# If we replaced the Moon with Ceres, how close would Ceres have to orbit to cause the same tides?

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?

• 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. Jun 21 at 20:25

Tidal forces 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.

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 with the following illuminating graphic:

• The Ceres moon would have a pretty short month (~3.5 days?), so we'd get spring & neap tides much more frequently. Jun 21 at 22:23
• @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. Jun 22 at 2:21
• @DavidHammen I share your sympathy but that's not how the badge works: the accepted answer needs to have a score of at least 11. Jun 22 at 7:11
• 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. Jun 22 at 11:28
• @CJDennis Yes, I included that below. Roche limit for Ceres on Earth is around 11175 km. Jun 22 at 11:28

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.

• @DavidHammen Ok, corrected and added Python graphs. Jun 22 at 11:19
• @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. Jun 22 at 14:55
– John
Jun 22 at 15:50
• @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
Jun 22 at 22:04

## 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: The Earth rotates once every 23.935 hours and the Moon orbits the Earth every 655 hours. Ceres would need to be orbiting at $$\approx 90,000\rm\,km$$ to exert the same pull (as shown by Connor's answer) – this would a give it an orbital period of 73 hours.

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:

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)
• 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)
• 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
Jun 22 at 15:57
• Jun 22 at 17:05
• @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. Jun 22 at 17:38
• 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. Jun 23 at 18:06