Mars has an inactive mantle and core and thus little magnetosphere to speak of. Assume that a lunar mass object is placed in Mars orbit at 300,000 kilometers out (one light-second). Given the tidal forces at play, how long would it take to bring the magnetosphere up to a strength where it could divert solar wind and radiation to a degree seen here on Earth?

Bonus points if someone can manipulate some of these hypothetical parameters to make the effect presentable in around one thousand years (orbital distance, orbital shape, lunar mass, ???).

  • $\begingroup$ So you want to use the moon's tidal forces to re-liquefy the core? $\endgroup$
    – Vince 49
    Jan 13, 2018 at 23:25
  • $\begingroup$ @Vince49 If that's what is required for a "magnetosphere up to the strength where it could divert solar wind and radiation to a degree seen here on Earth," then yes I think so. $\endgroup$
    – user44399
    Jan 13, 2018 at 23:28
  • $\begingroup$ I'll completely defer to the astrophysicists on this site, but I'm pretty sure the gravitational difference between (a) the moon and the surface of the planet and (b) the moon and the core of the planet is no where near great enough to overcome the friction between the core and its mantle. In other words, any lunar mass strong enough to create the necessary gravity difference would simply cause the moon to tidally lock with the planet and then nothing would happen. I'll be interested in any answers, but I'm thinking you can't do what you're suggesting. $\endgroup$
    – JBH
    Jan 13, 2018 at 23:48
  • $\begingroup$ @JBH, our moon, Luna, is tidally locked, but still produces a tide. I think one would need to know more about the mechanical properties of Mars' interior to calculate the tidal heating via inelastic deformation. Also, I think one would need to know, at least approximately, what the current temperature of the core is in order to calculate how much enery is required to get it molten again. $\endgroup$
    – Vince 49
    Jan 14, 2018 at 0:04
  • $\begingroup$ @Vince49, I'm probably using the wrong term. The moon itself is tidally locked (one face toward the earth), but it's orbit is not locked to Earth's orbit, which is why we see tides. My suspicion is that gravity that heavy would lock the orbit. Perhaps "geocentrically locked orbit?" $\endgroup$
    – JBH
    Jan 14, 2018 at 0:13

1 Answer 1


Calculation by comparison

I have not been able to find resources for calculating the tidal heating of a primary planet from the effects of a moon. However, there is available literature for calculating the tidal heating of the moon itself in its orbit around Earth. The equation for heating in a tidally locked moon with an eccentric orbit is

$$\dot{E}_{tidal} = k_2\frac{21}{2}\frac{R^5n^5e^2}{G}$$ where $k_2$ is the Love number of the moon (0.0222), $R$ is the moon's mean radius (1738100 meters), $n$ is the mean orbital motion in radians per second ($2.662\times10^{-6}$), $e$ is the eccentricity of the moon's orbit (0.0549), and $G$ is the universal gravitational constant ($6.67\times10^{-11}$ m$^3$kg$^{-1}$s$^{-2}$).

Plug all those numbers in and we get $2.2\times10^{W}$, or 22 GW. That is about the electrical power output of the Three Gorges Dam or the Saturn V rocket at launch.

So what can we conclude from that?

First off, the tidal heating is strongly controlled by having a fast eccentric orbit. This is why Io has such strong tidal heating; its orbit is days, with moderate eccentricity. Re-running the numbers for Io, I get 138 TW; more than 500 time what the moon is getting! For a planet, there is nothing to orbit. You need something massive like Jupiter pulling you around. Earth is barely big enough to cause any heat to the moon, so a moon sized object isn't going to do much to Mars.

So if we don't have any means to increase the heating, that level of heat generation is far, far to low to ever heat up the core of Mars. The Earth, for example, loses 44 TW from its interior, so tidal heating is orders of magnitude too low to even balance outflow from a hot core.

Lastly, lets say Mars was orbiting Jupiter, in the same orbit as Io. Due to its bigger size, its heating would be 3 PW. Now we're getting somewhere! Lets use the same assumptions about Mars' core as I used in this question. That means the heat capacity of Mars' core will be about $1.9\times10^{26}$ J/K. Given our input of $3\times10^15$ J/s; it will take us $6.3\times10^{10}$ seconds, or 2000 years to increase core temperature by 1 Kelvin, not even counting heat losses through transfer to the core.


The tidal heating on the moon from Earth is trivial compared to the losses from a hot core. Therefore if an Earth sized object cannot appreciably heat the moon, it is unlikely that a moon sized object will appreciably heat Mars.

Even if you put Mars in a perilously close orbit of Jupiter, it would still take thousands of years to heat up the core appreciably.

  • $\begingroup$ I am very far away from anyting i could claim having real knowledge about. But something makes me think, that the same force that heats the moon should heat the planet, too? Or am i on a completely wrong track there? $\endgroup$
    – Burki
    Jan 16, 2018 at 15:41
  • 1
    $\begingroup$ @Burki Yes and no. The same force applied to the moon will be applied to the Earth as well. But what material the force gets applied to matters. The mass of the ocean is 1/10 the mass of the moon, so a significant portion of the moon's gravity on Earth acts on the ocean (which we know, because of tides). But water's viscosity is low, so it slides around without generating much heat energy. On the other hand, the moon has no ocean, so the Earth's gravity twists its iron core, which generates a lot more heat friction. So composition matters a lot to heat generated, hence the Love number. $\endgroup$
    – kingledion
    Jan 16, 2018 at 15:45
  • $\begingroup$ Doesn't the moon being tidal locked to Earth mean it receives less tidal forces on it than the Earth? So just saying the moon does not receive much heating from the Earth isn't looking at the whole picture. $\endgroup$ Jan 16, 2018 at 17:41
  • $\begingroup$ @A.C.A.C. Since gravity depends on mass and distance, the work done by gravity will be relatively constant over time. Until a body is tidally locked, the energy from gravity will be exerted in changing the rotational energy of the body until it does tidally lock. For example, given enough billions of years, the moon will tidally lock the Earth to it. Once the body is tidally locked, no more energy is 'wasted' in changing its rotational energy, and so (relatively) more energy goes to tidal heating. $\endgroup$
    – kingledion
    Jan 16, 2018 at 18:06

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