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

Question

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.

Answer 1

Using the formula:

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

where:

• M1 and M2 are the masses of the two bodies
• G is the Universal Gravititional Constant ($6.6743 × 10^{-11}m^{3}kg^{-1}s^{-2}$)
• a is the Semi-Major Axis,

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.