
# If we replaced the Moon with Ceres, how close would Ceres have to orbit to cause the same tides?

(Asked by: username_0 on 2021-06-21)

How close to Earth would Ceres have to be to cause tides of the same strength as by the Moon, above the region it orbits? Ceres has 1.3% the Moon's mass, but that doesn't mean it must be at 1.3% the Moon's distance, right? Is it even possible for Ceres or is Ceres too low-mass to cause similar tides?

**Tags:** tidal-forces, ceres

> **Comments:**
> **username_1**: Note that, even with the same tidal force, the actual tides may be different. Actual tides are a complex harmonic and resonance effect, driven by the sun and moon, but acting on the flows of water around the coastline.  With different driving frequency from a smaller, closer moon, the resonances would be different.  
> **username_2**: I am amazed that the phrase “cube root” appears nowhere on this page!  Because the strength of tide is inversely proportional to distance cubed, the number you want is the cube root of 0.013, which is 0.235.  

---

## 3 answers

### Answer by username_8 on 2021-06-21. Score: 22 | Accepted answer

As is nicely put on the Wikipedia page about tidal forces, the tidal force is given by
$$T=Gm\frac{2r}{d^3}$$
where $T$ is the tidal force (see below), $G=6.67\cdot 10^{-11}\rm\,\frac{m^3}{kg s^2}$ is the gravitational constant, $r$ is the radius of the Earth, and $d$ is the distance between the centers of the two objects. This is not force in correct sense (in Newtons), but is given by $\frac{m}{s^2}$, like acceleration. You said, that the tidal forces must be the same, so:
$$T_M=T_C$$
$$GM_M\frac{2r_E}{d_M^3}=GM_C\frac{2r_E}{d_C^3}$$
$$\frac{M_M}{d_M^3}=\frac{M_C}{d_C^3}$$
$$d_C=d_M\left(\frac{M_C}{M_M}\right)^\frac{1}{3}\approx 89000\rm\, km$$
The height of the center above the surface would thus need to be around 82500 km.
But note that the tidal forces of Earth on Ceres would be large. Would Earth destroy Ceres? To answer this, we need an equation for Roche limit
$$d=R_m\left(2\frac{M_M}{M_m}\right )^\frac{1}{3}$$
where $R_m$ is the radius of secondary, $M_M$ is the mass of primary, and $M_m$ is the mass of secondary. With inserting the data for Earth and Ceres, we get
$$d=11175\rm\, km$$
and $11175\rm\, km <89000\rm\, km$. Thus, Ceres wouldn't be destroyed.

But, as @JamesK said, how would the tides look like? By the equation:
$$a_c=a_g$$
$$\frac{v^2}{r}=\frac{GM}{r^2}$$
and
$$v=\frac{2\pi r}{t_0^2}$$
we get
$$t_0=\sqrt{\frac{4\pi^2r^3}{GM}}=\sqrt{\frac{4\pi^2 (8.9\cdot 10^7)^3}{6.67\cdot 10^{-11}\cdot 6\cdot 10^{24}}}\rm\, s=3.052\rm\, d$$
So, Ceres would make one sidereal rotation period in 3.052 days. If Ceres rotates clockwise, then the one synodic period around the Earth is $((1\rm\, d)^{-1}+(3.052\rm\, d)^{-1})^{-1}=0.7532\rm\, d$. But if Ceres rotates counter-clockwise, then the one synodic period around the Earth is $((1\rm\, d)^{-1}-(3.052\rm\, d)^{-1})^{-1}=1.4873\rm\, d$. The two graphs for total tidal forces are drawn below (with their Python code):

Python code:
import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yCeres = []
ySun = []
y = []
for i in x:
    yCeres.append(np.sin(i * 4 * PI / 1.487) * 1.098e-6)
    ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
    y.append(yCeres[-1] + ySun[-1])

plt.title('Total tidal force: Sun and Ceres (counter-clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show()



Python code:
import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yCeres = []
ySun = []
y = []
for i in x:
    yCeres.append(np.sin(i * 4 * PI / 0.7532) * 1.098e-6)
    ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
    y.append(yCeres[-1] + ySun[-1])

plt.title('Total tidal force: Sun and Ceres (clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show()


Compare it to the real current tides:

Python code:
import matplotlib.pyplot as plt
import numpy as np
PI = 3.14159265358979
x = np.arange(0, 20, 0.01)
yMoon = []
ySun = []
y = []
for i in x:
    yMoon.append(np.sin(i * 4 * PI / 1.035087719) * 1.098e-6)
    ySun.append(np.sin(i * 4 * PI / 1) * 5.05e-7)
    y.append(yMoon[-1] + ySun[-1])

plt.title('Total tidal force: Sun and Moon (counter-clockwise)')
plt.xlabel('Time [d]')
plt.ylabel('Tidal force [ms^-2]')
plt.plot(x, y)
plt.show()

Which is pretty natural because the tides are at their highest on full and new moon.

> **Comments:**
> **username_8**: @DavidHammen Ok, corrected and added Python graphs.  
> **username_10**: Thanks for the correction. I've deleted my negative comments and have retracted my downvote.  
> **username_10**: As a side note, the Roche limit pertains to objects consisting of a fluid or particles that are held together by gravity and nothing else. Multiple nations and multiple corporations have placed lots and lots of objects in low Earth orbit, which is well below the Roche limit. Those objects don't break up because they are held together by chemical bonds that are multiple orders of magnitude stronger than gravitation. This would apply to Ceres as well if it was in low Earth orbit.  
> **username_10**: Your initial answer was quite wrong. I myself have provided incorrect answers to questions, so that your initial answer was wrong didn't bug me that much. What did bug me, at the fingernails dragged across a chalkboard level, was that people upvoted your answer. What bugged me even more was that the OP accepted your incorrect answer. To me, that was the equivalent to everyone in the room coming to the chalkboard to simultaneously drag their fingernails across it. Thanks again for correcting your answer.  
> **username_8**: @DavidHammen If he asked the question, he couldn't know the correct answer, so he thought that this is the one. That's why he accepted it (and no other answer was given at the moment.) And people trusted his checkmark, so they upvoted (people love such simple hypothetical questions). But the answer was wrong, so this was just an unfortunate sequence of events. I often wait 1 or 2 days before I accept an answer because often many great answers emerge.  
> **username_8**: @DavidHammen About your first comment: thus the Roche limit for Ceres would be a bit lower, but in no case higher => Ceres would be safe.  
> **username_0**: @DavidHammen By the time I accepted username_8's answer, it already had upvotes and there was no other. Their answer looked reasonable to me. When I discovered Garcia's one (that was the last time I checked) his had none while the above had 4 upvotes, so I trusted username_8. The upvotes were quite understandable too, as their answer looked reasonable. I corrected which answer I'd accept eventually.  
> **username_12**: @username_0 which is exactly why you wait and don't immediately accept the first and only answer (and why people shouldn't upvote "reasonable" looking answers if they have no actual clue if they are correct).  
> **username_0**: @username_12 On meta someone asked the question why their answers remain un-accepted. But better not to click that button at all. Sometimes there are multiple good answers. Upvoting suffices.  
> **username_13**: Maybe tidal forces wouldn't destroy Ceres, but how tight could the orbit get before atmospheric drag dropped it out of the sky?  

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### Answer by username_3 on 2021-06-21. Score: 47

[Tidal forces](https://en.wikipedia.org/wiki/Tidal_force) generated are proportional to $m/r^3$, where $m$ is the mass of the Earth satellite and $r$ is the semi-major axis (we assume circular orbits for simplicity). The derivation of this relationship is performed nicely by [Butikov](http://butikov.faculty.ifmo.ru/TidesOD.pdf).

So $m\_m/r\_m^3 = m\_c/r\_c^3$, where

$m\_m \approx 7.3 \times 10^{22} \rm\, kg$ is the mass of the Moon,

$m\_c \approx 9.1 \times 10^{20} \rm\, kg$ is the mass of Ceres,

and $r\_m \approx 385,000 \rm\, km$ is the semi-major axis of the Moon's orbit around the Earth

**Solve for the Ceres orbit distance $r\_c$ to get about $89,000\rm\, km$ to exert the same tidal effects as the Moon.**

**Note:** Tidal forces are due to a *difference* or *gradient* of gravitational forces across a body instead of simply the magnitude of the gravitational force on the body. That is why the lunar tides on Earth are more powerful than the solar tides, even though the Sun exerts 177 times more gravitational pull on the Earth than the Moon exerts on the Earth. [NOAA has an excellent description of this](https://oceanservice.noaa.gov/education/tutorial_tides/tides02_cause.html) with the following illuminating graphic:
[![enter image description here](https://i.sstatic.net/dadOX.gif)](https://i.sstatic.net/dadOX.gif)

> **Comments:**
> **username_4**: The Ceres moon would have a pretty short month (~3.5 days?), so we'd get spring & neap tides much more frequently.  
> **username_5**: @DavidHammen:  I don't think the gold badge cares if the answer is correct, it just cares if people vote for it (teasing for the second "correct" in your sentence").  I agree with the sentiment and have contributed my vote.  
> **username_6**: @DavidHammen I share your sympathy but that's not how [the badge](https://astronomy.stackexchange.com/help/badges/39/populist) works: the accepted answer needs to have a score of at least 11.  
> **username_7**: I was wondering if that distance was within the Roche limit, but for the Moon that's about 9500 km. So Ceres would likely also survive at 90,000 km.  
> **username_8**: @CJDennis Yes, I included that below. Roche limit for Ceres on Earth is around 11175 km.  
> **username_8**: @DavidHammen Here you have it, he got the golden Populist badge.  

---

### Answer by username_9 on 2021-06-22. Score: 7

## It can't cause tides of the same strength

Though you could have Ceres exert the same force on the ocean it has to be at a much smaller orbit. A smaller orbit means a shorter orbital period. This means that the Ceres tides can't be same as the current ones.

To calculate the wanted equations, we need to use [sidereal periods](https://www.britannica.com/science/sidereal-period): The Earth rotates [once every 23.935 hours](https://en.wikipedia.org/wiki/Sidereal_time#Comparison_to_solar_time) and the Moon orbits the Earth [every 655 hours](https://community.dur.ac.uk/john.lucey/users/lunar_sid_syn.html). Ceres would need to be orbiting at $\approx 90,000\rm\,km$ to exert the same pull (as shown by [Connor's answer](https://astronomy.stackexchange.com/a/44460)) – this would a give it an orbital period of [73 hours](https://keisan.casio.com/exec/system/1224665242).

The comparitive periods mean that over the period of rotation Ceres has moved much further than the Moon would have done and so the net angle of pull over a day differs.

Under an assumption that Ceres would rotate in the same direction as Moon (counter-clockwise), we can set an example: After 12 hours, the Earth has completed half a rotation and the ocean on this side of the Earth is now experiencing a pull in the opposite direction. The Moon has moved ~2% of its rotation whereas Ceres has moved ~16% of its orbit. The Moon pulls in the exact opposite direction on the ocean after ~12.5 hours compared to Ceres' 17.9 hours.

To get the above difference in time periods I just did:
$$
T\_{\text{Earth-Moon}}\omega\_{\text{Earth}}-T\_{\text{Earth-Moon}}\omega\_{\text{Moon}} = \pi
$$
and
$$
T\_{\text{Earth-Ceres}}\omega\_{\text{Earth}}-T\_{\text{Earth-Ceres}}\omega\_{\text{Ceres}} = \pi
$$
under an assumption that Ceres rotates counter-clockwise: in the same direction as the Moon.

$$
T\_{\text{Earth-Moon}} = \frac{\pi}{\omega\_{\text{Earth}}-\omega\_{\text{Moon}}}=\frac{1}{2(\frac{1}{23.935}-\frac{1}{655})}\rm\, h=12.42\rm\, h
$$

$$
T\_{\text{Earth-Ceres}} = \frac{\pi}{\omega\_{\text{Earth}}-\omega\_{\text{Ceres}}}=\frac{1}{2(\frac{1}{23.935}-\frac{1}{73})}\rm\, h=17.81\rm\,h
$$
On the image:
[![Image](https://i.sstatic.net/28NoZ.png)](https://i.sstatic.net/28NoZ.png)

---

Other effects:

* The Moon would be between the Earth and Sun more often, meaning the pull of Moon and Sun together creates a greater pull and larger tides (larger difference between [neap and spring tide](https://oceanservice.noaa.gov/facts/springtide.html))
* The Moon would more often be at right angles to the Sun-Earth line – a point at which the pull on the oceans is smaller and the tides are smaller (smaller difference between neap and spring tide)

> **Comments:**
> **username_0**: Frequency doesn't matter for the answer, just how high the tides are (the distance Ceres needs to be so that they're about equally high).  
> **username_9**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/126772/discussion-between-lio-elbammalf-and-user123).  
> **username_10**: @username_0 Frequency **does** matter. The Moon is currently receding from the Earth at significantly more than the average recession rate over the last several hundred million years. One key reason is the current shape of the North Atlantic. The shape is just right to enable resonances and hence greater losses with the M2 tide (frequency = one cycle per 12.42 hours).  Replacing the Moon with Ceres at ~89000 km from the Earth's center most likely would not result in the huge tides that are currently seen in the Bay of Fundy, or similar seen high swings elsewhere in the North Atlantic.  
> **username_11**: This answer should probably mention that resonances are important for tide height.  Ignoring that, a longer interval between low tides give more time for the water to "slosh" further to the bulges. So in parts of the world that don't resonate significantly, you could maybe match the tide height with somewhat smaller tidal forces.  The finer details like spring/neap would of course be different, along with the interval itself not being close to syncing with 24 hours, of course.  

---

