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So, imagine a team of space marines, fighting some sort of space bug zombies on a planet's moon. Things are going pretty bad for them, and the last surviving protagonists are running for their lives. The thing is, with each step, they soar into the air, higher and higher, until the point where the planet is directly overhead. At that point, the moon's low gravity is completely counteracted by tidal forces and the space marines' jumping forces, and the marines float out into space, where they are picked up by their ship (or, alternatively, where they get caught in the planet's gravity and burn up on re-entry, depending on if this is a comedy or tragedy).

Now, I already know this is possible; if the moon weighs 1kg it wouldn't be too hard to reach escape velocity, especially when aided by the gravitational pull of an Earth-sized planet nearby. What I'm wondering is how big I can make the moon and still achieve this effect.

For the parameters, let's say the planet is Earthlike, and the moon is right at the edge of its Roche limit (which should maximize tidal forces). An average human has to be able to reach escape velocity when the planet is directly overhead, just by jumping. Assume the moon has no atmosphere and is made of similar materials to those you'd find on our Moon.

Let me know if you have any further questions.

EDIT: The human(s) used in this question can be assumed to not be augmented in any way (aside from space suits, which I'm fine with removing for the sake of simplicity), and need not actually be able to 'run' across the surface of the moon in a way resembling how they would on Earth. Perhaps in another question, I can ask about how to quickly move across the surface of a body with little to no gravity, but that is beyond the scope of this question.

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    $\begingroup$ Or in rocket science terms, How small can a moon be that you can jump with enough speed difference to deorbit from the planet's orbit. $\endgroup$ – ratchet freak Sep 30 '15 at 14:31
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    $\begingroup$ Obligatory XKCD with specific interest to Deimos. $\endgroup$ – Green Sep 30 '15 at 17:45
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    $\begingroup$ The problem is not with escaping the moon - several known asteroids are small enough that you can escape them by human power (including the smallest moon of Mars) - but you will still remain in orbit around the planet - just a slightly different orbit than the moon. $\endgroup$ – vsz Oct 1 '15 at 6:02
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    $\begingroup$ Why did you ask here rather than on Physics? $\endgroup$ – curiousdannii Oct 1 '15 at 12:56
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    $\begingroup$ "if the moon weighs 1kg"... Then that isn't really the moon, it's just a rock bouncing away from the real moons: the space marines $\endgroup$ – binaryfunt Oct 1 '15 at 16:51
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No. You can't jump from a moon in stable orbit to the planet. This is because the orbital velocity of the satellite is sufficient to keep the satellite in orbit, the jumper starts with that same orbital velocity, and given how large orbital velocities generally are jumping is not going to make a difference. If the orbital velocity was that close to unstable, the orbit would not be stable enough for the moon to exist. Jumping will essentially just change your orbit by amount too small to allow you to escape the gravity of any moon large enough to really "jump" from.

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    $\begingroup$ and don't forget that if the orbit is that unstable, the force of your jumping that's exerted on the moon may well be enough to push the moon's orbit over the edge (either push it out of orbit or stabilise its orbit so the next person's jump can't be enough). $\endgroup$ – jwenting Oct 1 '15 at 6:06
  • $\begingroup$ Just wondering what, if the Earth-Moon were actual Double planets? Maybe tidally locked and spinning around each other? Would something like that be possible? $\endgroup$ – Ivo Beckers Oct 1 '15 at 9:07
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    $\begingroup$ +1, though the last sentence isn't quite right. You can escape the moon's gravity if its mass is small enough, it's just that you'll end up in an independent orbit around the planet similar to the moon's one. If tidal forces are insignificant you will collide with it again a half-orbit later, but you can miss it if they're strong. You might add that tidal forces aren't important apart from that - they don't really help with jumping off the moon in the first place, no matter how strong they are. $\endgroup$ – Nathaniel Oct 1 '15 at 9:47
  • $\begingroup$ @Nathaniel Why would you collide a half orbit later? As far as I can see to acomplish this you would need to not only cancel the moons orbital speed, but do this once again to get that speed only in the opposite direction. $\endgroup$ – Taemyr Oct 1 '15 at 10:17
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    $\begingroup$ @Taemyr it doesn't quite work like that. Both orbits are ellipses, and you move faster than the moon while you're inside its orbit but slower when you're outside - so they cancel out unless the differences in the shape of the orbits is really quite significantly different, which they won't be if the initial velocity difference is only a few metres per second. $\endgroup$ – Nathaniel Oct 1 '15 at 12:26
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As mentioned in a couple other answers, the problem here isn't just the escape velocity of the moon, but the orbital velocity of the planet as well.

Escape Velocity
So first, you need to escape the moon's gravity. Wikipedia says the fastest human sprinting speed is 12.4 m/s. Let's assume that's a pretty good number for jumping speed also. So we need a moon whose escape velocity is less than that. I happen to have a paper here which says the escape velocity is given by $v_{escape}=\sqrt{2Gm\over r}$. Notice that it depends on both mass and radius, so there isn't a one-size-fits-all approach. We could have a massive but large moon, or a lighter but smaller moon, and get the same escape velocity.

So let's say the upper limit to density is Earth's density, $\rho_E=5495 {kg \over m^3}$, and the lower limit is around the density of a comet, $\rho_C$$={0.3 g\over cm^3}$$=300 {kg\over m^3}$. The density of the moon is in the middle at $\rho_M=3343{kg\over m^3}$ We can re-arrange density to solve for mass. $\rho={m\over V}$$\leftrightarrow m=V\rho$. The volume of a sphere is $V={4\over 3}\pi r^3$, so $m={4\over 3}\pi r^3\rho$.

Ok, so we can plug density and our mass substitution into the escape velocity equation: $v_{escape}=\sqrt{2G({4\over 3}\pi r^3\rho)\over r}$$=2r\sqrt{{2\pi\over 3}G\rho}$. From here, we can re-arrange to solve for radius. $r={v_{escape}\over 2\sqrt{{2\over 3}G\pi\rho}}$.

$r(\rho_C)$$={12.4 {m\over s}\over 2\sqrt{{2\pi\over 3}6.673\cdot 10^{-11}{N\cdot m^2\over kg^2}300{kg\over m^3}}}$$={12.4\over 2\sqrt{{2\pi\over3}6.673\cdot 10^{-11}\cdot 300}}{{m\over s}\over \sqrt{{kg\cdot m\over s^2}{m^2\over kg^2}{kg\over m^3}}}$$=30279m$$=30.3km$

$r(\rho_M)$$=9070m$$=9.1km$

$r(\rho_E)$$=7737m$$=7.7km$

$r(\rho)$$={524447\over\sqrt{\rho}}$

So our moon's radius needs to be less than 30.3 km if it's a comet-like object, less than 9.1 km if it's moon-like, and less than 7.7 km if it's Earth-like. You can use the last equation for an arbitrary density.

De-orbit Velocity
But now we're just hanging out in space. We need to fall onto the planet. So we need our jump from the moon to leave us with enough velocity to cancel out our orbital velocity. The equation for orbital velocity is the equation for escape velocity. Let's say we're orbiting the parent planet at the Earth-Moon distance, 365,542 km. We can solve for the required mass of the planet.

$v_{orbital}=\sqrt{2Gm\over r}$$\leftrightarrow m=v_{orbital}^2{r\over2G}$$=12.4^2{m^2\over s^2}{365,542,000 m \over 2\cdot 6.673\cdot 10^{-11}{N\cdot m^2\over kg^2}}$$=4.225\cdot 10^{20}{m^2\over s^2}m{s^2\over kg\cdot m}{kg^2\over m^2}$$=4.225\cdot 10^{20}kg$.

The Earth has a mass of about $5.972\cdot 10^{24}kg$, which is about 14000 times the mass our planet needs to be. So with a tiny planet and a really tiny moon, you could jump from the moon to the planet.

If you want to play with different distances and planetary masses, you can use the following equations, remembering mass is in kilograms and distance is in meters.

$m_{planet}$$=1.1558\cdot 10^{12}\cdot r_{orbit}$
$\leftrightarrow$$r_{orbit}$$=8.6518\cdot 10^{-13}\cdot m_{planet}$.

Positioning
An important note here is that you can't jump towards the planet. That just gives you an eccentric orbit. You need to jump when the planet is on the horizon, and it needs to be backward compared to the moon's orbit.

Jump Backwards From Moon to Fall to Planet

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  • $\begingroup$ Keep in mind that the moon's gravity will still be pulling on you after you have jumped. jumping 'not quite hard enough' would have the effect of the moon's gravity slingshotting you away past the planet into an excentric orbit. $\endgroup$ – Timothy Groote Oct 1 '15 at 6:29
  • $\begingroup$ @TimothyGroote That effect is already accounted for in overcoming the escape velocity of the moon. $\endgroup$ – Taemyr Oct 1 '15 at 8:14
  • $\begingroup$ I'm not one for nitpicking, but technically speaking the slingshot scenario could also count as overcoming escape velocity. ;) $\endgroup$ – Timothy Groote Oct 1 '15 at 8:20
  • $\begingroup$ @Micael It's not that you get an eccentric orbit when you jump straight towards the planet. - Even when you jump retrograde you get an eccentric orbit (perigee in the planet, apogee level with the moon). The issue is that when you jump towards the planet you don't change the orbital energy, hence you are not only lowering your perigee you are also increasing your apogee. $\endgroup$ – Taemyr Oct 1 '15 at 8:20
  • $\begingroup$ @TimothyGroote I think you misunderstood me. What I meant to say was that the gravitational interactions with the moon is fully accounted for once the escape velocity is subtracted from your final velocity. (Baring a new encounter with the moon one or more orbital periods later.) $\endgroup$ – Taemyr Oct 1 '15 at 8:24
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According to Wikipedia, the Roche limit of the Earth and Moon is around 10,000 km. At 10,000km from Earth, gravity to the Earth is 1.48m/s^2. According to Google, the gravity of the Moon is 1.6m/s^2. So this scenario, at least at first, seems somewhat plausible; it does seem like there could exist a case where someone could jump off of a moon with help from tidal forces.

That is, until you consider what the Roche limit actually is. Essentially, it's the very thing you're looking for: the point at which tidal forces cancel out the gravitational pull of the orbiting body. If you have a moon hanging on just barely at the edge of this limit, things like rocks and dust are going to simply float away, perhaps forming a planetary ring. If the moon gets any closer, it may become unstable and tear apart. On the one hand, this is exactly how your space marines are going to be able to escape just by the force of their legs, but on the other hand it makes the chances of your moon surviving for a few more million years very low.

Thus, the question really shouldn't be how big this could get, but how safe you want to play it. Essentially, you want tidal forces + force of jumping = gravity of the moon. With a smaller moon, your jumping forces are going to play a larger part in this equality, and you can keep your moon at a safe distance. For a larger moon, you're going to need to achieve nearly zero-g conditions for this to work, which means your moon's going to get torn apart.

As a fun sci-fi thriller alternative, perhaps your moon is being de-orbited. Thus, at some point in time, this scenario has to work, even if you're seconds away from hitting the planet.

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  • $\begingroup$ Unfortunately you would be unable to utilize tidal forces if you wanted to jump from the moon to the earth. In fact if the goal was to reach the earth the tidal force would work to prevent you from leaving the moon. Tidal forces pull away from the body you are on along the line through the two bodies, on lines perpendicular to this line tidal forces pull you towards the body. In order to get to the earth you will need to impart a retrograde impulse. $\endgroup$ – Taemyr Oct 1 '15 at 8:06
  • $\begingroup$ The Roche limit is, as you say, pretty irrelevant. What's important is the jumping velocity and the dV required to reach and cross the lagrange point (to be more specific than talking of orbital / escape velocity.) You might want to clarify that at the beginning. My first glance made me think you thought Roche limit was the right approach. $\endgroup$ – Level River St Oct 1 '15 at 10:21
  • $\begingroup$ @Taemyr The lagrange point represents the lowest enegy required to escape the orbit of the moon and enter into orbit of the planet. It's lower than the calculated escape velocity of the moon. The contours at en.wikipedia.org/wiki/Lagrangian_point should make it clear (they show a planet and star, but it's the same principle as moon and planet.) Also, the number of contours crossed from moon to lagrange is much less than from lagrange to planet, yet another way of showing that escaping the moon will just put you in orbit around the planet, not allow you to collide with it. $\endgroup$ – Level River St Oct 1 '15 at 10:43
  • $\begingroup$ @Taemyr I see you deleted your comment, but I'll leave my response up in case someone else has the same question. $\endgroup$ – Level River St Oct 1 '15 at 10:45
  • $\begingroup$ @steveverrill That was the argument I expected you to make. However, it is insufficient to show that your new orbit will intersect the planet. The problem is this - In order to reach the L1 point you will need to have some lateral velocity, because the L1 point will have moved since you left the moon. If you have to much lateral velocity your orbit is not sub-orbital. $\endgroup$ – Taemyr Oct 1 '15 at 10:55
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How is everyone missing the big picture?

I don't care if you can jump off the moon or not, a successful jump gets you into orbit but you have no way to get from orbit to the planet. If the moon were so far out that it's orbital velocity was within what someone could produce by jumping it would be far outside the hill sphere of the parent body and would have wandered off long ago.

Thus to actually accomplish this maneuver you need a tiny, tiny moon in a very distant orbit about a rogue planet deep in interstellar space. Now you have two problems:

1) The fall time is going to be considerable--you'll likely run out of life support.

2) A planet deep in interstellar space is going to be cold. Incredibly cold. You won't have any atmosphere to speak of and thus no aerobraking and no parachutes. If you have enough delta-v in your armor to land on the planet the whole point became moot as it was far more than you needed to get off an ordinary sized moon.

Not to mention that once you jump you're a sitting duck if they have any sort of ranged weapon.

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  • $\begingroup$ You can have a less distant orbit if it's highly eccentric. $\endgroup$ – Taemyr Oct 2 '15 at 12:01
  • $\begingroup$ @Taemyr No. What counts is your orbital velocity at the point you jump. Something in an eccentric orbit that's currently near it's parent is actually moving faster than something in a circular orbit of the same distance. $\endgroup$ – Loren Pechtel Oct 3 '15 at 3:03
  • $\begingroup$ You jump when the moon is furthest from the planet. $\endgroup$ – Taemyr Oct 3 '15 at 12:54
  • $\begingroup$ @Taemyr That will cut the needed velocity a bit but it's not going to make much difference. $\endgroup$ – Loren Pechtel Oct 4 '15 at 2:35
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Not practically possible no matter which values you use.

For a moon measuring too less in mass (and length), it would be impossible to have space bugs chasing a whole team of marines as the place is simply too small for them all. Furthermore, you cannot jump successfully and then land back on that moon if it is too little. Your jump alone will launch you into space. That would be an asteroid sized body. Even for a body measuring 50 cubic km and made of earthly rock, there are few chances you would land back if you jumped with full power. Also notice you wouldn't be able to run full speed on a low-gravity object as your feet wouldn't be getting enough friction to strongly grip the ground and let you use your thigh power to launch you forward. You would only be able to jump upwards and then you'd be lost in space.

If the body is large enough to let you jump and play chase games, then it means the body is sufficiently large to not let mortals escape it with mere jumping around. You would have to have grasshopper-like strong legs to get escape velocity on such a body.

Something like a mini-mini Io versus a planet double the size of Jupiter comes to mind where it might have been possible. Good luck with the jump though. You won't want to enter the atmosphere of a gas giant ...

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  • $\begingroup$ With a robotic exoskeleton integrated in the suit of the marines, it is definitely possible for a fit, fully grown soldier to do this on a moon. It is not specified against by OP. $\endgroup$ – the_OTHER_DJMethaneMan Sep 30 '15 at 14:17
  • $\begingroup$ @DJMethaneMan I didn't think I had to specifically specify 'no exoskeletons'. Though I will admit that running was more of a soft requirement, and I by no means intended it to factor into any answers. $\endgroup$ – DaaaahWhoosh Sep 30 '15 at 14:35
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especially when aided by the gravitational pull of an Earth-sized planet nearby.

No. The "tidal" pull will be negligible compared to a gravitational pull great enough to allow the astronauts run.

If the moon is locked in place, then you might be able to justify the permanent tidal bulge giving them enough additional elevation to be able to "jump" off the moon.

Again, making the "moon" small enough for this to be possible may also be reducing the local gravitational force below that needed to be able to run.


One possible solution to your scene would be to have the moon rotating very fast (both rotation and revolution in the same direction) The point nearest the planet will also be the point on the moon where the moon's rotation is offsetting the moon's revolution. Add a mountain along the moon's equator will give a bit of extra oomph.

The arbitrarily high rotation will give the extra force needed to enter orbit and the will help "de-orbiting" away from the moon and fall closer to the planet (not rapidly though).

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  • $\begingroup$ My apologies, the 'running' part was meant more for context, and less for the actual scope of the question. I don't actually need the planet to have enough gravity for conventional running; assume space marines are crawling, or locomoting in some way that allows them to remain close to the surface of the moon until they jump. $\endgroup$ – DaaaahWhoosh Sep 30 '15 at 14:37
  • $\begingroup$ Regardless, the "tidal" pull is negligible compared to local gravity. If they can "jump" free of the moon at the point closest to the planet, they can jump free elsewhere. The point opposite the planet may actually be (very) slightly easier if the direction of rotation matched the direction of revolution, and the rotation is fast enough. That does bring to mind one possible solution for your scene. I'll add that to my answer. $\endgroup$ – Michael Richardson Sep 30 '15 at 14:46
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I'm afraid that is not possible.

In the example where you state that the moon is 1kg, the marine would weight what? 70, 80 or 90 times more? then the marine would be the one to attract the moon, not the other way around (of course none of the two would apreciatly attract each other). If you go adding weight to the moon you will add gravity, but don't forget the gravity of the planet. If you for some strange reason reach a point where the planet would attract the marine (just jumping and not using external devices to escape moon gravity) then that same gravity would have attracted the moon to the planet long ago and no moon should exist.

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