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Due to tidal deceleration, Phobos is predicted to be destroyed by Mars in the next 30-50 million years either by collision or by breaking up in a planetary ring. If Phobos falls on Mars in one piece could the impact could be comparable with Chicxulub meteorite? Since Mars is smaller that Earth, would the effects of the impact cause a global restart of volcanic activity and temporary warm the planet? Would this be sufficient to make Mars more friendly to life for a few thousand years?

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The impact of Phobos, even in one piece (less likely), would be different from Chicxulub. Phobos is probably a little larger than the Chicxulub impactor, but much slower, and comes in almost perfectly tangential. The impact energy would be less than a 10th of the Chicxulub impact, and the energy would be distributed over a large region around the Martian equator. It wouldn't start volcanism, and it wouldn't warm Mars in a relevant way. It wouldn't make Mars more friendly to life by warming. The only noteworthy contribution to habitability would be additional organic matter which Phobos is presumed to contain.

Phobos will (probably) desintegrate due to tidal forces, before its fragments will collide with Mars. Any loose rocks, dust, regolith on the surface will separate from Phobos below the Roche limit of $$d=1.26~R_M\cdot\left(\frac{\rho_M}{\rho_m}\right)^\frac{1}{3},$$ with $R_M=3389.5\mbox{ km}$ the radius of Mars, $\rho_M=3.9335 \mbox{ g/cm}^3$ the mean density of Mars, and $\rho_m=1.887\mbox{ g/cm}^3$ the mean density of Phobos. This results in a Roche limit of $$d=1.26\cdot 3389.5\mbox{ km}\cdot\left(\frac{3.9335 \mbox{ g/cm}^3}{1.887\mbox{ g/cm}^3}\right)^\frac{1}{3}=5455.6\mbox{ km}.$$ That's for Phobos' center about 2059 km above the equatorial surface of Mars (taking 3396.2 km as equatorial radius.) Therefore desintegration will start long before the impact. Loose rocks will collide with Phobos and unlock more material. Whether there will be a considerable solid remnant to finally cause a big (tangential) impact is difficult to predict.

The orbital speed of the fragments at the time of collision will be about $$v_0=\frac{v_e}{\sqrt 2},$$ with $v_e=5.03\mbox{ km/s}$ the escape velocity of Mars. Phobos' inclination of 1.093° (relative to Mars's equator) is almost neglectable for the calculation of the impact velocity. Mars' equatorial rotation velocity is about 241.17 m/s.

The impact velocity will hence be about $$\frac{5030\mbox{ m/s}}{\sqrt 2}-241.17 \mbox{ m/s}=3316\mbox{ m/s}.$$ Phobos' mass is about $1.0659\cdot 10^{16}\mbox{ kg}.$ This results in a kinetic impact energy of $$\frac{1}{2}mv^2=\frac{1}{2}\cdot 1.0659\cdot 10^{16}\mbox{ kg}\cdot (3316\mbox{ m/s})^2=5.86\cdot 10^{22}\mbox{ J}.$$

The heat of fusion of water is about 334 kJ/kg, the heat of vaporization of water at 0°C is about 2504 kJ/kg, together 2838 kJ/kg are needed to evaporate water ice of 0°C. Hence the impact energy of Phobos would be sufficient to evaporate $$\frac{5.86\cdot 10^{22}\mbox{ J}}{2.838 \cdot 10^{6}\mbox{ J/kg}}=2.06\cdot 10^{16}\mbox{ kg}$$ of water ice of 0°C, corresponding to $2.06\cdot 10^{13}\mbox{ m}^3$ of liquid water.

Mars' surface area is about 144,798,500 km², or $1.447985\cdot 10^{14}\mbox{ m}^2$. Equally distributed, we could get a layer of water ice corresponding to $$\frac{2.06\cdot 10^{13}\mbox{ m}^3}{1.447985\cdot 10^{14}\mbox{ m}^2}=0.142\mbox{ m},$$ (correspoding to 14.2 kg/m² of liquid water) evaporated by the impact energy.

With a surface gravity of 3.711 m/s², this would correspond to a pressure of $14.2 \mbox{ kg/m}^2\cdot 3.711 \mbox{ m/s}^2= 52.8 \mbox{ N/m}^2=0.528\mbox{ hPa}$.

The current atmospheric pressure on Mars is about 6.3 hPa, with diurnal, seasonal and regional changes. Thus we could get a maximum of 10% short-term pressure increase in the Martian atmosphere by water vapor, in an ideal case.

At the end, an impact of Phobos, or its fragments, would result in a regional reworking of the Martian surface in the equatorial region, but not in a long-term global change.

Just melting water ice would lead to a global ocean with a depth of about 1.2 m. But this ocean would boil due to the low atmospheric pressure, and freeze out again. Same with local lakes due to the uneven topography.

Model calculatios with a release of carbon dioxide would cause some temporary greenhouse effect, but in the equatorial region a release of carbon dioxide is even more unlikely than a release of water vapor.

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I always like your answers, Gerald. – Jeremy Apr 30 '14 at 20:14
Thanks! Lot of calculations, this time, to get an intuitively accessible scenario. – Gerald Apr 30 '14 at 20:21
I am intrigued and I wish to subscribe to your newsletter. – dotancohen May 1 '14 at 12:25

I noticed an important error in Gerald's otherwise excellent calculation above. A layer of water 0.142 m thick per meter squared has a mass of 142 kg, not 14.2 kg. This means Gerald's results should be multiplied by 10:

14.2 kg/m2 x 3.711 m/s2 = 528 N/m2 = 5.28 hPa

This is comparable to the current average Martian surface pressure of ~6.3 hPa.

A side note: The orbit at impact could also be an ellipse with current orbit as apoapsis and the Martian surface as periapsis. This hypothetical change in orbit would increase the impact speed to ~4.3 km/s (first equation at ). However, this should probably be achieved artificially and has nothing to do with the natural decay of the moon.

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This is more of a comment, or perhaps a suggested edit to an already existing answer, than an answer on its own. Could you please at least provide some reference or calculation for the last part, that Phobos impact would almost double the current Martian atmosphere? Describing the mechanism how Phobos increases orbital eccentricity while keeping its apoapsis would also be welcome. Thanks! – TildalWave Sep 27 at 10:50
No, a single dimension doesn't weigh anything. I guess you mean a 14.2 cm thick layer of water of 1 meter square surface, i.e. a volume of 142 liters. At which point we get to what weight is, before we can get to pressure. Say, 142 liters of water on Mars (at surface gravity of 3.711 m/s²) weigh 526.96 N or about 53.73 kg, not 142 kg. Also, please edit additional explanation into your answer, that's what the purpose of my comment was in the first place. See How to Answer. Thanks! – TildalWave Sep 27 at 11:28
Dear Tidalwave, if that was your purpose, you should have pointed those things out in your first comment. Now my original answer is changed accordingly. Thanks, and bye. – Jarmo Sep 27 at 11:42

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