The Physics of jumping

[Willoughby]

I know that this is an old topic, but I just thought I would throw my two cents in.  The notion that someone can jump 21 times higher on the moon than they could on Earth is not wrong in theory.  It is reasonable - assuming ALL OTHER variables are exactly the same (they aren't in reality, but for the sake of argument, let's assume they are).  I can demonstrate this, and I apologize for how convoluted this will be, but those saying that the physics doesn't add up are not really correct.  Allow me to show why.

Consider an astronaut with a mass of 80 kg who can jump 18 inches on Earth (0.457 meters) with his full range of motion.  Some force is exerted (on the ground really) for some amount of time.  In other words, imagine the astronaut bends over - ready to take his jump, and as soon as he starts his jump, start this clock.  Stop the clock when his feet leave the ground.  This is the amount of time his force is applied (this will be important later).  Once his feet leave the ground, he has some velocity upward.  High enough that he reaches 0.457 meters.  I've actually timed myself jumping and consistently get between 0.25 and 0.35 seconds for the time here, so I'll just use 0.3 seconds.  First, calculate his "initial" velocity - that is, his velocity the moment his feet leave the ground and a force is no longer being applied.  This can be calculated as follows :

where g = 9.8 m/s² (since we are talking about a jump on Earth here - in Earth's gravity)
h = height

v = √(2gh) = 2.99 m/s

I've estimated a reasonable time for the amount of time a force was applied - this was about 0.3 seconds.  We can then calculate the acceleration :

a = dv / dt = 2.99 / 0.3 = 9.97 m/s²

From this, we can now calculate the force :

F = ma = 80kg * 9.97 m/s² = 797.6 Newtons.

This is actually the NET force (which is probably where a lot of people go wrong when they are trying to calculate jump heights between the Earth and moon).  Obviously, we know that there is a gravitational force on the astronaut as well, so any acceleration is the result of a NET force - that is, this 797.6 Newtons we just calculated is that force which is opposite in direction to gravity, but IN EXCESS of it, so we need to know the gravitational force :

W = mg = 80 * 9.8 = 784 Newtons.

This means the actual force the astronaut exerts on his jump is 1,581.6 Newtons - 784 N of which overcomes gravity, the remaining 797.6 N is the net force upward.

This force will be the same on the moon.  Astronauts do not get stronger or weaker.  Assuming the astronaut jumped as hard as he could on Earth, his hardest is the same on the moon, so we need to know how this force propels him on the moon.  Again, we need to know the NET force on the moon, so how much of this 1,581.6 Newtons is in excess of lunar gravity?

W = mg = 80 * 1.6 = 128 Newtons

So this means that the net force upward on the moon for our astronaut is 1,453.6 N.

How does this net force accelerate our astronaut (I'm not considering other factors - like the space suit - just yet).

a = F / m = 1,453.6 / 80 = 18.17 m/s²

With full range of motion, our astronaut is able to apply his force for 0.3 seconds, so the initial velocity on the moon (again, his velocity the moment the force is removed and his feet leave the ground) is :

v = at = 18.17 * 0.3 = 5.5 m/s

How high will something go in lunar gravity with an initial velocity of 5.5 m/s?

h = v² / 2g = 30.25 / 3.2 = 9.45 meters.

So, this is actually pretty simple physics - just considering all the factors a lot of people leave out, but if you assume all other factors are the same, an astronaut who can jump 0.457 meters (18 inches) on Earth can potentially jump 9.45 meters (31 feet) on the moon.  This is close to 21 times higher as the original comment references.  It's sound physics.  It's just not very good at considering all factors. 

One MAJOR factor not considered here is the extra 90 kg of mass the astronaut has on his back when he is on the moon!  I'm not sure how you can forget about this, but if someone has done the math right and concluded they should be jumping 21 times higher, they forgot to add this mass into the equation.  Let's do it (starting from the acceleration on the moon using 170 kg instead of 80 kg :

a = F / m = 1,453.6 / 170 = 8.55 m/s²

And the initial velocity now :

v = at = 8.55 * 0.3 = 2.6 m/s :

Now, the height :

h = v² / 2g = 2.11 meters.

Already, we are getting closer to observed reality, but there's more factors to consider (more than I will mention - these are just the big ones).  We are assuming that the astronaut has the same range of motion while wearing that cumbersome suit as he did on Earth.  This is most certainly not the case.  So what does this change?  Well, he can't bend over as far while wearing a suit.  If he can't bend over as far, then clearly the distance he travels between his fully crouched position and fully extended position is shorter.  Since the force is only applied during the time it takes to go from crouched to extended, then clearly, the time his force is being applied is shorter.  How much shorter is certainly up for debate, but there's no doubt that it would be significantly shorter.  Let's suppose it is 2/3rds of what it would be on Earth.  I think this is actually quite generous.  I think it would be at least half the original time, but let's use 0.2 seconds - adjusting from the point where we calculate the initial velocity above (through our adjusted mass calculations as well) :

v = at = 8.55 * 0.2 = 1.71 m/s

Now, the height :

h = v² / 2g = 0.91 meters. 

Now, we are getting into something which is more consistent with what we see.  There are more factors, but the point of this exercise was just to point out that any claims that the way the astronauts are jumping on the moon is in any way inconsistent with physics is nonsense.  Even when those making the claim understand the physics concepts enough to calculate something accurate, it doesn't mean they have considered all the factors - and the person who says they should be jumping 21 times higher certainly wasn't considering all the factors.

I think it's been mentioned, but all of this assumes that the astronauts are going around jumping as high as they can with every jump, and that is not the case at all, so that right there might be the biggest factor into why they do not jump as high as some expect.  One last thing to consider - which I believe Jay brought up (that I was unaware of - thanks, Jay!) is that the joints in the suits themselves would provide resistance, so this would further reduce the jump height as not only would the astronauts' jump force need to overcome lunar gravity, but whatever resistance the joints in the suits themselves provided as well - reducing our upward net force that much more.

EDIT (because I wanted to add this for fun!) :

One other thing I didn't consider in my original comparison was that our original astronaut who accelerated upward at 9.97 m/s² on Earth - who would in turn accelerate upward at 18.17 m/s² on the moon (without a suit and with the same range of motion), would not actually jump as high as 21 times higher on the moon simply because the higher acceleration would make him leave the ground sooner, so the force would be applied for less than 0.3 seconds.  Assuming that the distance traveled during the force is about 0.5 meters (I measured on myself), at 18.17 m/s², this distance is traveled in just 0.23 seconds (not 0.3), so to adjust that -

v = at = 18.17 * 0.23 = 4.2 m/s

h = v² / 2g = 5.5 meters.

So, in reality, an 80kg person who has a 0.457 meter vertical on Earth has the potential to jump 5.5 meters, so just 12 times higher.  I hope someone enjoys this mess!

[bknight]

Well done!  Seems like you have accounted for most of the variables and matched the observational values.   The CT's won't budge either.

[ka9q]

You can simplify the analysis even further so you can do it in your head. Gravity changes weight but not mass. As you leave the ground, all your energy is kinetic, and that depends only on mass. Assuming that your legs work equally well in both places, you should have equal amounts of kinetic energy and therefore equal velocities as you leave the ground.

Yes, that's a first order estimate that ignores the distance you have to lift your weight as you straighten your legs. But we're assuming a quick jump where most of the energy goes into accelerating your mass upward.

After your feet leave the ground, your kinetic energy is converted to potential gravitational energy and back into kinetic energy as you fall. Potential energy is proportional to gravity and distance, so a given amount of kinetic energy will therefore take you six times higher on the moon, no more.

On top of that, on the moon you're wearing a heavy suit/PLSS that weighs about as much as you do, and it's so stiff from inflation that your legs can't possibly work as well as they do on the earth without a suit.

[nomuse]

Thank you, ka9q.  I wanted to mention the mass is still constant, so the better formula is probably something like f = ma minus gravitational loss to arrive at initial velocity. But i'm home sick and can barely handle logic today, much less math.

[Willoughby]

...you should have equal amounts of kinetic energy and therefore equal velocities as you leave the ground....

...so a given amount of kinetic energy will therefore take you six times higher on the moon, no more.

On top of that, on the moon you're wearing a heavy suit/PLSS that weighs about as much as you do, and it's so stiff from inflation that your legs can't possibly work as well as they do on the earth without a suit.

You're only accounting for one aspect the reduced gravity changes : acceleration due to gravity.  It also changes the net force of your jump.  If all other factors are the same, you could jump MUCH higher than six times higher on the moon.  However, as you point out (and I as well in my original post), all factors are not the same.  The restriction in movement (limiting the amount of time you can apply your force) and the fact that there is an additional 90 kg of mass is the difference.

If you have initial velocity (which you would have to if your jump leaves the ground), then that velocity is a result of a NET force upwards.  This is the sum of all forces - including gravity - which is six times stronger on the Earth.  On the moon, with the exact same force of your jump, the NET force would be much larger - therefore your initial velocity much higher.

For example, if your mass is 80 kg and you accelerate your body upward at 10 m/s², then the NET force on your body is 800 Newtons.  This is not the force of your jump - it is the force of your jump IN EXCESS of Earth's gravity.  To calculate the total force of your jump, you have to account for gravity - which on Earth would be 784 Newtons.  If the NET force is upward at 800 Newtons, then the total upward force would be the 784 Newtons it takes to overcome gravity plus the 800 Newtons in addition to it - for a total of 1,584 Newtons.  On the moon, this force of 1,584 Newtons is only countered by the moon's gravity - so 128 Newtons on your 80kg body.  The NET force then is 128 Newtons to overcome the moon's gravity, and the rest - 1,456 Newtons is a NET force upwards, resulting in an acceleration of 18.2 m/s².  You must also account for the fact that since your acceleration is higher, you will leave the ground faster (so the force is applied for less time), but the initial velocity is still higher.  In my original comment, I have an edit where I do a rough calculation on this - and came to the reasonable conclusion that if all factors are the same with regards to range of motion and mass, a person could potentially jump about 12 times higher on the moon.  You must account for the change in gravity both for the rate at which something falls AND the counter force to your jump.  I understand that my original post was pretty convoluted, and simplification is better for a lot of people (and I would tend to be one of those people), but if your simplification arrives at the conclusion that six times higher is the absolute highest anyone could jump on the moon, then your simplification is so simple, it's leaving something out. 

[Apollo 957]

Has anyone accounted for the fact that the astronauts may not have wanted to jump as high as they could?

[Willoughby]

Has anyone accounted for the fact that the astronauts may not have wanted to jump as high as they could?

Yes, this is another - probably more relevant - reason most of the "jumps" are not as high as hoaxers expect.  I'm not using those jumps as a basis for my comparison though. There is at least one clip of astronauts jumping as high as they can on Apollo 16.  Charlie Duke and John Young estimate their jumps to be about 4 feet which is "not bad for a 380 pound guy" (addressing his earth weight including suit).  Which is consistent with my estimation that one could potentially jump 12 times higher on the moon.  This translates to a 4 inch vertical on Earth with all that mass and restricted movement vs 8 inches if you are only able to jump 6 times higher - which I have a hard time believing they could achieve on Earth.  The vast majority of "jumps" on the moon aren't jumps, but hops and are just the way astronauts figured put was the best way to move on the moon since gravity was lower, but their inertia was not.  Walking normally was too hard, and you're right - those hops are not an example of how high the astronauts could jump on the moon.

[ka9q]

If you have initial velocity (which you would have to if your jump leaves the ground), then that velocity is a result of a NET force upwards.  This is the sum of all forces - including gravity - which is six times stronger on the Earth.

You are correct. In the limit of a very sudden, instantaneous jump, the momentary upward force of your muscles swamps your downward weight on both the earth and moon, so you'd rise 6 times as high on the moon. But you are correct that when you consider the upward force to be of finite duration, some of it is required to overcome weight, so the difference would be greater.

It's the same thing as "gravity loss" in a rocket, the reason why lots of liftoff acceleration is much better than just a little more than that required to overcome gravity, even if the total impulse (thrust times time) is the same.

I had to explain this once to my significant other after she put an low thrust engine in one of my rockets at a launch event. Despite having the same total impulse as a correctly sized engine, it valiantly struggled off the pad and made it to only maybe 100' before burning out and nosing over into (into) the earth before the ejection charge could fire...