I have decided to change the numbers to make it easy on Hohmannfan. These two planets, for all intents and purposes, are identical to Earth, with he moon replaced by another Earth. So now the question is, how close can these two earth's be without being tidal locked?

Since it seems necessary to say it, the planets must be stable for about a trillion years.

  • $\begingroup$ If this is a duplicate, I can just as easily delete it. I just was wondering what the answer would be. $\endgroup$ – Xandar The Zenon Feb 11 '16 at 22:03
  • $\begingroup$ You mean, without crashing? Because if you don't mind, the answer is 0. If you want this to be somehow stable, however... $\endgroup$ – Mołot Feb 11 '16 at 22:08
  • 1
    $\begingroup$ Related: physics.stackexchange.com/questions/12541/…. $\endgroup$ – HDE 226868 Feb 11 '16 at 22:08
  • 1
    $\begingroup$ @Mołot They can have zero distance between them without tidally locking? $\endgroup$ – Xandar The Zenon Feb 11 '16 at 22:08
  • 1
    $\begingroup$ You would need to specify radius to get mass. Just use Newton's law of gravitation at the surface. You have $$g=\frac{GM}{R^2}\to M=\frac{gR^2}{G}$$Set $g$ to 0.8 times its value on Earth. $\endgroup$ – HDE 226868 Feb 11 '16 at 22:13

For the Eath clones, we can use the approximate time until tidal locking from the wiki.

An estimate of the time for a body to become tidally locked can be obtained using the following formula:

$$t_{\text{lock}} \approx \frac{w a^6 I Q}{3 G m_p^2 k_2 R^5}$$

$w$, is the initial spin rate (radians per second)
$a$, is the semi-major axis of the motion of the satellite around the planet (given by average of perigee and apogee distances)
$I$, $\approx 0.4 m_s R^2$ is the moment of inertia of the satellite.
$Q$, is the dissipation function of the satellite.
$G$, is the gravitational constant
$m_p$, is the mass of the planet
$m_s$, is the mass of the satellite
$k_2$, is the tidal Love number of the satellite
$R$, is the mean radius of the satellite.

Solving for a with $t=$ a trillion years (70 times the age of the universe...), we get a required distance of 3,300,000 km.

Keep in mind that this is not very accurate because many of the variables are poorly known.

Next section only relevant for initial question:

Of those parameters, we only know the gravitational constant! Even their masses are unknown because the only data you have given is surface gravity, that is not entirely dependent on mass. (for instance, Mars and Mercury has the same surface gravity, but different mass.)

  • $\begingroup$ @XandarTheZenon I can make one useful estimate, working on it right now. Just wait a little $\endgroup$ – SE - stop firing the good guys Feb 11 '16 at 22:17
  • $\begingroup$ I made the planets Earth, because I didn't know there were so many variables. I guess it's not as simple as it seemed. Sorry if I wasted your time, trying to estimate owe values. $\endgroup$ – Xandar The Zenon Feb 11 '16 at 22:51
  • $\begingroup$ That makes 2,050,525 miles. That is a long ways. About 260 times the diameter of earth. $\endgroup$ – Xandar The Zenon Feb 14 '16 at 23:51
  • $\begingroup$ @XandarTheZenon But 1 trillion years is a really long time. $\endgroup$ – SE - stop firing the good guys Feb 14 '16 at 23:57
  • $\begingroup$ Well, this system is created by silicon based life-forms who want to study the progression of carbon based sentient life, so they'll keep the system together for a long time. The silicon based life-forms are pretty high up on the kardashev scale. $\endgroup$ – Xandar The Zenon Feb 15 '16 at 5:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.