At what speed can a person jump in zero g?

In summary: 500 lbs? Exactly. Force limited. SO ... the same force applied to 20 lbs will result in a larger acceleration than if it is applied to...500 lbs?
  • #1

DaveC426913

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Relating to Stratus' thread in this forum, I'm trying to figure out initial velocity of a jump in zero g.

Lets start in 1g. In one g, a person might be able to jump 1m. From this, using : v^2 = u^2 - 2as, we can calc his initial u as 4.45m/s.

Is it a safe assumption that, were an astronaut to push off the floor in zero g, he could achieve no more ghwn that same velocity of 4.45m/s, or is there some weight factor no longer present that would make it higher?
 
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  • #2
DaveC426913 said:
Relating to Stratus' thread in this forum, I'm trying to figure out initial velocity of a jump in zero g.

Lets start in 1g. In one g, a person might be able to jump 1m. From this, using : v^2 = u^2 - 2as, we can calc his initial u as 4.45m/s.

Is it a safe assumption that, were an astronaut to push off the floor in zero g, he could achieve no more ghwn that same velocity of 4.45m/s, or is there some weight factor no longer present that would make it higher?
Crouching would be impossible in zero ##g## unless you have something to hold on to.

If you do have some way to pull your mass down, then there is no gravitational resistance during the jump. In theory, your take off speed would be more than under ##1g##.

That said, you'd need a proper biomechanical analysis to resolve the question fully.
 
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  • #3
This relates to a Incendus' space station thread.

His scenario is actually under 0.2g, so enough to squat. I'm just curious if I can assume a constant initial jump speed under varying g's.
 
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  • #4
Just common sense says that the leg muscles could achieve a greater initial velocity under low g and less under high g, all assuming a squat to start with. Think about this, Dave: How fast is your initial velocity just jumping up here on Earth. Now suppose you have exactly the same body and same musculature but you weigh 500 lbs. How high to you think you could jump then and with what initial velocity (if you could even jump at all)? Isn't it obvious to you that if you only weighted 20lbs you could jump a lot higher and with a greater initial velocity?
 
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  • #5
DaveC426913 said:
His scenario is actually under 0.2g, so enough to squat. I'm just curious if I can assume a constant initial jump speed under varying g's.
There is, of course a way to test this with a cell phone recording you launching off the blocks...:wink:

1671132027837.png

http://www.swimbetterhq.com/2017/09/11/50-freestyle-optimal-starts-and-breathing/
 
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  • #6
How about laying on your back on a skateboard?

Marcus asked how high you could jump on Ceres a few years ago and mfb referenced an older thread.
It went on for 2 years! I just thumbed through it and I'm not sure anyone came up with a definitive answer.
The math involved is apparently so peculiar that even Dale initially boogered it. I've been here 15 years and have NEVER seen Dale make a mistake. As far as I can tell, this is simply an extension of that problem, with the gravity taken to zero.
So I think we need someone with a kid with a skateboard to do this experimentally. Just have them jump and record the height difference. Then have them push off from a wall on the skateboard and record their speed. Maybe get their friends to do it so we have more data.

I would do it myself, but just thinking about a skateboard makes my hip hurt. And simply bending over makes me dizzy these days.
 
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  • #7
I think it reasonable to assume that the leg (at one g) in this circumstance is limited by the force it can supply (and not the speed required). For instance if it were speed limited, then fat people would jump as high as thin ones.
If force=Fmax is limiting, then the work done by your leg extending through length l $$W=F_{max} l $$ remains the same regardless of local g and $$mgh_{earth}=F_{max}l=\frac { mv^2} 2$$
This also implies that on the moon you can jump (sans spacesuit) to a height of ##6 h_{earth}## before asphyxiating
 
  • #8
OmCheeto said:
Marcus asked how high you could jump on Ceres

As far as I can tell, this is simply an extension of that problem, with the gravity taken to zero.
I am not interested in how high someone can jump, only the initial velocity.
Because in the scenario I am looking at, someone can jump infinitely high, so...
 
  • #9
My point was that the speed will be the same I believe. Because your legs are force limited not speed limited.
 
  • #10
hutchphd said:
My point was that the speed will be the same I believe. Because your legs are force limited not speed limited.
Exactly. Force limited. SO ... the same force applied to 20 lbs will result in a larger acceleration than if it is applied to 500 lbs. See post #4.
 
  • #11
hutchphd said:
This also implies that on the moon you can jump (sans spacesuit) to a height of 6hearth before asphyxiating
So if on the earth the leg muscles can supply only 1 g force, just enough to raise the body from squat, the jump height hearth is 0.
It would seem odd that the same person would jump only 0 height on the moon, by the formula given.
 
  • #12
I jump on Earth. The acceleration I apply to my body is (legs - gravity). I get little height, so the accel is a little more than gravity. Let's say it is 10m/s^2.

Next we need to know the amount of time that this acceleration is applied. We don't know that yet, but we do know that this acceleration was applied for a distance of one meter. My calculus is terminally rusty so I can't solve this. My guess is that the answer is two meters per second. A better jumper could get twice as much acceleration and get three meters per second.

Of course this is all an estimate because the force varies in time in some complicated ergonomic way.
 
  • #13
hutchphd said:
My point was that the speed will be the same I believe.
The initial acceleration phase also has a lift in the centre of mass, and that'll consume energy => reduce speed.
I think an energy-based assumption (the potential energy difference between low and top point in standard gravity all goes to motion energy in case of zero gravity) would be more accurate.

Ps.: there will be some inaccuracy due the reduced force exerted at higher speeds in the acceleration phase, so it's still just kind of an upper limit...
 
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  • #14
hutchphd said:
My point was that the speed will be the same I believe. Because your legs are force limited not speed limited.
I don't think so. If you raise gravity, you eventually get to a point where you simply can't jump.

This is a fairly entertaining problem. I've been to half a dozen websites and they all seem to agree with you. One problem I discovered at the last one I visited was that their equation stated that you can jump on the sun. Even though I would weigh the equivalent of 2200 kg they claim that I could jump 1.5 cm with the same 3 m/s jump velocity as every other body in the solar system.

I calculate 22,000 newtons weighing me down, and 2000 newtons available for jumping. I don't even see how I'm going to stand up.
 
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  • #15
256bits said:
So if on the earth the leg muscles can supply only 1 g force, just enough to raise the body from squat, the jump height hearth is 0.
It would seem odd that the same person would jump only 0 height on the moon, by the formula given.
Not what I meant. $$mgh_{earth}=F_{max}l=\frac { mv^2} 2$$ says nothing about ##h_{moon}##. That will be determined from v in the usual manner using ##g_{moon}=g/6##. Clearly if v=0 singularities arise unless care is taken.
I should have stipulated v>0 I guess. moon##.
 
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  • #16
Here's my attempt: I assume the legs produce a constant force during the jumping motion
Hopefully I did not make any mistakes/typos
I left out some of the algebra
--------------------------------------------------------------------------------------------
m: your mass
ge earth gravity acceleration
gm somewhere else gravity acceleration
L standing center of mass
αL crouching center of mass
βL earth jump height center of mass
ve velocity leave ground on earth
vm velocity leave ground somewhere else
mf=m(a+ge) constant force(assumption) produced by legs
--------------------------------------------------------------------------------------------
½mve2=mgeL(β-1)
ve=(2geL(β-1))½
--------------------------------------------------------------------------------------------
ve=(f-ge)t=at
L=αL+½(f-ge)t2=αL+½at2

a=ve2/(2(1-α)L=ge(β-1)/(1-α)
--------------------------------------------------------------------------------------------
L=αL+½(f-gm)t2=αL+½(a+ge-gm)t2
t=(2L(1-α)/(a+ge-gm))½

vm=(f-gm)t=(a+ge-gm)t
=(2L(1-α)(a+ge-gm))½
= (2L(1-α)(ge(β-α)/(1-α)-gm))½
Notice that vsun would be an imaginary number, i.e. no jumping on the sun.
If gm=0 then vm=(2Lge(β-α))½
 
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  • #17
Frabjous said:
Here's my attempt: I assume the legs produce a constant force during the jumping motion
Hopefully I did not make any mistakes/typos
I left out some of the algebra
--------------------------------------------------------------------------------------------
m: your mass
ge earth gravity acceleration
gm somewhere else gravity acceleration
L standing center of mass
αL crouching center of mass
βL earth jump height center of mass
ve velocity leave ground on earth
vm velocity leave ground somewhere else
mf=m(a+ge) constant force(assumption) produced by legs
--------------------------------------------------------------------------------------------
½mve2=mgeL(β-1)
...
Can you explain your first equation. It looks to me like you are multiplying L and β, which are both distances which ends up yielding ke = mgh2,
which I do not believe is correct. I also don't understand why you are subtracting 1 from β.
 
  • #18
OmCheeto said:
Can you explain your first equation. It looks to me like you are multiplying L and β, which are both distances which ends up yielding ke = mgh2,
which I do not believe is correct. I also don't understand why you are subtracting 1 from β.
##\alpha## and ##\beta## are non-dimensional. I am essentially doing ##\Delta##KE=-##\Delta##PE

I assume that the legs apply constant force between ##\alpha##L (v=0) and L (v=ve). The person reaches a maximum height of ##\beta##L under gravity alone, a distance of (##\beta##-1)L.

I then use the SUVAT equations on earth to determine a.
I then use the SUVAT equations somewhere else to determine vm
 
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  • #19
Instead of using SUVAT, I could have used energy expended at the point one leaves the ground giving

##\frac 1 2##mve2+mgeL(1-##\alpha##)=##\frac 1 2##mvm2+mgmL(1-##\alpha##)

ve2+2geL(1-##\alpha##)=vm2+2gmL(1-##\alpha##)

vm2=ve2+2(ge-gm)L(1-##\alpha##)

vm2=2geL(##\beta##-1)+2(ge-gm)L(1-##\alpha##)

vm2=2geL(##\beta##-##\alpha##)-2gmL(1-##\alpha##)

which gives the same answer

L is more accurately your tippy toe COM.
 
  • #20
Frabjous said:
##\alpha## and ##\beta## are non-dimensional. I am essentially doing ##\Delta##KE=-##\Delta##PE

I assume that the legs apply constant force between ##\alpha##L (v=0) and L (v=ve). The person reaches a maximum height of ##\beta##L under gravity alone, a distance of (##\beta##-1)L.

...
This still doesn't make sense to me. Perhaps if you would tell me what values you've assigned to L & ##\beta## , I can figure it out myself.

PF Frabjous beta minus one 2022-12-16 at 14.22.14.png
 
  • #21
Using your values of ##\beta##L, L , ##\alpha##L gives ##\beta##=.85m/.65m=1.31 and ##\alpha##=.3m/.65m=.46 and L= .65m for initial conditions.
I get ve=1.99m/s
For zero g environment, I get vm=3.29m/s
--------------------------------------------------------
I get a=5.6m/s2 so f=15.4m/s2
So using m=80kg gives F=1232 newtons
---------------------------------------------------------
PE=mgeL(##\beta##-1)=80kg*9/8m/s2*.65m(1.31-1)=158joules
Note: my potential energy is calculated using h=.85m-.65m=.2m because only gravity applies in this regime which allows me to calculate ve which is the velocity at .65m which is the point feet leave the ground.

edit:corrected my accelerations
 
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