# In this binary, how far would the secondary planet orbit the primary planet and still have its own day-night spin?

A scenario I've been exploring lately:

Orbiting a binary of G0 stars (each one 105% as wide, 110% as massive and 126% as bright as our sun) from a distance of 2.065 AUs is a planetary binary. The primary one is 7,520 miles wide and 90% as massive as Earth, identical to Venus, but its atmospheric density is identical to that of Earth, so this smaller planet gets to keep a little extra heat and moisture.

Orbiting the primary is the secondary planet in the binary, and it is even smaller--5,866 miles wide and 65% as massive as Earth, yet its atmospheric density is still as much as Earth's, allowing for extra heat, extra humidity and extra usefulness for flying creatures.

The secondary planet orbits the larger primary, but not so close as to be tidally locked (in other words, one side always facing the parent forever.) The secondary has its own spin, its own day-night cycle. How far must the secondary orbit the primary without being tidally locked?

• (1) And I asked how close you think the components of the binary star can possibly be while allowing both stars to remain yellow dwarfs. (2) Why would they be stuck being eternally bright on one side and eternally dark on the other? They are tidally locked between them, not with the stars. (Consider our Moon: it has days and nights, altough it is tidally locked to Earth.) Jan 9 at 20:20
• The Moon spins. Of course it spins. If it didn't spin we would be able to see all its surface from Earth, and we don't -- we always see the same side. Which means that it spins with the same period as its revolution around Earth. Its solar days are 29 days, 12 hours and 44 minutes long. This is, in fact, what being tidally locked means. And the distance between the stars is relevant because it may (or may not) affect the illumination of the planets, hence the days and the nights. Jan 9 at 20:30
• If the Moon didn't spin, as it goes around the Earth it would show opposing sides when it is in the first quarter and when it is the last quarter. But it shows the same side. Which means that while it completes one half of the revolution around Earth it also spinns 180° around its axis. Jan 9 at 21:44
• I am at a loss for words. The Moon always shows the same face towards Earth. It also revolves around the Earth. How can it always show the same face towards Earth as it revolves around the Earth if it doesn't spin? Try making a drawing with the Moon on two opposing point on its orbit around Earth and not spinning and check which side is towards Earth and which side is away from Earth at the two points. Jan 9 at 22:35
• To have one side always dark and one side always bright the planet must spin, and specifically spin with a rotation period equal to its orbital period. If it didn't spin, then the light from the star would fall on different parts of the planet as the planet revolves around the star. Jan 9 at 23:06

Probably too far away to be stable, is the way these things usually go, but it isn't so implausible that you couldn't just squeeze it into the realms of plausibility.

I think it is OK to model the binary stars as a single mass, for the purposes of simplifying things a bit. That gives the larger of your two planets a Hill radius of ~2.3 million kilometres. Stable orbits are likely to be within a third of this, so lets say ~765000km.

Tidal locking timescale can be approximated by $$T_{lock} \approx {\omega a^6 I Q \over 3Gm_p^2 k_2 r^5}$$ where $$\omega$$ is the satellite's rotational rate in radians per second, $$a$$ is the satellite's orbital semimajor axis, $$I$$ is the moment of inertia (which is the satellite's moment of inertia factor x the mass of the satellite x the radius of the satellite squared), $$Q$$ is the dissipation function, $$G$$ is the gravitational constant, $$m_p$$ is the mass of the larger world, $$k_2$$ is the Love number of the satellite and $$r$$ is the satellite's radius.

That's an unfortunate number of unknowns, some of which (like $$Q$$ and $$k_2$$) are not really well known for other planetary bodies. With a bit of handwaving, you can set $$Q$$ to be 100 (if wikipedia is to be believed) and $$k_2$$ for your smaller world (again, using wikipedia's suggested approximation) will be something like 0.94. That's a little high for the tidal locking approximation (which wants $$k_2 \ll 1$$) but we can throw caution to the wind and try it anyway (FWIW, your smaller planet's $$k_2$$ is about ten times higher than the our own moon). I'll use the approximate moment of inertia factor of the Earth (.33) and use a 24 hour day.

Throwing all those numbers in gets a tidal locking timescale of about 100 million years. That's far too short by the standards of planetary evolution, and it seems very likely that your worlds will be tidally locked to each other long before interesting things could evolve on them. They can't be mover farther apart (increasing the $$a$$ term which dominates rapidly as distances increase) because their co-orbit is likely to become unstable, and they'll fall out into separate orbits around the central primary.

Now, this is only a very rough approximation, and a lot of handwaving went into the many different unknowns. It is very likely to be out by at least one order of magnitude.

If it is out by two orders of magnitude, then there's room to squeeze in the rotation that you wanted. That seems unlikely to me, but not totally beyond the realms of possibility. If your secondary was spinning much faster initially, for example, it might be possible to have it still spinning in your setting's "present". I'm not sure what a plausible rate of rotation is, but "full rotation in under 3 hours" seems like a stretch for such a big body, though it might be dense enough to survive such a situation intact and would give you at one of the two orders of magnitude you needed (and the second could be handwaved in given the number of wild guesses in the parameters!) Tidal forces would have slowed that earlier dizzying rate to something much more sedate.