# Would it be possible to "fall off" Phobos? [closed]

Phobos (one of Mars' moons) has the following dimensions: 27 × 22 × 18 km. It has low mass, and little or no gravity/atmosphere.

Let's just assume that it's possible to walk around on Phobos without dying or getting hurt. With it being so small, you should be able to just walk to one of the edges of Phobos.

This brings me to my first questions: Since there is so little gravity/atmosphere and it is so small, would you be able to "fall off" Phobos ?

This concept is hard for me to grasp because on earth, its hard to imagine to "fall off" the earth, because you're never physically on an edge (since the earth is round). Phobos however doesn't seem to have a smooth round shape (and is alot smaller).

Question number two: When you're standing on of the edges of phobos, would you be able to "stare down" into the empty vacuum of space?

I compare this in my head with standing on the edge of a chair or table, looking down to the floor. Only in this case, you're staring straight down into space. Does this comparison make sense, or am I looking at this the wrong way?

• I am not sure how you imagine the "edge" of phobos. That rock might be small and not quite round, but its gravity is still pointing to its center, no matter where you stand on it. Mar 13 '14 at 20:13
• @Philipp well its mainly because Phobos is so small that I get the idea that it would be possible to stand on the "edge". kinda like you would be able to go to the edge/corners of a football field, and go "out of bounds" so to speak. Mar 13 '14 at 20:17
• Last time I checked, football fields aren't round and aren't in an orbit around a planet. Mar 13 '14 at 20:20
• Let's put it that way: No part of Phobos is falling down to mars. So no matter on which side of Phobos you stand, you aren't falling down to Mars either. Mar 13 '14 at 20:29
• Well, technically, all of Phobos is falling down to Mars. However it's moving sideways so fast, it keeps missing Mars. Mar 14 '14 at 4:57

In order to leave a gravity well, you need to exceed the escape velocity. The escape velocity depends on the mass of the body and your current distance from its center. The escape velocity on the surface of Phobos is approximately 11.4 m/s or 41 km/h. It likely varies depending on where you are on Phobos because the moon is quite irregular-shaped. You would expect it to be less when you are on top of a mountain than inside a crater, but it's hard to tell because we don't know much about Phobos' internal structure. When its internal density is as unsymmetrical as its external shape, its gravity field might have considerable local variations. But you can expect the escape velocity to be roughly about 40 km/h anywhere on Phobos' surface. When you manage to run or jump so fast that you exceed that speed, you would escape from Phobos and then end up in an orbit around Mars.

Reaching such a high speed using just your own body strength, especially while you are wearing a heavy space suit, is unlikely. But with a bit of technical help it might be possible. Riding a bicycle on Phobos might be enough (remember that bicycle-across-the-moon scene from E.T.? It would look a bit like that).

• This is not correct. You are ignoring the tides exerted by Mars. Mar 13 '14 at 20:15
• @DavidHammen Can you educate us how the tides exerted by Mars would affect the escape velocity from Phobos? Mar 13 '14 at 20:21
• @DavidHammen By my calculations, the tidal acceleration on the edge of Phobos is on the order of 10% of the moon's gravity. So unless you have close to escape velocity, you're falling back to the moon. Mar 13 '14 at 20:23
• @AlanSE - See my answer. Tidal accelerations are more than a 1/3 of Phobos gravity. You don't need to reach escape velocity to escape. You just need enough to take you out to the Hill sphere. 3 m/s should suffice. Mar 14 '14 at 13:20
• @DavidHammen Our approximations aren't much further off than what one would reasonably expect. However, one might read your comment and answer to say that the Mars-facing side requires 3 m/s whereas the side face will be 11.4 m/s. It's closer to 6.2 m/s versus 11.4 m/s, even using your approach. I believe you forgot a factor of 2 in the Wolfram alpha link, and the radius of Phobos is about 2 km larger than what's employed for the escape velocity number. Mar 14 '14 at 13:49

would you be able to "fall off" Phobos ?

Right now, no. In a few million years, yes. Phobos is slowly spiraling in towards Mars. In a few million years it will be close enough to Mars that tidal forces from Mars will tear Phobos apart. The very rocks that loosely comprise Phobos will fall off of Phobos.

Phobos is 27 × 22 × 18 km. Your "edge of Phobos" is that point on Phobos that is the furthest from the center of Phobos, or about 13.5 km. I'll use this figure of 13.5 km as the basis for my answer.

Currently, Phobos's orbit about Mars brings the center of Phobos to with 9234 km of the center of Mars. At this distance, the tidal acceleration toward Mars at a distance of 13.5 km from the center of Phobos is 1.47 mm/s2.

That's very small, but then again so is the gravitational acceleration toward Phobos. An overly simplistic calculation using $GM/r^2$ gives a value of 3.9 mm/s2 at the furthest most point of Phobos from its center of mass. Since Phobos isn't spherical, the gravitational acceleration toward Phobos will be a bit more than this at that distance. But it's not going to be a whole lot more than that.

The tidal forces currently aren't enough to make you "fall off" Phobos. At least not yet.

They will be as Phobos spirals ever inward. The tidal forces from Mars will rip Phobos apart when Phobos falls to within less than about 7000 km from Mars. That will happen in a few million years.

Another way to look at it: Phobos's Hill sphere is currently 16.37 km. That's just a tiny bit more than the 13.5 km figure used above. Phobos's Hill sphere will shrink as Phobos spirals inward. It will become less than 13.5 km when Phobos periapsis distance drops to about 7600 km. At that point Phobos will cease to exist since Phobos apparently is a rubble pile.