# Controversy over heights achieved at moon jumpings

1. Jul 15, 2009

### eroock

There was an article in a German online magazine debunking the theories of moonlanding deniers. One of the issues was revolving around the question why the astronauts didn't seem to reach the heights expected when trying to jump off the ground.

The logic: Since moon's gravity is 1/6 of earth's they should be jumping as high as 6 times of what they would be doing on earth. So if we assume them to jump 20 cm on earth (suit included) they should effortlessly be doing 1 m on moon's surface which apparently wasn't happening.
Now the counter argument of the magazine was that these folks were confusing mass with gravity and gravity is not the only factor in the equation. While jumping you first have to overcome initia which refers to an objects mass and doesn't change whether you are staning on earth or the moon. They didn't get deeper into the topic as it was a non-science acticle.

I wanted to know if there's some truth to it and researched the net. What I found is this formula to calculate the height of a jump:

Hmax = (Ftake-off power / (m * g) - 1) * hacceleration distance

with
F = 1,800 N
m = 160 kg
g = 9.81 (earth) and 1.62 (moon)
h = 0.3 m (downward bending of your knees)

I arrive at a ratio of even 1:10 when inserting g for earth vs. moon. Unfortunately I cannot cite an English source for the equation above but I could go into more details if need be. For the time being, what is your general take on the mass/gravity impact on the experiment.

Thanks,
Edgar

2. Jul 15, 2009

### cepheid

Staff Emeritus
Welcome to PF Edgar!

I'm not going to comment on your equation, but to me it seems generous to say that one could jump nearly a foot on Earth while wearing a bulky astronaut suit. If one jumped as hard as he could, then sure, maybe. But can you think of a reason why an astronaut would jump upwards as hard as he could on the moon?

I imagine that many of the hops that were seen on the moon were the result of an attempt at a normal forward motion and would not have even been hops if the same movement had been carried out on Earth.

3. Jul 15, 2009

### mgb_phys

I think it's rather more the case that they were trying NOT to jump too high.
They were (mostly) trying to get work done and floating around in mid (not)air waiting to come back down isn't terribly efficent.
The suits were also clumsy and difficult to balance in - you don't want to fall over and possibly damage the equipment you are setting up or risk the suit or hitting the face plate on a rock.

Last edited: Jul 16, 2009
4. Jul 15, 2009

### Staff: Mentor

E=mgh should be enough... energy of the jumper is identical to the work done when jumping and you can assume it is identical in both cases - that will give you ratio of heights.

However, this assumption (about identical work) - is wrong. Just because they could jump high doesn't mean they did. As far as I remember they have trained how to move on the Moon not jumping too high for safety reasons. You don't want to fall on the Moon and risk hole in your spacesuit.

Edit: to similar answers give while I was composing my post... so I will add that as far as I remember one of the astronauts has actually fallen down whe doing some stunt (golf shot? or was is something else?) - and nobody was happy about that.

5. Jul 15, 2009

### cepheid

Staff Emeritus
Since all three of us are in agreement, let me now play devil's advocate. Leaving likely astronaut behaviour aside for a minute, isn't this assumption also wrong for another reason?

If an astronaut weighs less on the moon, then doesn't this mean that if he exerts the same effort during a jump on the moon as he does on the Earth, the NET force propelling him upward during this type of jump is actually greater (due to his reduced weight)? Therefore, won't he actually do more work executing the same type of jump, and therefore have a higher launch speed and reach a max height that is higher by more than a factor of six?

Again, this is just a thought experiment. I'm not saying that I would have expected to see the Astronauts reach heights > 1 m in practice. We have all given good reasons why this probably did not happen.

6. Jul 15, 2009

### diazona

The reduced weight would also correspond to a reduced normal force, though. And maybe those two effects would cancel out?

I'm having a hard time thinking through this properly to decide whether they would or not.

7. Jul 15, 2009

### Staff: Mentor

No idea how muscles behave in such situation.

Assuming they work just like a spring - amount of energy stored before jump is constant, so the height depends only on the energy conservation.

This assumption is probably wrong, but I have no better idea how to approach the problem.

8. Jul 15, 2009

### pallidin

The suits are specially designed. They are not meant for rigorous activity.
So, there are some activity constraints.
Make any sense?

9. Jul 15, 2009

Staff Emeritus
"Right idea, Mr. Bond." "But wrong pussy?"

The normal force is what it is. If you push down on a moon-sized body with your legs, the moon will push back up with all the force you can provide. And that's the thing everyone has neglected - nobody has asked "what happens when one jumps?"

When you jump, you bend your knees, and rapidly straighten them. As you are straightening, your muscles are pushing down on the ground, and the normal force is pushing up, and that force accelerates your center of mass upwards. So long as you are touching the ground, that force continues to accelerate you upwards, which is why jumpers rotate their feet to keep the ground in contact with their toes as long as they can.

Imagine a gym on the moon. The height of the jump is given by conservation of energy: mgh = Fd, or h = Fd/mg. When one goes to the moon, g is 1/6 as big, but d is also smaller, because the moment when one's feet leave the ground is sooner. So instead of h being 6 times as large, it's maybe only 2 or 3 times as large.

Now, instead of a gym, put on a bulky suit. The A7L weighed about the same as an astronaut. So m doubles, so now the maximum jump height is only slightly above what one could do in gym shorts on the earth.

10. Jul 16, 2009

### Staff: Mentor

d is the distance?

And if I understand correctly you mean that astronaut was fast enough to leave the ground and loose contact with it before his legs were straight... That means other assumptions - about the speed at which legs get straightened at knee joint. I don't think we can move ahead without some hard data about mechanic properties of the system built of muscles/joints/tendons/bones.

11. Jul 16, 2009

Staff Emeritus
Yes.

You're right that one needs more information to come up with a more accurate estimate. F, for example, is not a constant, it's a function of time. And while F(t) on the earth is probably pretty close to the optimal for 1g, it may not be optimal for the moon: I can at least imagine better profiles: pushing up just hard enough to keep your feet on the ground until you are almost fully extended and then pushing as hard as you can.

That said, I can come up with some half-baked arguments - estimates, really - that the height per unit mass goes as one over the square root of the force of gravity, which suggests astronauts jumping about 20-25% higher than they would on the earth. That's not too far off from what we see on films.

12. Jul 16, 2009

### eroock

Hi guys,

thanks so far for the input. Let's try to sort this thing out...

Admittedly, austronauts are not the best example to prove or disprove my point. As mentioned before, there's unwanted problems like
- safety issues
- underperformance due to regressed muscle tissue
- lack of firm ground

So better let's go with the suggested thought experiment of a person in a gym (earth vs. moon base). We assume the person is fully trained, wearing casual clothes and has unlimited headspace for the jump. Also we assume that the leap is performed from a crouch at a constant F throughout the upwards movement while the muscle contraction speed is not affected by gravity at all. F is known.

Under this premise my big question would be:
Is there a straight relation between gravity and the height you can jump, suggested by a simple formula like h = F/mg (g being the only variable). Common sense will tell you this.

Now, from what I read up on the internet the issue might be more complicated than one thinks. The argument:
When performing the jump, the leap height depends on the upward speed v0 of your body at the time your legs are fully stretched (and you leave the ground). The claim is that in lower gravity (lower weight Fg) you gain lower speed. I don't actually get that idea because to me lower weight means higher a:
Realized a is yielded by leap power F reduced by the counteracting weight Fg divided by m:

a = (F - Fg) / m

with Fg = m * g

a = F / m - g (so a lower g means higher a, right?)

I also recommend checking those websites for more insight
http://www.nyskies.org/articles/pazmino/moonjump.htm
http://www.moonhoax.lipi.at/leaps.htm [Broken]

The second one has an example with a weightless hammer. I don't exactly get how this plays into the whole problem but maybe somebody here can put this puzzle together for me - my head is spinning, need some rest =)

Last edited by a moderator: May 4, 2017
13. Jul 16, 2009

### Lok

The first equations i don't really understand as if m too big then value gets negative so no more analyzing that.

The whole problem is as always energetic. if a person can jump 0.3 m on earth than on the moon he will jump 9.81 / 1.62 = 6 times higher that is 1.8 m ( not much ) but he has a doubled mass so the final height is 0.9 m which is even less than interesting ( if I am not wrong somehow )

Eearth=mgh = 235.44 J ( m=80kg)

hmoon= Eearth/2*m*gmoon = 0.9 m ( 2*m -> means twice the mass )

I think they reached this value on the moon if I remember correctly.
How much did they jump in the films anyway ...

14. Jul 16, 2009

### Bob S

mgv phys might have the right suggestion. A slight error in jumping posture could result in some unintended angular momentum, and with gravity being only 1/6 as strong and air time longer, could result in an unintended somersault, or worse. Remember, the astronauts had very little training in moon jumping.

15. Aug 3, 2009

### eroock

For whom it may interest, I found a thourough analysis of the problem done by a scientist, Sigrid Thaller, posted in https://www.blogger.com/comment.g?blogID=3722233&postID=117313251765556787&pli=1"

"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 www.uni-graz.at/sigrid.thaller/Model.htm [Broken] .
There is a simulation program of a squat jump, but unfortunaltely it is only in German (www.uni-graz.at/sigrid.thaller/modelling.html [Broken]).

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."

Last edited by a moderator: May 4, 2017
16. Aug 4, 2009

### eroock

Sorry, but the entire thread was to disprove the 1/6 relation.
A spring for example follows Model 2 above and would jump approx. 11 times as high on the moon.