How high can you jump on the Moon?

[theteacher]

It has been asked a couple of times, if you can actually jump 6 times as high on the Moon as on Earth, and some have asked for definitive answers in the form of equations to prove the given fact.

My answer is: You can jump much higher than 6 times what you can jump on Earth, and though you are so heavy, that you cannot possibly jump on Earth, you will still be able to jump on the Moon. In both instances the 6 times notion is irrelevant.

I have made an attempt to figure out how high one can jump on the Moon under different circumstances supporting the answers with "equations", though I will not guarantee that the answers given here are definitive :-)

So: We use an average test person 1.80 meters high weighing 75 kg jumping 0.5 meters vertically, which is a little above average. We assume the individual squats to parallel thighs, which yields an acceleration path of approximately 0.4 meters. We also assume that the acceleration is uniform, which is of course an idealization. Calculations are in metric (SI) system.

It is necessary to calculate the jump in two different phases: The acceleration phase and the free-flight phase. In the acceleration phase there are two forces on the person acting in opposite directions: The propelling force from the muscles and gravity. In the free-flight phase there is only one force: Gravity. These principles are the same on the Earth and on the Moon.

Kinetic energy at take-off = potential energy at peak height = work done.

On Earth



We now move the experiment to the Moon

On the Moon approximately 5/6 of the gravity is removed. This has consequences for both the acceleration phase and the free-flight phase. To begin with the removed gravity works as a contribution to the upward acceleration:



So the test person can actually jump 10 times as high on the Moon as on Earth.

But why can't the astronauts do that then?

We suit the test person up as an astronaut on the Moon

The Apollo spacesuit with PLSS weighs around 85 kg. The astronaut will then have 75 + 85 = 160 kg to accelerate. Furthermore the suit is restrictive, so he is not able to squat to parallel thighs. We assume that his acceleration path is only 0.2 m, but we assume that he exerts the same force, namely the 920 Newton.



This jump height corresponds fairly well with what we see as absolute maximum jump height on video from the Moon as far as I can tell, though it is probably even too optimistic. The very high jump that Armstrong performed, was done with the aid of a grab in the railing of the ladder, so it doesn't really support a claim of the possibility of the astronauts being able to jump considerably higher than calculated here.

What do you think?

[ka9q]

I had always based the 6x figure on your muscles imparting equal amounts of kinetic energy to your body at the moment your feet leave the ground, but I suppose that's not necessarily a good model. Come to think of it, it's possible your leg muscles are more or less efficient when working against different amounts of resistance, and that may be an even larger unmodeled factor.

I suppose one could look at your analysis as analogous to the gravity loss that affects rockets taking off vertically from a planet or moon; some of the thrust is used merely to overcome gravity so less is available to accelerate the rocket upward.

[ka9q]

I think a really complete analysis requires more than basic physics; it requires an understanding of physiology that I don't have.

A somewhat different way to approach the problem is to see that during the jump on earth, your legs are applying not only the force that is accelerating your body upwards, but also the force to support it against gravity. If you're accelerating 75 kg of body mass upwards at 12.26 m/s^2 over that 40 cm, your muscles are having to generate an additional 75 kg * 9.8 N/kg = 735N just to support your weight, for a total of 920 + 735 = 1655N. Now if your leg muscles could generate the same 1655N of upward force during a jump on the moon, with only 75 kg * 1.63 N/kg = 122.5 N of weight to overcome, 1532.5N of force is available for the upward acceleration. This would last until your legs straighten after 40 cm, for an imparted energy of 1532.5 N * 0.4m = 613J of upward kinetic energy, the same number you get that turns into a 5m altitude.

But that's probably a rather simplistic view of how muscles and legs work. A more sophisticated model would take into account the lengths, masses and moments of inertia of your upper and lower legs, the forces produced by each of the muscles, the decreasing weight components supported by the muscles as your legs straighten, shifting gravity to a compressive force, and and the heights that each of those masses are lifted and/or rotated. It could get pretty complicated.

[trebor]

There is a published paper available as well.

[theteacher]

Nov 23, 2010 4:04:12 GMT -4 ka9q said:

I had always based the 6x figure on your muscles imparting equal amounts of kinetic energy to your body at the moment your feet leave the ground, but I suppose that's not necessarily a good model.

Would you stretch it so far as to saying that it is necessarily not a good model?

Come to think of it, it's possible your leg muscles are more or less efficient when working against different amounts of resistance,...

Yes. I guess you're definitely right here. The shorter time, you have to make the acceleration, the more power (Watts) must be exerted to convey the same energy. Therefore the lower the gravity or the lighter the person - and vice versa - the less valid the model. One could get the thought that Earth gravity is the gravitational environment, where the most power can be exerted when it comes to jumping, simply because the human body through evolution has been optimized to that environment.

... and that may be an even larger unmodeled factor.

Larger than?

In this respect the crucial figure must be the take-off speed. On Earth 3.13 m/s, on Moon 4.04 m/s, on Moon suited up 1.52 m/s, but would have been 2.14 m/s with same mobility as in track uniform.
Take-off speed on Earth and Moon under same conditions apart from gravity are not that different.

I suppose one could look at your analysis as analogous to the gravity loss that affects rockets taking off vertically from a planet or moon; some of the thrust is used merely to overcome gravity so less is available to accelerate the rocket upward.

I'm not much into rocket science, but I get your point.

[theteacher]

Nov 23, 2010 12:49:16 GMT -4 trebor said:

There is a published paper available as well.

Online or elsewhere?

[theteacher]

Nov 23, 2010 4:32:22 GMT -4 ka9q said:

I think a really complete analysis requires more than basic physics; it requires an understanding of physiology that I don't have.

Absolutely. But basic physics is what I can come up with :-)

A somewhat different way to approach the problem is to see that during the jump on earth, your legs are applying not only the force that is accelerating your body upwards, but also the force to support it against gravity. If you're accelerating 75 kg of body mass upwards at 12.26 m/s^2 over that 40 cm, your muscles are having to generate an additional 75 kg * 9.8 N/kg = 735N just to support your weight, for a total of 920 + 735 = 1655N. Now if your leg muscles could generate the same 1655N of upward force during a jump on the moon, with only 75 kg * 1.63 N/kg = 122.5 N of weight to overcome, 1532.5N of force is available for the upward acceleration. This would last until your legs straighten after 40 cm, for an imparted energy of 1532.5 N * 0.4m = 613J of upward kinetic energy, the same number you get that turns into a 5m altitude.

Yes, that might be a more correct/fundamental approach to the physics involved.

But that's probably a rather simplistic view of how muscles and legs work. A more sophisticated model would take into account the lengths, masses and moments of inertia of your upper and lower legs, the forces produced by each of the muscles, the decreasing weight components supported by the muscles as your legs straighten, shifting gravity to a compressive force, and and the heights that each of those masses are lifted and/or rotated. It could get pretty complicated.

I have based my calculations on a simple empirical approach, where I have calculated "backwards" from a table value taken from this site, and I chose 0.5 meters out of convenience. It could of course be more or less. And I am not convinced, that the physio-kinesiological analysis is relevant. When the jumper leaves the ground at a certain speed, it is no more important how he reached that speed. The whole body moves at a certain speed independent of the complicated body movements, that led to it - I think.

Maybe I should have posted in a hoax thread instead, because the point I have been trying to make is, that 6 times is definitely wrong and far too less. So when HBs say, that you should be able to jump 6 times as high, I guess it has a surprising effect to go with them and say that it is much more than that, and then WAIT - and see, how they react. On that background it may be easier to explain, why the astronauts can't do that - because you have their attention.

[trebor]

Nov 24, 2010 18:29:39 GMT -4 theteacher said:

Nov 23, 2010 12:49:16 GMT -4 trebor said:

There is a published paper available as well.

Online or elsewhere?

It is online, although I don't have the link to hand.
I do have a pdf of it saved and can send it to you if you want.

[theteacher]

Nov 26, 2010 4:42:41 GMT -4 trebor said:

Nov 24, 2010 18:29:39 GMT -4 theteacher said:

Online or elsewhere?

It is online, although I don't have the link to hand.
I do have a pdf of it saved and can send it to you if you want.

If you will have the trouble I'll be grateful. Thanks.

[ka9q]

Nov 24, 2010 18:54:27 GMT -4 theteacher said:

And I am not convinced, that the physio-kinesiological analysis is relevant. When the jumper leaves the ground at a certain speed, it is no more important how he reached that speed.

But that's assuming a lot -- that the body leaves the ground at a certain speed, isn't it?

I don't know much about physiology, but it seems to me that any given muscle can produce a given maximum force, sustaining it for some maximum period of time until the ATP (the internal energy store) is depleted and has to be replenished.

So estimate those forces and apply them to a system of masses and levers that model the jump. As you reduce gravity, you spend less of the muscle force overcoming it and more accelerating the body upwards, probably leaving the ground before the ATP is depleted. So I would expect the jump energy to be a nonlinear function of gravity.

[tedward]

Must admit this is above me but looking at the arrangement of muscles and bone and where they appear to be attached, evolution has gone for practical rather than efficient. Or should it be the overall design is as efficient as it can be, at least as in as far as we have got. There would also be a difference in what the same muscle can achieve in a mix of people of a similar size?

[theteacher]

Nov 28, 2010 5:43:54 GMT -4 ka9q said:

Nov 24, 2010 18:54:27 GMT -4 theteacher said:

And I am not convinced, that the physio-kinesiological analysis is relevant. When the jumper leaves the ground at a certain speed, it is no more important how he reached that speed.

But that's assuming a lot -- that the body leaves the ground at a certain speed, isn't it?

It depends of what you assume that I in fact assume :-)
For a given person at a given level of gravíty, the take-off speed will determine the height of the jump. Alternately the time of free-flight will also determine the jump height, as it is one of the ways to measure jump heights accurately, as it can be done with the aid of a timing mat. I guess it's possible to set up a series of tests simulating different levels of gravity and thus work on a strictly empirical basis like for instance the tests done at Langley.

I don't know much about physiology, but it seems to me that any given muscle can produce a given maximum force, sustaining it for some maximum period of time until the ATP (the internal energy store) is depleted and has to be replenished.

I guess we can count ATP-depletion out, as we talk about one single contraction of the involved muscles.

So estimate those forces and apply them to a system of masses and levers that model the jump. As you reduce gravity, you spend less of the muscle force overcoming it and more accelerating the body upwards, probably leaving the ground before the ATP is depleted. So I would expect the jump energy to be a nonlinear function of gravity.

Yes, I agree and I would expect that too. But I didn't expect the non-linearity to have the great effect it obviously has.

I have studied the film clip from Langley, that trebor provided along with the report, he was kind enough to send to me, and the findings herein contradict my calculations. 6-7 times and 10 times are far apart, so it is obvious that my simple model is not usable for the purpose.

I have found this article in "American Journal of Physics" dealing with the vertical jump using a force platform. It might hold a key to the problem, but I'm into other matters at the moment, so maybe some of the experts in physics can dig into it.

I hope though that I have problematized 6 as the conversion factor and shown that an astronaut, who cannot jump on earth, will be able to do so on the Moon.