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?

  • 1
    $\begingroup$ I am almost sure such an arrangement isn't sustainable. This should result in a collision, or both escaping their system. $\endgroup$ Jul 14, 2016 at 21:19
  • 2
    $\begingroup$ What makes you think this can be done? $\endgroup$
    – Mast
    Jul 14, 2016 at 21:24

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Jul 14, 2016 at 21:38
  • $\begingroup$ 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. $\endgroup$ Jul 14, 2016 at 21:39
  • 1
    $\begingroup$ @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". $\endgroup$
    – HDE 226868
    Jul 14, 2016 at 21:43
  • 1
    $\begingroup$ @CharlesRockafellor I added in the calculations for that. $\endgroup$
    – HDE 226868
    Jul 14, 2016 at 21:47
  • $\begingroup$ 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? $\endgroup$ Jul 15, 2016 at 0:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .