# Day Length on Double Planets

Assume two similarly sized bodies tide-locked to one another orbiting a barycenter between the two. That barycenter orbits a star. Since the two are tide-locked, their sidereal rotational period and sidereal orbital period (about one another) are the same (right?). There's no axial tilts or eccentricities to screw with things either for the sake of simplicity.

What would I base the length of a conventional day on for a society living on these worlds?

My gut's telling me to calculate the sidereal period based of the barycentral semi-major axes (or would I use the separation between the two worlds still?) for each and then compute the synodic periods and just use those, but my brain's too stupid to tell whether that's going to actually be anything close to a "solar day". I don't need perfect - just close enough.

• At least part of this might be better on the World Building SE. We could discuss exactly what an observer on the surface would see in regards to any sort of solar period. But a convention of a hypothetical society on a hypothetical world is very much the domain of World Building. They may not have any concept of "day" like ours. Asimov's Nightfall story (or the novel version) involves a society on a planet orbiting 6 stars (only four of which are actually named/visible, I think) that has at worst twilight conditions (except for the rare plot event the story is based on). – zibadawa timmy Jun 5 '15 at 22:13
• @zibadawatimmy yeah I think you're right – Nephatrine Jun 6 '15 at 5:59
• Am I the only one who thinks those two bodies would not orbit each other or their barycenter, but collapse on each other due to their mutual gravity? They would have to be extremely distant from the star for the star not to affect their mutual orbit and make it unstable. – Pierre Paquette Jan 27 at 0:31

Yes, if the planets are tide locked to one another, their orbital period and sidereal day are the same. That's assuming their orbital plane about each other isn't much inclined to their orbital plane about the star.

From the Wikipedia article on orbital period:

$T_d=2\pi*\sqrt{a_p^3/(G*(M_1+M_2))}$

Where G is gravitational constant, $M_1$ is mass of 1st body and $M_2$ mass of 2nd body. $a_p$ is semi major axis of elliptical orbit of the planets about each other.

For the planet's year you'd use a very similar equation.

$T_y=2\pi*\sqrt{a_s^3/(G*(M_0+M_1+M_2))}$

Where $a_s$ is the radius of their orbit about the sun and $M_0$ is the mass of the sun. If you like, you can leave off $M_1$ and $M_2$ as their masses are probably neglible in comparison to the sun's mass.

I am guessing by "conventional day" you want the time an inhabitant sitting on his front porch would measure between one high noon and the next.

Solar day = sidereal day * (1 + (sidereal day/year))

For example our solar day is about 1/365 longer than a sidereal day.

• Why's Mathjax not working for me? – HopDavid Jun 6 '15 at 0:53
• You made some typos is why. You have code such as M_1_ when it should be M_1. – zibadawa timmy Jun 6 '15 at 0:56
• @zibadawatimmy d'oh! Thanks for the edits. – HopDavid Jun 6 '15 at 0:57
• Maybe I'm missing something, but why would tidally locked planets have bodacious tides? I think tidally locking would significantly reduced tidal effects? – userLTK Jun 6 '15 at 5:18
• Was asking how to determine day length for the double planet system, not how to make the period 1 Earth day. Also, since they're mutually tide-locked, wouldn't the tides be effectively stationary? Not sure that would really have much effect on hypothetical surfers. – Nephatrine Jun 6 '15 at 5:59