# At what approximate distance and what size should binary planets be for a day to be no more than 30-32 hours long?

For the story I'm trying to find a more or less plausible description of the planet.

Initially I was considering planet size like Mars or mid-size between Earth and Mars. Аnd this planet is in tidal lock. The living conditions are more or less close to those on Earth. (Magnetic field, atmosphere, etc. - I was guided by the conditions as they describe the hypothetical terraforming of Mars).

So I was thinking about the habitable planet in a situation like Pluto and Charon or, perhaps, the planet is actually a satellite of a major planet so is technically a habitable moon.

At what distance should the second celestial body be and what size for length of day on the planet no more than 30-32 hours? Or is this basically impossible? (As far as I understand, in the situation with Pluto and Charon, this would be too long a day for life like that on earth)

Is it possible that with such a day length, the second celestial body could be small enough (not a gas giant, but a rocky planet) for both planets to be tidally locked, as in the Pluto example above?

Very strict accuracy of calculations is not important. I'll be grateful if you could suggast at least a plausible example.

• Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
– Community Bot
Commented Jan 15 at 7:07
• Please ask 1 question per post. And check the content: it looks like you duplicated most of the text.
– L.Dutch
Commented Jan 15 at 7:09
• Also, planets do not typically tidally lock other planets. Tidal locking only occurs between a primary and its satellite, i.e. a star and planet or a planet and moon. Commented Jan 15 at 7:10
• What about Pluto and Charon? Perhaps I do not fully understand the correct terminology, but I am talking approximately about this case. Thank you for pointing out that there was a technical error and the text was duplicated. Commented Jan 15 at 7:18

Using the formula:

$$T_{binary} =2π\sqrt{\frac{a^3}{G(M_1+M_2)}}$$

where:

For $$M_1 = 1M_{Earth}$$, $$M_2 = 0.5M_{Earth}$$ and Time = 31 hours

this calculator returns a semi-major axis of 57,349km.

As to tidal locking, this is an imprecise calculation, which can be approximated by the following formula (because I can't be bothered using the more complicated formula on the linked Wikipedia page):

$$t_{lock}\approx 6\frac{a^6R\mu}{m_s m_p^2}\times 10^{10}\mathrm{years}$$

Where:

• $$m_s$$ is the satellite mass,
• $$m_p$$ is the planetary mass,
• $$R$$ is the mean radius of the satellite,
• $$\mu$$ is newtons per square metre, roughly $$3\times10^{10}Nm^{-2}$$ for rocky objects or $$4\times10^9Nm^{-2}$$ for icy objects.

Using

• $$m_s$$ = $$2.986\times10^{27}$$kg,
• $$m_p$$ = $$5.972\times10^{27}$$kg,
• $$R$$ = $$5056670$$ m (the radius of a world with half the volume of earth),
• $$a$$ = $$57349000$$ m (from the earlier calculation)

we get $$3\times10^{-8}$$ years.

This is an approximation, of course, but the low value suggests that the two bodies would become tidally locked very quickly.

• Thank you! That is, theoretically, at this distance the larger celestial body could be a rocky planet and both would be tidally locked. Commented Jan 15 at 9:48
• Thanks for the link, I tried to use the calculator myself, it showed 53238 km. But I probably haven’t fully figured it out. Commented Jan 15 at 9:59
• Can you define μ more explicitly? It has the units of pressure, but it's not obvious why. Commented Jan 16 at 20:44
• @AntonSherwood I believe that μ is there because of the torque applied to surface protrusions. Read the linked Wikipedia article on Tidal Locking... Commented Jan 17 at 3:12