<p>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.</p>

<p>From the <a href="http://en.wikipedia.org/wiki/Orbital_period#Two_bodies_orbiting_each_other" rel="nofollow">Wikipedia article on orbital period</a>:</p>

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

<p>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.</p>

<p>For the planet's year you'd use a very similar equation.</p>

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

<p>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.</p>

<p>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.</p>

<p>Solar day = sidereal day * (1 + (sidereal day/year))</p>

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

CC BY-SA 3.0
2015-06-06T00:53:30.253Z
Why's Mathjax not working for me?
HopDavid
CC BY-SA 3.0
2015-06-06T00:56:08.710Z
You made some typos is why.  You have code such as `M_1_` when it should be `M_1`.
zibadawa timmy
CC BY-SA 3.0
2015-06-06T00:57:05.277Z
@zibadawatimmy d'oh! Thanks for the edits.
HopDavid
CC BY-SA 3.0
2015-06-06T05:18:25.240Z
Maybe I'm missing something, but why would tidally locked planets have bodacious tides?   I think tidally locking would significantly reduced tidal effects?
userLTK
CC BY-SA 3.0
2015-06-06T05:59:12.807Z
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
CC BY-SA 3.0
2015-06-06T14:50:48.513Z
@userLTK You're right. There'd be a substantial tidal bulge but since it remains fixed with regard to the planet, no surfer's paradise. I deleted that comment.
HopDavid
CC BY-SA 3.0
2015-06-06T15:34:15.143Z
@Nephratine What do you mean by "conventional day"? I had taken that to mean in the neighborhoold of 24 hours. Perhaps you're looking for the solar day? The time an inhabitant on his front porch would measure between 1 high noon and the next?
HopDavid
CC BY-SA 3.0
2015-06-06T16:28:32.803Z
@HopDavid Surfers don't rely on the tide for waves, they rely on the wind, which is caused by heat from the sun. Therefore, the surfing is the same on those planets.
LDC3
CC BY-SA 3.0
2015-06-06T18:40:11.670Z
@LDC3 I don't know that much about surfing. But casual Googling seems to indicate tides are important to surfers. For example http://www.surfoutlet.com/guides/how-to-choose-the-best-tides-for-surfing
HopDavid
CC BY-SA 3.0
2015-06-06T18:50:53.750Z
@HopDavid OK, so the tides influence the depth of the water, which determines when the waves break as they approach the shore. The size of the waves are determine by the wind as it blows across the water. So on a planet without tides, the location where the waves break is well defined. http://en.wikipedia.org/wiki/Wind_wave
LDC3
CC BY-SA 3.0
2015-06-07T00:52:45.440Z
@LDC3 Googling was surprised to find the Great Lakes enjoy good surfing. And winds blowing from the right direction seems to be the main factor. I believe you're right. You've changed my opinion.
HopDavid
CC BY-SA 3.0
2015-06-06T00:52:01.990Z
2015-06-06T15:54:26.760Z
2015-06-06T15:54:26.760Z
HopDavid
<p>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.</p>

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

<p>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.</p>

CC BY-SA 4.0
2021-01-27T00:31:41.077Z
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
CC BY-SA 3.0
2015-06-05T22:13:22.867Z
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
CC BY-SA 3.0
2015-06-06T05:59:44.937Z
@zibadawatimmy yeah I think you're right
Nephatrine
<orbit><time>
Day Length on Double Planets
Nephatrine
# Day Length on Double Planets

(Asked by: username_0 on 2015-06-05)

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.

**Tags:** orbit, time
> **Comments:**
> **username_2**: 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).  
> **username_0**: @zibadawatimmy yeah I think you're right  
> **username_1**: 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.  

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## 1 answers

### Answer by username_3 on 2015-06-06. Score: 1

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](http://en.wikipedia.org/wiki/Orbital_period#Two_bodies_orbiting_each_other):

$T\_d=2\pi\*\sqrt{a\_p^3/(G\*(M\_1+M\_2))}$

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.

> **Comments:**
> **username_3**: Why's Mathjax not working for me?  
> **username_2**: You made some typos is why.  You have code such as `M_1_` when it should be `M_1`.  
> **username_3**: @zibadawatimmy d'oh! Thanks for the edits.  
> **username_4**: Maybe I'm missing something, but why would tidally locked planets have bodacious tides?   I think tidally locking would significantly reduced tidal effects?  
> **username_0**: 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.  
> **username_3**: @userLTK You're right. There'd be a substantial tidal bulge but since it remains fixed with regard to the planet, no surfer's paradise. I deleted that comment.  
> **username_3**: @Nephratine What do you mean by "conventional day"? I had taken that to mean in the neighborhoold of 24 hours. Perhaps you're looking for the solar day? The time an inhabitant on his front porch would measure between 1 high noon and the next?  
> **username_5**: @HopDavid Surfers don't rely on the tide for waves, they rely on the wind, which is caused by heat from the sun. Therefore, the surfing is the same on those planets.  
> **username_3**: @LDC3 I don't know that much about surfing. But casual Googling seems to indicate tides are important to surfers. For example http://www.surfoutlet.com/guides/how-to-choose-the-best-tides-for-surfing  
> **username_5**: @HopDavid OK, so the tides influence the depth of the water, which determines when the waves break as they approach the shore. The size of the waves are determine by the wind as it blows across the water. So on a planet without tides, the location where the waves break is well defined. http://en.wikipedia.org/wiki/Wind_wave  
> **username_3**: @LDC3 Googling was surprised to find the Great Lakes enjoy good surfing. And winds blowing from the right direction seems to be the main factor. I believe you're right. You've changed my opinion.  

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