Mars has two moons: Phobos and Deimos. Both are irregular and are believed to have been captured from the nearby asteroid belt.

Phobos always shows the same face to Mars because of tidal forces exerted by the planet on its satellite. These same forces causes Phobos to drift increasingly closer to Mars, a situation that will cause their collision in about 50 to 100 million years.

How I can calculate, given appropriate data, the estimated time at which Phobos will collide with Mars?

  • $\begingroup$ Do you understand the nature of the tidal transfer of angular momentum? Why it happens and therefore what factors play into it's strength? $\endgroup$ Commented Sep 1, 2011 at 3:06
  • $\begingroup$ You might be able to do this easily by measuring the radiating temperature of Phobos, assuming it's all light non-radioactive elements (I don't think such a small object has significant internal heating). The temperature will tell you the energy lost to tidal friction. $\endgroup$
    – Ron Maimon
    Commented Sep 5, 2011 at 20:56
  • 2
    $\begingroup$ @Ron: It will tell you the energy that tides on Phobos lose to friction. I would think that would be zero, because Phobos always turns the same face to Mars, and so has no tides. The energy lost because of the tides Phobos induces on Mars is going to be much too small to be inferred by measuring Mars's temperature. $\endgroup$ Commented Nov 25, 2011 at 4:57

4 Answers 4


First, you state a few things that aren't quite right in your question. While the view that's generally talked about is that Phobos and Deimos are likely captured asteroids, dynamically it's a pretty difficult problem (you generally need a third (in this case fourth?) body to take away the extra energy, and it's hard to get a circular orbit around the equator). See for a bit more on that.

In terms of Phobos' demise, there are two things that make this problem very difficult to estimate. First, Phobos' orbit evolves as it orbits around Mars, so you can't just take a linear approach and say, "It's moving towards Mars at 18.3 cm/year so it's going to hit in about 50 million years." It's more complicated and non-linear.

But besides that, there's the Roche Limit to consider, whereby the moon will break up due to tidal forces before it would actually hit. The problem there is that Phobos is already within the Roche Limit, meaning that it's only being held together now by the physical strength of the rock it's made of. And since we don't really know what it's made of inside (though we can make educated guesses and I'm sure there are models out there for its strength), these unknowns make it somewhat difficult to estimate.

  • $\begingroup$ Since Phobos is within the Roche Limit, that implies that there can't be any loose material on its surface, at least not near the points pointing directly towards and directly away from Mars. Any such material would be pulled off the surface by the tide. It also means that the local effective gravitational acceleration is negative. (But it appears to be covered with at least 100 meters of regolith, which is a bit of a mystery.) Reference. It would be fascinating to see a plot of the local acceleration over the surface. $\endgroup$ Commented Oct 12, 2011 at 2:05
  • $\begingroup$ According to the articles I've read, the Phobos Roche Limit has not been reached. It will reach it when the moon is at around 7000 km above the center of Mars (or 3,620 km above its surface). I am curious where you got that information. $\endgroup$
    – David
    Commented Feb 16, 2021 at 5:41
  • $\begingroup$ Roche Limits are computed for fluid bodies, which are bodies that have no internal strength, like a rubble pile asteroid. Phobos is inside that limit. The opposite extreme, the rigid body, is where you assume it's solid rock. Phobos is outside that limit. We don't know the internal structure of Phobos and its cohesion, so it is within a Roche Limit, outside of another, and so could tidally disrupt "at any time" based on where it lies on that continuum. What I wrote, "it's only being held together now by the physical strength of the rock it's made of," is a reference to this rigid body limit. $\endgroup$ Commented Feb 17, 2021 at 1:39

Another symptom of Phobos being so close to mars is that objects on its surface are not all in zero G. On average, object at the same altitude will weigh about 0.285 kg*mm/s less per kg on the day side than on the twilight one, with a gravity sometimes dipping under 2 mm/s^2, this could have noticeable effect on escape velocity.

Additionally, Phobos escape is not really necessary to escape Phobos, a jump above Stickney crater that would launch you all of 3 centimeters up on Earth would be enough to get to the Lagrange L1 point between Phobos and Mars. So you would than be in zero G, and never fall back down (staying in mars orbit) (although this jump would take around half an hour on the way up).

So, an astronaut trying to move around on said crater would be nightmarish without some sort of rocket pack, Let's say he bends over to pick up a rock, he gets back up over the course of 2 seconds, moving his center of mass 1 meter, he will now be moving a minimum of 1/2 meter/s away from Phobos, easily enough to reach L1 and go into a Mars orbit. Let's say that (s)he tries to walk quickly, his/her center of mass moves at several tenths of a meter per second upward, in addition to his m/s forward, therefore, he moves upward and drifts again into Mars orbit. Throw a 5 kg rock downward? go into orbit, throw it upward? still go into orbit, as does the rock, Here is the scary part, people on Phobos might lift not just that rock, but there lander if they try very hard at all. Oh, and heaven help you if you thought firing a gun downward was a good idea! less than a half-second of assault-rifle rounds into the crater will send you to Mars orbit. Or what if you puncture your suit? You might think you could just hurry up and be inside your pressurized spaceship within 2 minutes and you'd be fine, but nope, if it is facing upwards, air will rush out at 340 m/s, if it is escaping through a 1 cm^2 hole (widened by the pressure), you would lose 34 liters per second of air at 1/3rd sea-level pressure, even if your tank holds hundreds of times that, you still emit about 12 grams/s of air, accelerating you at 0.0408 m/s^2, enough to get to get to L1 with around 7 seconds of air released.

Oh, and don't even think about using golf to demo low gravity. 2 kg of metal flying back up at an appropriate speed could Easily send you into mars orbit.

As could a poorly balanced washing machine (send anything on top of it).

Diet Coke and Mentos rockets could reach escape velocity. Forget diet coke and Mentos, Diet coke by itself might be able to Reach L1.

It would take only a few nukes to deorbit Phobos into mars, creating an explosion that obliterates whole martian regions.


Dynamical models over a likely timescale (say $10^6$ to $10^9$ years) would have significant error bars as mentioned above, and therefore one off predictions about individual moons have little validity.


Because Phobos is already within the Roche Limit, within which it should disintegrate due to tidal forces, "appropriate data" would have to include quite a bit of detail about Phobos's structure and composition (information which we currently lack), which would let you determine not so much the "time at which" it will collide, but the period over which bits of Phobos would.

  • $\begingroup$ This answer(v1) is similar to Stuart Robbins' answer over at Astronomy.SE, cf. physics.stackexchange.com/q/14212 $\endgroup$
    – Qmechanic
    Commented Oct 25, 2011 at 21:16
  • $\begingroup$ @Qmechanic: Yes it looks like there's also much additional detail there. Should I incorporate his; link to it — what's the correct etiquette? $\endgroup$
    – orome
    Commented Oct 25, 2011 at 21:26

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