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Imagine a universe in which Deimos and a human are the only objects in the universe. It doesn't matter if the human jumped even 1 nanometer per second slower than escape velocity, they would still eventually be drawn back to Deimos even if they traveled vast distances during that time.

But in our universe Deimos happens to be pretty close to Mars. Imagine a human standing at the sub-Martian point on Deimos so that Mars is directly overhead.

If said human jumped perpendicular to the ground at slightly less than escape velocity, even though they should theoretically fall back to Deimos, they would instead come close enough to Mars to the point where they will either crash into it or enter an orbit.


My question is: What is the maximum velocity in which a human jumping towards Mars at the sub-Martian point on Deimos could be going and still end up falling back to Deimos rather than get overwhelmed by Mars' gravity?

NOTE: Feel free to pretend the human, Deimos, and Mars are the only bodies in the universe and that they're all spheres of uniform density to eliminate any complications. This is less of a question about Deimos/Mars in particular and more about the real escape velocity of a very small satellite orbiting close to its planet like Deimos.

Also don't worry about how a human can somehow achieve speeds near Deimian escape velocity with a simple perpendicular jump. Assume they're superhuman ;)

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An object moving along the line that passes from Mars and Deimos, will be attracted towards Deimos or towards Mars depending on its location with respect to the L1 Lagrange point. If Mars' mass is $M_M$, Deimos' is $M_D$ and Deimos' orbital radius is $d$ (Deimos' orbit has a very small eccentricity), the location of the L1 point can be found from: $$r_{L1} = d \sqrt[\leftroot{-1}\uproot{4}\scriptstyle 3]{{M_D \over 3M_M}} \approx 21.5 \ \text{km}$$

$r_{L1}$ is also the radius of the Hill sphere of Deimos.

Deimos has an average radius $R_D = 6.2 \ \text{km}$. The energy required to go from the surface to the L1 point is given by the difference of the gravitational potential energies

$${1 \over 2} v^2 = {GM_D \over r_{L1}} - {GM_D \over R_D}$$

Which gives $v \approx 4.7\ \text{m/s}$.

You may have noticed that I have omitted the contribution of Mars' potential energy. The reason is that Deimos is orbiting around Mars and therefore, in the vicinity of Deimos, Mars' attraction is cancelled by the centrifugal force. The centrifugal force has been also taken into account when using the above formula for the position of the L1 point.

It should be also noticed that, when jumping from the surface of Deimos directed straight towards Mars, you are moving radially and therefore the Coriolis effect will move you sideways. The result is that you will not reach the L1 point because your trajectory will be curved. The easiest way to fall onto Mars will be to jump in Deimos' retrograde direction, so that when you find yourself outside Deimos' sphere of influence, you have the least angular momentum with respect to Mars. Unfortunately, in order to be sure to fall onto Mars, you will have to negate precisely all the angular momentum in the Deimos' orbit. This means that you will have to jump in the retrograde direction at about the same speed of Deimos, which is 1.3 km/s.

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  • $\begingroup$ Nice answer +1, better than mine. But should we also consider aerobraking as well as an elliptical orbit tangent to the surface instead of a suicide burn to zero velocity? $\endgroup$ Sep 8 at 14:03
  • $\begingroup$ @fasterthanlight Thank you! Let's say that you have to reduce your angular momentum to the point you actually hit the planet. I agree with you that it isn't necessarily zero angular momentum. Anyway, I feel that this is not the definitive answer yet. I was thinking of doing a numerical simulation to find the actual smallest velocity to leave Deimos and the direction towards one should jump, but I don't know if I'll find the time. Maybe you will have the chance to do it $\endgroup$
    – Prallax
    Sep 8 at 14:18

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