# Tides on a planet with 5 moons? [duplicate]

What would be the effects on an earth-like planet with 5 moons, one the size of Earth's moon and the others approximately 1/10th its size? How massive would the tides be when they align?

• Tides depend on gravity. Gravity depends on mass and distance. So you need to give the distances of these additional moons. Aug 12, 2022 at 15:42
• Also, if one starts with the Earth/Moon system as it currently exists, and stick 4 more moonlets into the equation in a position where they'd be visible from Earth's surface, I'd be doubtful as to the stability of the system. Earth's moon is pretty big, as rocky planet moons go. The Earth/Moon barycentre is actually just a quarter of the way into the Earth. This question has been addressed - stability is a problem. Aug 12, 2022 at 18:57

# Just do the math

You failed to give the orbital distances of each moon, so I'll assume they are at roughly equal distances as our Moon. You stated 1 moon the size of Earth's, and 4 moons 1/10th the size. Tidal forces are dependent only on mass and distance. Therefore, the total tidal forces when they all align (which technically isn't possible if they are all at the same distance from the planet) is 1 + 4 × 0.1 = 1.4 times the Moon's tidal pull on Earth (which is $$1.1\times 10^{-6}\hspace{0.5em} m\cdot s^{-2}$$).

Of course, these are probably not orbiting your planet at similar distances to the Moon or each other. To calculate the tidal force of all of the moons on the Earth when they are aligned, you'll need to use the following formula:

$$F=2GR\cdot\sum_{i=1}^{5}\left(m_{i}r_{i}^{-3}\right)$$

Where:

• $$G$$ is the gravitational constant $$6.6743\times 10^{-11}\hspace{0.5em} m^{3}\cdot kg^{-1}\cdot s^{-2}$$
• $$R$$ is the radius of the planet (at sea level)
• $$m_{i}$$ is the mass of moon $$i$$
• $$r_{i}$$ is the distance from moon $$i$$ from the Earth

You also may want to account for the tidal forces of the planet's star. The Sun's tidal acceleration on the surface of Earth is $$5.2\times 10^{-7}\hspace{0.5em} m\cdot s^{-2}$$. This of course will be different for a different planet around a different star, so you'll use the same formula as before: $$F=2GRmr^{-3}$$, using the star's mass and distance for $$m$$ and $$r$$.