If it were possible to replace the Moon with Mars, how distant would it have to be to essentially create the same oceanic tides as the Moon currently does? Mars seems to be roughly 3 times the mass of the Moon, so does that mean it should be 300% the distance?

Furthermore, would the increased distance cause any issues with Earth's orbit around the Sun or potentially cause some sort of situation like Charon and Pluto where the centre of mass is between them?

  • $\begingroup$ What do you mean by "same tides"? Any two bodies orbiting each other are going to cause "tidal effects", so the answer might be doesn't matter, all orbiting bodies will cause tidal effects like the moon. But if you are looking for a system that causes the same amplitude, or the same time frequency, or the same distribution of diurnal/semi-diurnal tides; or if you you asking about tidal effects on just oceans or other systems — or all of the above — working out the orbital mechanics becomes a bigger challenge. What is it you are trying to solve for? $\endgroup$ Jun 5 '14 at 14:12
  • $\begingroup$ Sorry, I did mean oceanic tides, I fixed it. I'm not trying to solve for anything in particular, just my own curiosity. $\endgroup$ Jun 5 '14 at 17:27
  • $\begingroup$ A curious tidal effect is that, assuming equal density, the size of the object in the sky determines the strength of the tides. Mars is almost twice the radius of the moon, so to cause equal tidal bulge, it would need to be about twice as far, which as pointed out, wouldn't be a stable orbit for very long. Mars being slightly denser than the moon accounts for the slightly greater than twice the distance in the answer below. $\endgroup$
    – userLTK
    Mar 11 '16 at 18:56

The Tidal acceleration between 2 bodies is calculated with this formula:

$$ a_T = \frac{2GM_MR_1}{Dm^3} $$

Where $M_M$ is Mar's mass, $R_1$ Earth Radius and $D_m$ the distance to the Moon. If you equal this to Moon's Tidal acceleration you will get $D_M$ as distance to Mars to get the same Tidal acceleration having $\frac{M_M}{M_m}=8.73328184501$:

$$ D_M = \sqrt[3]{8.73}D_m\simeq2.06Dm $$

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    $\begingroup$ ...went that far, should finish with cube-root-of-3 is about 1.44. Or about 1.5 times the current moon's distance. $\endgroup$ Jun 5 '14 at 14:55
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    $\begingroup$ Note that, at 0.791 million km, this is a bit outside the usual region of orbital stability, though still within Earth's Hill sphere, so Mars would not likely stick around in the very long term. $\endgroup$ Nov 11 '15 at 19:19
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    $\begingroup$ Now that would be an impressive sight in the night sky :) $\endgroup$ Feb 8 '17 at 12:31

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