I don’t know about Phobos, but apparently you could escape Deimos with a bike and a ramp.
Another interesting tidbit, by the way: The orbital period for a low orbit around an object depends only on the object’s density. Since mineral objects (like Earth, Mars, or Phobos) all have around the same density, they also all have around the same orbital period. So just as an orbit around the Earth takes about an hour and a half, if you jumped into orbit around Phobos, it would likewise take you about an hour and a half.
That is very cool! Why is density and not mass the determinant of orbital frequency?
Strictly speaking, mass and orbit radius determine the orbital frequency.
The object’s radius (which is about the same as orbit radius in a low orbit) is determined by its density for a given mass.
xkcd has a good to-scale representation of gravity wells and force needed to escape objects in our solar system. Not a direct answer to OP’s question, but relevant to the discussion. Often, I’ve noticed, visual representation furthers understanding significantly.
Orbital velocity scales as v[sup]2[/sup] ~ M/r : The more massive the object you’re orbiting, the faster, and for a given mass, the closer you’re orbiting, the faster. You can get this just from conservation of energy. And the orbital period is inversely proportional to velocity, but of course also depends on how far you have to go, which is proportional to r: P ~ r/v, or P[sup]2[/sup] ~ r[sup]2[/sup]/v[sup]2[/sup] . Substituting in the previous expression for v[sup]2[/sup], we have P[sup]2[/sup] ~ r[sup]2[/sup] * r/M , and since M/r[sup]3[/sup] is proportional to density, P[sup]2[/sup] ~ 1/rho, or P ~ rho[sup]-1/2[/sup]. Thus, the orbital period P depends only on density and some constants of proportionality.