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I have two planets that are the exact same size and mass of Earth. They orbit each other, and have an orbital period of one day. How far apart are they?

The planets both orbit a star the same as the Sun, and the center of mass for the two planets is on Earth's orbit.

These planets can also be totally locked.

EDIT - I finally found some previous questions which would be useful for answers.

What is the Roche limit for these two planets?

Could two planets be tidally locked to each other so close they share their atmosphere?

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Xandar The Zenon
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Using Kepler's third law, I get $$a=\sqrt[3]{\frac{P^2G\times(M+M)}{4\pi^2}}=5.32\times10^4\text{ kilometers}$$ where $a$ is the semi-major axis, $P$ is the time it takes the planets to orbit each other, $G$ is the universal gravitational constant, and $M$ is the mass of one of the planets - one Earth mass. The result is a semi-major axis of about 10 times the radius of Earth; the separation is twice this amount. I'd call this possible, though there would be very strong tidal effects.

Specifically, the tidal acceleration $a_t$ is treated as $$a_t\propto\frac{M}{D^3}$$ where $D$ is the separation, i.e. $2a$. The Earth is about 100 times the mass of the Moon, so that's an increase of about 100. Furthermore, the Moon orbits Earth at a distance of about 3.85$\times$105 kilometers, about 3.85 times this separation. Therefore, we get an increase in tidal acceleration of about 5700.

That's a lot.

HDE 226868
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  • Are you sure that is right? A quick search of your link reveals that Keepers third law covers planets orbiting stars of much greater mass, so it might not work if the two planets are orbiting their shared center of gravity. – Xandar The Zenon Jul 14 '16 at 21:38
  • With the Earth-Luna system being ≈384,500 km, I wonder how great the resulting tidal differences would be. These planets' separation is twice that of Luna's from us, but the co-orbiting planet's mass is 2 magnitudes that of Luna's. – Charles Rockafellor Jul 14 '16 at 21:39
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    @XandarTheZenon I should have been more explicit. The $2M$ is really $M_1+M_2$; in these cases, the masses are identical, so this is simple $2M$. However, in a planet-star system, the mass of the planet is neglected. That said, I made a typo; "separation" should be "semi-major axis". – HDE 226868 Jul 14 '16 at 21:43
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    @CharlesRockafellor I added in the calculations for that. – HDE 226868 Jul 14 '16 at 21:47
  • Just a quick check on your result. The Earth orbits the Sun at much bigger distance than the Earth orbits the Sun, the mass of the Sun is much bigger than Earth's, and Earth's mass is much bigger than that of the Moon. Why would two bodies with the same mass of the Earth need to orbit each other at a shorter distance than the Earth-Moon system? – Luís Henrique Jul 15 '16 at 00:34
  • @LuísHenrique It's not that they would need to orbit at a shorter distance; it's that given the period of their orbit around each other, this is the distance as predicted by Kepler's laws. – HDE 226868 Jul 15 '16 at 00:40
  • But M1+M2 in the case of two planets with the same mass as the Earth would be 2ME, and in the case of Earth and Moon it is 1.012 ME; so, since G is the same in both formulae, and P is the same as required by the OP, the distance between the hypothetical planets should be bigger than that between Earth and Moon. Or what am I getting wrong? – Luís Henrique Jul 15 '16 at 00:55
  • @LuísHenrique $P$ is the time it takes one planet to orbit another - in this case, one day. That's roughly 30 times faster than it takes the Moon to orbit Earth. – HDE 226868 Jul 15 '16 at 00:56
  • Yeah, I see. For some reason I took the day as the period of the Moon, instead of the month. The mass of this hypothetical system is twice the mass of the Earth-Moon system, but the orbital period is 1/29.5. This means that P squared is about 1/900. Which more than neutralises the increase in mass. – Luís Henrique Jul 15 '16 at 01:09
  • @LuísHenrique Correct. – HDE 226868 Jul 15 '16 at 01:10
  • But since 21/900 = 1/450, and the cubic root of 1/450 is almost 8, I suppose that the proportion between the orbital distances of each system is quite bigger than 3.85 (which indeed seems to be the case since (3.8510^5)/(5.32*10^4) is about 7.2. – Luís Henrique Jul 15 '16 at 01:22
  • I added a little to my question which I would appreciate you looking at when you have some time. – Xandar The Zenon Jul 17 '16 at 20:01
  • @XandarTheZenon Do you mean the tidal locking? We'd need to know something about the initial conditions to determine how the angular momentum would be transferred and what the final configuration would be. – HDE 226868 Jul 17 '16 at 21:30
  • To confirm what HDE 226868 said, these two planets must be much closer than the Earth and the Moon because the orbital period is so short.  (Reality check #1: what’s the orbital period of Mercury? 88 Earth days.  Mars? 1.88 Earth years. [source])  Reality check #2: this situation is (somewhat) like that of a satellite in geosynchronous orbit, which has a radius of 42,164 km [source]).  And $\sqrt[3]2$ is $1.26$, so $\sqrt[3]2×42164$ km $≈53100$ km, which is within 0.2% of the above answer. – Peregrine Rook Aug 03 '16 at 03:56