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I just had this idea of orbiting a planet just by jumping and then flying upon it on its orbit kind of like superman. So,

Would it be theoretically possible or is there a chance of that small body to be & remain its unity?

Orbiting an asteroid by jumping

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I want to see an answer! – m0nhawk Dec 8 '12 at 23:00
Since asteroid comes in all sizes right down to grains of dust the answer is trivially "yes". Now, is there a particular one that has been identified will take a little calculation and going through catalogs. – dmckee Dec 8 '12 at 23:04
THIS PICTURE. +1 – Dylan Sabulsky Dec 8 '12 at 23:13
Of course running on surface of that thing is going to be difficult, especially as you approach the orbital speed getting any friction against the ground is going to be hard. – SF. Dec 8 '12 at 23:22
Other problem: finding a perfectly spherical asteroid. Irregular shapes will make this a lot more difficult ! – FrenchKheldar Dec 9 '12 at 2:40
up vote 48 down vote accepted

Let's assume mass of the person plus spacesuit to be $m_1$=100kg

Asteroid density: $\rho=$2g/cm$^3$ (source) that is 2 000kg/m$^3$

15km/hour is a good common run. That's roughly v=4m/s

The orbital height is negligible comparing to the radius, assume 0 over surface.

Linear to angular velocity (1): $$ \omega = {v \over r } $$ Centripetal force (2): $$ F = m r \omega ^2 $$ Gravity force (3): $$ F= G \frac{m_1 m_2}{r^2} $$ Volume of a sphere (4): $$ V = \frac{4}{3}\pi r^3 $$ Mass of a sphere (5): $$ m_2 = V \rho = \frac{4}{3}\pi r^3 \rho $$ Combining (1),(2),(3), reducing: $$ { m_1 r v^2 \over r^2 } = G { m_1 * m_2 \over r^2 } $$ $$ r v^2 = G m_2 $$ Combining with (5) $$ r v^2 = G \frac{4}{3}\pi r^3 \rho $$

$$ r^2 = \frac{v^2}{\rho G \frac{4}{3}\pi} $$

$$ r = v ({\frac{4}{3}\pi G \rho})^{-{1 \over 2}} $$ Substituting values: $$ r = 4 ({1.33333*3.14159* 6.67300*10^{-11} * 2000})^{-{1 \over 2}} $$

That computes to roughly 5.3 kilometers

More interestingly, the radius is directly proportional to the velocity,

$$ r[m] = 1.337[s] * v [m/s] = 371.51[h/1000] * v[km/h] = 597[m*h/mile] * v[mph] $$

So, a good walk on a 2km radius asteroid will get you orbiting.

Something to fit your bill would be Cruithne, a viable target for a space mission thanks to a very friendly orbit.

Note, while in rest on Cruithne, the astronaut matching the m_1=100kg would be pulled down with force of 4.5N while not in motion. That is like weighing about 450g or 1lbs on Earth.

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While this is a great answer (+1), it should be pointed out that walking would be very difficult (perhaps impossible?) on a body with such low gravity, and that makes it hard to imagine how you could get up to 15km/hour in order to make the jump. – Nathaniel Dec 9 '12 at 0:52
@Nathaniel: Probably with help of a small (handheld?) jet propulsion engine. You'd want some means of exiting the orbit anyway. – SF. Dec 9 '12 at 2:17
Shouldn't it be 371.51 mph and 597 km/h? – Eugene Seidel Dec 9 '12 at 10:50
@SF.: According to the "Shell Theorem", any spherically symmetrical body is equivalent to a point source of the same mass, if you are outside of it. (Inside, the apparent mass of the point source changes.) – Dietrich Epp Dec 9 '12 at 14:09
@Beska: You can always use it to keep yourself pressed to the surface instead, and just cut it off when you're running fast enough. – SF. Dec 9 '12 at 16:01

No, not by jumping. Jumping gives you an acceleration only from the location on the surface. As soon as you leave the surface, you have no way of adjusting your orbit. Either you reach escape velocity, or you will return to your initial location after exactly one orbit.

enter image description here

The only way to prevent this would be to have an additional acceleration once you have departed from the surface. Spacecraft use rockets to do this. A tiny acceleration may be enough — though I wouldn't like approaching a planet with high speed only to move 5 cm over its surface with high speed!

Edit: A different way would be jump from a ladder, as Claudius pointed out in the other answer.

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Another approach: Take a rock with you, and throw it directly backwards when you're half way around. This should give you enough extra velocity so your orbit doesn't intersect the surface at your starting point. – Keith Thompson Dec 8 '12 at 23:57
You don't necessarily come back to your original location after your jump. Any other ballistic trajectory is possible if your speed is below the escape velocity. – FrenchKheldar Dec 9 '12 at 2:39
@FrenchKheldar - true, but I think that's a quibbling objection. Assuming it's a central force you would return to your starting point if the planet didn't get in the way. – John Rennie Dec 9 '12 at 9:47
+1 for this pic as well :) – Nix Jul 9 '13 at 10:44
"though I wouldn't like approaching a planet with high speed" — but if you ran/jumped to start on your orbit, then that's how fast you'll touch back down, so it's not like you'll actually be moving at a very high speed upon touchdown. It's physically impossible to end up moving faster than that if the surface is a perfect sphere. – romkyns Feb 2 '15 at 22:48

OK, I tried to do the math here. Something remotely resembling maths, at least.


  • It is possible to reach an orbital/horizontal speed of $v_O = 5\textrm{ ms}^{-1}$, for example by running.
  • The density of the object to orbit is similar to Earth's density, i.e. $\rho = 5500\textrm{ kgm}^{-3}$.
  • We want to orbit at a height of $2\textrm{ m}$ above the ground. You can get there with a ladder (Yes, you will have to start running on that ladder or something like about stilts?).
  • No atmosphere or other source of friction.


The basic idea is to link the orbital velocity $v_O$ to the radius $r$ of the object. The mass is given by $ M = \frac{4}{3} \pi r^3 \rho$ (God I hope I remembered this formula correctly).


We have

\begin{eqnarray} & v_O & = \sqrt{\frac{G M}{r+2\textrm{ m}}} = 5\textrm{ ms}^{-1} \\ \Rightarrow & M & = \frac{25\frac{\textrm{m}^2}{\textrm{s}^2} \left( r + 2\textrm{ m} \right)}{G} \\ \Rightarrow & 25 \frac{\textrm{m}^2}{\textrm{s}^2} r + 50 \frac{\textrm{m}^3}{\textrm{s}^2} & = \frac{4}{3} \pi G r^3 5500 \frac{\textrm{kg}}{\textrm{m}^3} \end{eqnarray}

which then should give us $r$. I used Mathematica for this because it is half past eleven in the evening and I don’t want to guess solutions to get a starting point for polynomial division, getting:

In:  Solve[-4/3 * Pi * 6.67384*10^(-11) * x^3 * 5500 + 25 x + 50 == 0, x]
Out: {{x -> -4031.33327417391}, {x -> -2.00000049201392}, {x -> 4033.33327466592}}

That is, if you found an asteroid of $r \approx 4\textrm{ km}$, your dream might come true. However, if it is mostly ice (rather than molten iron, which I imagine would be a pretty good reason to stay in orbit), you will have to correct the 5500 up there to the density of ice, say, 930, and would then need an asteroid of $r \approx 9.8\textrm{ km}$.

Note that the assumption that $m_{\textrm{Human}} \ll m_{\textrm{Object}}$, encoded in the expression for orbital velocity, is fulfilled relatively well in these cases (five orders of magnitude).

Nevertheless, feel free to point out mistakes :)

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Are any asteroids of this size remotely spherical? – gerrit Dec 8 '12 at 23:37
Beat me to the answer. I get a similar result (about 3.5 km) phrasing the question as "can a single jump from a typical person impart enough energy to reach escape velocity). – Chris White Dec 8 '12 at 23:37
@gerrit No, but asphericity helps if you start on the bulge. – Chris White Dec 8 '12 at 23:38
@gerrit Just build a spherical one. 4 km is not that much :-) – Claudius Dec 8 '12 at 23:38
@Claudius well, that would be awesome! :D if they could place it on of the lagrangian points between moon and earth, it would also be easy to get there for a superman-like vacation :D – Ahmet Yıldırım Dec 8 '12 at 23:56

Since the calculations are already in others' answers, I'll just refer to this great, classic xkcd. Deimos and Phobos, the two small moons of Mars, match (or almost match) the criteria SF and Claudius derive.

As Munroe points out,

xkcd 861

(The diagram is a representation of the gravity wells of both moons, represented by their height at constant Earth surface gravity.)

Based on that I think you really should be able to run yourself into orbit using a smallish ramp and a fire extinguisher to stabilize your orbit on the other side (to avoid the pitfall gerrit mentions).

Deimos is between 10 and 15 km across and its escape velocity is about 20 km/h. At low altitudes, and since circular-orbit velocities are lower by $\sqrt{2}$ than escape velocities, you'd need to run up to some 15 km/h to orbit. Thus you'd do about one lap every three hours, whizzing along this ~city sized object at about bicycle speeds.

Picture of Deimos

On the other hand, it's unlikely that you will last very long in that orbit. The reason for this is that orbits are elliptical only around perfectly spherical planets, and any irregularities in the body you're orbiting will tend to perturb and even destabilize your orbit. Even on the Moon, low orbits are unstable and end up crashing into the surface, as was the fate of a subsatellite deployed during Apollo 16, which lasted only a month in orbit. With something as lumpy as the Martian moons, you would probably want to stay well away!

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+1 for the fire extinguisher. – Claudius Dec 10 '12 at 18:30
Thanks for the answer and graphics :) I'm also a game developer,After I see that this is physically quite possible,I'm quite impressed. Do you think a game based on travelling on asteroids by jumping & collecting stuff etc. would be fun? – Ahmet Yıldırım Dec 10 '12 at 18:53
Of course that would depend on the game and how smooth the handling was, but I'd definitely be quite willing to try it. It reminds me of A Slower Speed of Light somewhat. – Emilio Pisanty Dec 10 '12 at 18:57
I'm thinking, the part you collect bonuses etc. would be in 3d TPS-like view on asteroid. But jumping to another asteroid would need to switch to 2d to make it easier to interact for player. I would ask user to make the jump on a specific position for maximum jump. And after jump it would switch back to 3d and user would control yaw/pan/tilt to land. – Ahmet Yıldırım Dec 10 '12 at 19:05
Also, if you're thinking of jumping from one asteroid to another, then the details of the initial condition are crucially important and if you choose your problem wrong you'll have exponential dependence on them (i.e. meaning chaos, and a spectacularly hard gameplay where a hair's breadth in jump-off point lands you on the other side of the target asteroid). That is one reason the Apollo programme was as much about developing the onboard computer as it was about the rockets. – Emilio Pisanty Dec 10 '12 at 19:10

If you want an idea what this might actually be like have a look at Kerbal Space Program. This is a game currently in development by Squad. So not real life, but the orbital physics is accurately modelled (atmospheric flight not so much, yet). There are several small moons and asteroids in the Kerbin system where you can do essentially this jump to orbit maneuvre using nothing but EVA suit thrusters. You can see examples in some of Scott Manley's videos. Here is a video featuring an interplanetary trip with an EVA suit - a 49 day space walk!

(I'm not affiliated with KSP, Squad or Scott Manley in any way, and since the question has been properly answered already I thought this might just be a fun thing to share. Also, KSP and the similar game Orbiter are good ways of building intuition for orbital mechanics. :) Hope this doesn't break the rules.)

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