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

Question

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?

Answer 1

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:

Answer 2

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.

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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()

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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()

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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.

Answer 3

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:

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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)