# How much water would remain at the poles of these planets?

My binary terrestrial planets are roughly the size of earth and they are tidally locked with one another. They are 12,000 miles (19,312 km) apart. That distance is based off of this article, http://phys.org/news/2014-12-binary-terrestrial-planets.html, which suggests a likely distance of only three planet radii.

No axial tilt. The atmosphere is similar to Earth. I know the gravitational pull would cause these planets to bulge toward one another. My question is: how much, if any, water on these planets would remain at the poles?

• Depends on how much water there is on the planet, remember the rotation of the pair pushes the water back. Planets are too close though, your setup would collapse. – Twelfth Jun 22 '16 at 17:16

At 12,000 miles, this first reminded me of another question about two colliding planets (not sure how to link to related questions) and the effects on atmosphere and surface gravity. If we accept that these two bodies would orbit one another relatively stably, http://www.calctool.org/CALC/phys/astronomy/planet_orbit showed me that they would orbit one another once every 111 minutes. I went to http://keisan.casio.com/exec/system/1360312100 and plugged in a few combinations of masses and distances, with the following results:

• If our moon were moved in to a distance of 12,000 miles, the tide force would be more than 7,000 times as powerful as it is now.
• If the mass of the object at 12,000 miles were increased to that of the Earth, the tide force would be over 600,000 times as powerful as it is now.

Would there be any water left at the poles? It seems as though any liquid water would be piled up into Interstellar-level tides at the near and far sides of the planets. If we can imagine frozen water at the poles remaining through the "kissing" event that established this orbit (I can't imagine that - it sounds deeply apocalyptic) then it might remain frozen at the poles for a very brief time until tidal heating melted it all and it made its way to the party.

So...probably not.

• If planets are tidally locked and their orbit is circular, tides are just a memory of a distant past. – Pere Feb 9 '18 at 22:34

12 000 miles is only an Earth diameter and a half away. That kind of system cannot form. The masses would tear each other apart and form one solid planet soon enough.

• Using the equation for the Roche limit as $R=1.26R_M(M_M/M_m)^{1/3}$, with both masses equal, we get a separation distance of $1.26 R_M$. This is less than the sum of the radii of both planets, so they'll be stable at any separation. – ckersch Jun 22 '16 at 17:23
• @ckersch You are just taking an ideal case there... the simplest form. Read up on the Roche Limit. en.wikipedia.org/wiki/Roche_limit That system will not form. – MichaelK Jun 22 '16 at 18:14
• Any source supporting "That system will not form"? 12000 miles is outside the Roche limit for fluid bodies. – ckersch Jun 22 '16 at 18:40
• @ckersch I can give you one heck of a reason for it: you are hoping that two planets will form well away from of the bottom of the gravity well... that 12 million billion billion kilograms of star dust will climb out of the gravity well, in opposite directions, to an altitude of 6000 miles each and form levitating planets. Unless some god made these two planets, all ready and done... and then set them spinning like that, there is no way in hell you are going to get that. And since the tag is "science-based": no gods allowed. – MichaelK Jun 22 '16 at 18:48
• @MichaelKarnerfors They would need to have fairly low relative initial velocities, as mentioned in the article Manda cited. The central idea of that paper is that a grazing collision would have too much angular velocity to coalesce into a single planet, a hypothesis which was supported by their simulations of that process. – ckersch Jun 22 '16 at 19:38