Monday, March 05, 2007

Jumping in Space

A fun fact from a McDonald's Happy Meal bag.
You can jump six times higher in space.
What does "jump" or "higher" mean in space? Given the pictures on the bag I believe what they meant to say was
You can jump six times higher on the surface of the moon than on the surface of the earth.
Ask the Astronomer agrees since the ratio of gravity on the moon and the earth is about 1/6th. Is that correct? Not quite. In a 1973 Physics Teacher note, Van Neie fixed the mass M of a person and the force F the person exerts to get a height ratio of
(6F/Mg -1)/(F/Mg-1)
where g is the gravitational constant. This does approach six in the limit but only "if the force F is several times the individual's Earth Weight, an unrealistic assumption." If a person exerts twice his earth weight when he jumps, he will jump 11 times higher on the moon.

See what you can learn eating at McDonald's.


  1. Both may be "right". "Ask the Astronomer" fixes the energy expended while van Neie fixes the force exerted. I'm not sure which is the more appropriate assumption.

  2. anonymous:

    Given the assumption that the energy expended is identical, Ask the Astronomer is "correct," but this assumption is unreasonable.

    Since a jump on the Earth and the moon would go through roughly the same physical motion (since your legs are the same length in both places), the only way to affect the energy of a jump is to vary the force exerted during that motion. If the energy (and thus the net force) really is the same in both places, that means that a jumper on Earth needs to expend an extra force of 5/6 of their own body weight to get the same "net force" (and thus the same energy when they leave the ground).

    This much extra force is a lot: try holding a 100lb weight, and see how it affects your jumping height on Earth. Unless you are a serious weight lifter, you will be lucky to get off the ground at all (Note that this is a thought experiment, and I don't advocate anyone actually trying this, for obvious reasons).

    Now, on the other hand, we might reasonably expect a slight decrease in height on the moon versus the naive energy calculation (i.e. Lance's 11-times height, which is a reasonable approximation): because the speed of motion on the moon will be much higher, there may be a bottleneck due to the speed of muscle motion required, rather than the actual force. But (unless an anatomist can correct me?) it seems unlikely that this difference would be nearly enough to make up the 6x-to-11x gap, and thus "slightly less than 11x" is probably the closer estimate. (I'm basing this unlikeliness on the aforementioned thought experiment, which suggests that our jumps are primarily force- rather than speed-constrained.)

  3. I think what you need to consider is not "net force" but rather impulse. The duration of your contact with the ground would be much, much shorter on the moon. Also, as you mentioned, the bottleneck becomes the force your muscles are able to exert at speed (how fast they are able to accelerate), which is significantly less than at rest. There's a huge difference between "maximum force output" and "maximum speed" (look up plyometrics), and it varies widely from person to person.

    So obviously, neither assumption is very reasonable because of the differences in anatomy from person to person. If I had to make a bet, though, I'd say that you end up closer to the 6x end rather than 11x.

  4. Did you include the weight of the space suit in the calculation?

  5. anonymous 3:

    "The duration of your contact with the ground would be much, much shorter on the moon."

    Yes, that's why we're using energy calculations, which are only affected by the distance through which a force acts, and not the time -- the distance your legs move is roughly the same in both places, as I said, and so the energies can be compared directly by considering only the force exerted.

    I'm not sure how you're using "impulse" in a way that is different than force. Force is the instantaneous pressure being exerted at each moment, not the total energy.

    "There's a huge difference between "maximum force output" and "maximum speed""

    Of course... but the typical person (perhaps not the typical bodybuilder) on Earth is more force-constrained in their jumps than speed-constrained, so at least initially we expect their jump height to increase much more than a direct inverse of the gravity would suggest. I agree that may not carry over to a full 11x increase on the moon, but I doubt people are as close to being speed-constrained as you're saying...

    "So obviously, neither assumption is very reasonable because of the differences in anatomy from person to person."

    You haven't taken enough introductory physics classes. Obviously, the people involved in this problem are all spherical of identical radius :)

    anonymous 4:

    My hypothetical people were jumping from within an air-sealed permanent moon base, so a suit was unnecessary.

  6. why dont we just send someone up there and see how it turns out??

  7. David, thanks for the condescension. Thanks also for missing my point.

    My point was that your force output as a function of distance, F(d), would not be the same as on the Earth, because the duration of you pushing on the ground would be a lot shorter on the moon than on the Earth. This is because of the inability of muscles to accelerate very quickly -- "max force" vs. "max speed". To go back to your original thought experiment: while on the Earth you strap on an additional 100 lbs and try to jump. You'll find that F(d) is approximately constant over d (not quite, but let's fudge here). Now you try to jump on the moon without the extra weight. F(d) would drop significantly as you progress through your jumping motion. The amount that it drops depends on the condition of your muscles -- the plyometrics that I mentioned before -- their ability to accelerate.

    Therefore it really doesn't make any sense to say that integrating F(d) over d on the Earth would be equal to integrating F(d) over d on the moon. That's why holding energy constant is bogus -- most people just cannot accelerate their legs that quickly!

    I brought up impulse just to illustrate the importance of the duration of contact with the ground. Yes, I am aware that force is instantaneous pressure, not energy. I am also aware that impulse equals change in momentum.

    Anonymous #6: If by "up there", you mean "to a Hollywood movie studio", then sure.

  8. Whoops, David, sorry. What I thought was condescension was really not. I read through your post too quickly.

    "You haven't taken enough introductory physics classes. Obviously, the people involved in this problem are all spherical of identical radius :)"

    I originally read that as a put-down, but now re-reading it I understand your point.

    Here, why don't you try this: step away from the computer and go down into a squatting position. Then try to jump as high as you can from there. If you do things correctly, you should be off the ground before your legs are even fully extended. Also notice that most of the force exerted happens relatively early on.

    Now go pick up something heavy (say, 50 lbs). If you are able to get off the ground, notice that you are exerting more force over a longer distance (you still get off the ground before your legs are extended, but you're pushing more throughout the entire jumping motion). It's pretty clear from this that the amount of energy expended (integrating F(d)) is not equal.

    Also if you doubt that speed-constraint is too much of an issue, find a basketball player and a bodybuilder of about the same weight and compare how high each can jump. Basketball players train in order to build up the ability of their muscles to accelerate quickly for the specific purpose of being able to jump higher.

  9. The problem is discussed in detail, but unfortunately in german, in the article by Sigrid Thaller 'Biomechanik - Hochsprung auf dem Mond' ('bio-mechanics - high jump on the moon')

  10. I can't read German, but from the graphs it looks like it's saying what I said. Anybody know what the paper concluded?

  11. Addendum: I plugged it into babelfish and I pulled out some of the reasonably intelligible sentences.

    "If one considers however the different conditions of work muscles, then one encounters unexpected effects... If one draws only the forces and energies in consideration, as one would do it with a spring, then the thing is clear... [We?] want to consider however in the following one physiological effects... In the previous model we assumed that the strength exerted by the muscle on the soil is constant. A muscle produced however during a movement not always the same strength, surprisingly decrease they separate with increasing speed of the movement. The exact connection between strength and speed hangs of the individual muscle characteristics and with the Hill equation is described (see "the force development in the muscle", S. 89)... At the beginning muscle power is large in accordance with the Hill equation, because the speed of the movement is still small, however works the control of the muscle not yet completely, and the knee angle is also unfavorable. In the course of the jump conditions change: Muscle power becomes smaller, but the joint angles become more favorable for it. At Institut for sport sciences at the University of Graz we accomplished computer simulations of jumps in different gravitational fields [ 2, 3 ]. Through such investigations expects one explanation about it to gotten, which muscle characteristics astronauts on a longer space flight must train, in order to be optimally operational thereafter. The computations at the computer led to the following surprising results: A general statement about it, around like much the jump height on the moon is larger than on earth, is not possible! It depends substantially on the individual characteristics, as highly a person can jump. It can even occur that the sequence in the jump achievement turns around: Better Springer on earth does not have to be necessarily better Springer on the moon (table 1)...

    Summary: Around as much more highly humans on the moon than on earth jump, depend substantially on the individual muscle characteristics. A movement is determined by a complex interaction by muscle power, speed, control muscles and geometry of the joints. Therefore the size of the effects of a change of gravitation depends again on the individual characteristics. It can come even to a reversal of the sequence in the jump achievement. Then the better one of two Springer on earth on the moon is the worse."

    Conclusion: Both models are bogus.

  12. Second addendum:

    Hill's Equation

  13. I am very happy to see my paper discussed on this web page. As the translation by an automatic translating program is not so good I would like to summarize my paper “Hochsprung auf dem Mond”, Physik in unserer Zeit 34/2 (2003),87-89.
    There are three models of the jump on the moon (or mars) discussed. The first 2 models obviously are very simple and one can easily see why they are incorrect. Also the 3rd model is simple but it includes some of the physiological properties of muscles and the geometry of the joint.
    Model 1: Assume that at take off the legs are streched completely (which is not true in reality) and that the take off velocity v is the same on earth and on moon. Then we get for the height h of the center of mass: h = (v^2)/(2*g), g being the gravitational acceleration. (h = 0 at take off). Therefore, the height on the moon is 6 times higher than on earth. In this model the energy decreases with decreasing gravitation, i.e., on the moon the body needs less energy to jump (the kinetic energy at take off is by assumption the same and the potential energy at take off depends on the gravitation).
    Model 2: Assume that the force exerted on the ground by the muscle is constant and independend of the gravitation (which is not true in reality because at the beginning while standing in squat the force is exactly the weight of the person, then it increases, and decreases till take off where it is zero). In this model the energy is the same on earth and moon ( the same force applied on the same way (squat heigth to length of the legs)), the take off velocity on the moon is greater, v_take off = sqrt((f-mg)*2*(s_take off – s_squat)/m), and the heigth is approximately 11 times higher on the moon as on earth (assuming f being 2mg).
    Model 3: In this model we use the velocity dependence of the muscle force (Hill’s equation f_m = c/(v_m – b) – a, a,b,c denoting constants describing the properties of the musle in regard, f_m being the muscle force and v_m being the contraction velocity. In addition, we have to consider the activation of the muscle. It takes some time until the total muscle force can be used. We describe this by a function of time, S(t) = (1-exp(-A*t)), A being a parameter describing the speed of activation (which is constant and a property of a person’s muscle). Furthermore, we have to consider the geometry of the joints. For the same force outside you need different forces in the muscle, depending on the knee angle. If you are standing in a deep squat you need more muscle force than standing with straighter knee. These considerations lead to a highly nonlinear differential equation for the movement of the center of mass. The velocity at take off is greater on the moon, the jump height is higher (as expected ;-)), and the energy is less. The complete duration of the jump is longer on the moon, but the time until take off is shorter. Therefore you cannot pretend a moon jump in a movie by using slow motion. The height of the jump depends on the person, because the constants in Hill’s equation and for the activation and geometry are different for different persons. For example, for persons with slow activation the jump on the moon is so fast that there is take off before the muscle is fully activated. It may happen that if one of two persons jumps higher on earth he/she need not jump higher than the other person on moon.
    Additional question: What happens if an astronaut has so much equipment to carry with that his weight on the moon is the same as his weight on earth without equipment? Will he still jump higher? Model 3 says: he jumps higher on the moon, the take off velocity is less on the moon and he uses more energy on the moon.
    The equations of model 3 can be found on .
    There is a simulation program of a squat jump, but unfortunaltely it is only in German (
    Model 3 is still a very simple model. It describes only the center of mass and it does not consider the changes of muscle coordination caused by the changes of the movement conditions.
    I hope I could answer all questions.

  14. Children on Earth with less weight than adults have faster movements than adults it seems. So then why did the men on the moon move as in slow motion? With less weight on their legs and arms they should move faster NOT slower. Is this another evidence that the Moon landings were a FAKE?? If I am able to jump much higher on the moon will I land on an appreciable longer amount of time??

  15. To be clear on my question before: If I can jump 10 feet high on the Moon and if I were able to jump the same ten feet on Earth, will an observer who did not know where I was, be able to tell where each jump takes place by noticing the time it takes for me to land?