# Hypothetical Scenario: Planet with Two Moons

1. Jul 11, 2014

### Captainscar

Alright, so there's a RP website I'm on that has a fairly detailed description of their solar system. However, they have no details on their planet's moons. This is something I've been trying to work out, as I want to introduce a lunar calendar of sorts to a nation I RP there. I've managed to get pretty far, but I'm not sure if my math is correct.

Also I want to make sure that with the numbers I have, the tides would be similar to Earth's, or at least not at a small-tsunami kind of level. I don't know how to figure out tides, so that's why I'm asking you guys for help.

Can anyone help me?

The current set up looks like this:

Planet in question:
Diameter: 15,670 km
Rotation Period: 24 Hr
Orbital Period: 378 Days
Mass: 9.94E+24 kg
Density: 4.9359 g/cm3
Volume: 2.01E+12 km3
Number of Moons: 2
Diameter of Moons: First moon (5,417 km), second moon (2,330 km)

I've had to fill in a bunch of information from here, since they don't exactly get into that much more details on the moon (though they go into much more detail about their system). So here are the numbers I've come up with. I've tried to make sure the orbits are stable by plugging the numbers into Universe Sandbox and running a simulation.

First moon:
Mass: 1.525812 × 1022 kg
Diameter: 2,330 km (was provided for me)
Distance from planet: 350,000 km (they mention that the distance of the first moon to the planet is twice the distance of our moon to Earth)
Estimated Orbit time (according to Universe Sandbox): about 18.5 days
Tidal force from first moon: 3.88205 x 10^18 N to 3.52668 x 10^18 N

Second moon:
Mass: 1.257673 × 1023 kg (a bit smaller than Titan)
Diameter: 5,417 km (was provided for me)
Distance from planet (semi-major axis): 700,000 km (they say that the first moon is the closest moon and I'm worried that if I go any closer the first moon would zoom off into deep space)
Orbital Eccentricity: 0.05
Estimated Orbit time (according to Universe Sandbox): about 52 days
Tidal force from second moon: 4.4448 x 10^18 N to 3.29197 x 10^18 N

Last edited by a moderator: Jul 11, 2014
2. Jul 11, 2014

### Bandersnatch

Hi, Captainscar.

At a first glance, it looks alright.

The tides would be a tiny fraction of what we get from the Moon due to tidal acceleration being very strongly dependent on the distance between the interacting bodies.
A quick calculations suggests you'd get approx. 2.5% of Lunar from the smaller satellite, and 7% from the larger.

3. Jul 11, 2014

### Captainscar

Thanks! Since your reply I have changed the mass of the second moon to make it so it's density is closer to that of Mars. I'd imagine this would lower the tidal acceleration since the mass is going down, but if anyone sees problems please let me know!

4. Jul 11, 2014

### D H

Staff Emeritus
Density is pretty much irrelevant. It's mass and distance that count. The tidal force is proportional to mass and is more or less inversely proportional to the cube of the distance.

Your second moon's orbit is not stable. Assuming a star of one solar mass, that orbit of 2.7 million km is orbiting at a radius of about about 1.5 Hill sphere radii. The orbital radius needs to be less than 1/2 (and maybe less than 1/3) of the Hill sphere radius for the orbit to be stable.

5. Jul 11, 2014

### Captainscar

I've adjusted the parameters as you said, bringing the second moon to a semi-major axis of 550,000 km and the first moon to a semi-major axis of 245,000 km. In the simulations I've ran this is a stable orbit, but how would this affect the tides?

EDIT: NVM, found a calculator that calculates tidal force and my god the second moon has some massive tidal forces at that orbit.

Last edited by a moderator: Jul 11, 2014
6. Jul 11, 2014

### Captainscar

I found a calculator online that calculates tidal force (in newtons) so I used that to help me with this latest change. Is it any good? I really would like to know what the tides would look like as well.

Last edited by a moderator: Jul 11, 2014
7. Jul 12, 2014

### Staff: Mentor

Tidal forces in Newtons don't make sense. The effect on the planet is proportional to (mass)/(distance)^3, neglecting the effect from the slightly larger main body, so you can calculate the effect relative to our moon here.
You can reduce the effect of tides if you reduce the size or depth of water bodies on the planet.

Our moon is quite extreme in its mass and distance - going further out makes orbits problematic. It is not sufficient to calculate a few orbits, you are probably interested in the stability over billions of years. A second large moon won't help in that respect.

It would help to know the mass of the star or the distance planet/star (if you know one value then the other one can be calculated).

8. Jul 12, 2014

### Captainscar

I'm going on the assumption that the star isn't that different from our sun, and basing the distance off the fact that their year is 378 days (exactly).

I'd run some more calculations, but at the moment I'm nowhere near my PC.

EDIT: Sidenote, the tidal forces calculator I was using is this: http://keisan.casio.com/exec/system/1360312100

Is this even the right one to use?

9. Jul 12, 2014

### Staff: Mentor

Okay, assuming the star is sun-like, the Hill sphere radius is 1.8 million kilometers (used 1 AU, those 20 days difference in the year are negligible). 1/3 would be 600 000 km.
700 000 km is not so easy, but it could work.

I'm not sure if the absolute value of the tidal force from the calculator has a real meaning, but the relative values are certainly something you can use.