3
$\begingroup$

I write science fiction and am trying to figure out how often two (or more) moons would be on the same side of a planet when orbiting at different speeds, if their days to orbit are known. Is there a simple formula for this?

Thanks.

  • 0
    You might wanna check out the Worldbuilding SE. Also, wouldn't the moons have crashed a long time ago if they had different orbital velocities, but lay in the same orbit? – 2017-01-04
  • 0
    I didn't say they were at the same distance from the planet. :) – 2017-01-04
  • 1
    @Lovsovs: two satellites cannot be on the same orbit with different speeds. – 2017-01-04
  • 0
    @YvesDaoust That was exactly my point. But it seems that was not what OP meant anyways. – 2017-01-04
  • 0
    @Lovsovs: you get me wrong. I didn't say must not, I said cannot, by the laws of physics. – 2017-01-04
  • 0
    @YvesDaoust True. – 2017-01-04

2 Answers 2

2

I'm not sure exactly what you mean by "on the same side".

For simplicity let's suppose the moons are in circular orbits, so their motions are uniform on the circle, in the same plane and in the same direction, with periods $p_1 < p_2$. At time $t$ they are at angular positions $\theta_1(t) = \theta_1(0) + 2 \pi t/p_1$ and $\theta_2(t) = \theta_2(0) + 2 \pi t/p_2$. The angle between them is $\theta_1(t) - \theta_2(t) = A + B t$ where $A = \theta_1(0) - \theta_2(0)$ and $B = 2 \pi (1/p_1 - 1/p_2)$. The time between conjunctions (when they are in the same direction) is $$C = 2 \pi/B = 1/(1/p_1 - 1/p_2) = p_1 p_2/(p_2 - p_1)$$
Time $C/2$ after a conjunction, they are in opposition (in opposite directions).

  • 0
    Sorry, I mean such as when there's an eclipse on Earth, as in the exact same position in the sky between both celestial bodies. – 2017-01-04
  • 0
    OK, that's a conjunction. – 2017-01-05
  • 0
    Since I asked for a simple way, then to simplify your answer, I could just use (P1 * P2) / (P2 - P1). Is that right? – 2017-01-05
  • 0
    If I wanted to know when three moons are in conjunction, how would I modify (P1 * P2) / (P2 - P1)? – 2017-01-05
  • 0
    Three moons will typically never be exactly in conjunction. It requires too much of a coincidence. – 2017-01-05
  • 0
    I'm not worried about exactness. The tidal impact is a more mundance concern. This function allows me to know when the highest and lowest tides will be. – 2017-01-05
  • 0
    With no exact conjunctions of three moons, there will be no actual "highest" tides. There will be high tides when the three moons are almost in conjunction, and slightly higher tides when they are slightly closer to being in conjunction. BTW, don't forget tides are caused by the star as well. – 2017-01-05
  • 0
    Exactly my point. – 2017-01-05
1

Place yourself on the moon $1$, which revolves in $T_1$, i.e. with angular speed $\omega_1=2\pi/T_1$. The apparent angular speed of the moon $2$ is $\omega_2-\omega_1$.

This moon changes side every multiple of $$\frac \pi{\omega_2-\omega_1}=\frac{T_1T_2}{2(T_1-T_2)}.$$