7

In a comment on Is there a self-rounding celestial body from which an Olympian could jump into space? @jpa made a modest proposal.

What if we stacked a bunch of athletes on top of each other and each pushed up the next?

In the distant future of 2021, Nike and Warner Bros have cooked up a cross-promotion publicity stunt for Space Jam 2 and Nike Space Force Vacuum Max shoes. They've placed an army of LeBron James clones on the surface of Mimas. The goal is to launch one LeBron James away from Mimas.

The LeBron clones can do a vertical jump of 1 m on Earth. Each mass 110 kg and are 2 meters tall. Source. His extensive exposure to cartoon logic while filming Space Jam 2, plus his trusty Nike Space Force Vacuum Max shoes, allows LeBron James to survive in space without a heavy space suit.

Mimas has an escape velocity of 159 m/s.

The clones are standing on top of each other in a human pyramid. Each "stage" is holding onto a (for the purpose of the question massless) platform for the next stage to stand on. The first "stage" jumps pushing the rest of the clones up. At an optimal point the second "stage" jumps and the first lets go. And so on.

For example (not to scale).

               LJ                stage 5
              -----
              LJ LJ              stage 4
             --------
             LJ LJ LJ            stage 3
            -----------
            LJ LJ LJ LJ          stage 2
           --------------
           LJ LJ LJ LJ LJ        stage 1
       ====== Mimas ========

What does this stack of LeBron James clones like? How many are needed in each stage? Can it even work? Will the first stage of clones be crushed by the weight of the stages above them?

See also Could a human jump off Mimas without return?

Schwern
  • 8,006
  • 1
  • 30
  • 34
  • 3
    I think you fail at the second hurdle; jumper number 2's leg extension will mostly push jumper 1 down, due to the weight of the stack of LeBrons above him. I wonder if there's a better gymnastic arrangement that avoids that issue, though... – Starfish Prime Jan 25 '20 at 22:51
  • @StarfishPrime Thrust should still be generated due to the equal and opposite reaction. I'm assuming each stage has many LeBron James's. I'll clarify that. – Schwern Jan 25 '20 at 22:56
  • Heck, stage 1 can stand on their hands, and shove up stage 2 with their feet while stage 2 shoves down on the platform, for even more thrust.Stage 2 can do the same for stage 3, and so on. – notovny Jan 25 '20 at 23:16
  • @Schwern the problem still stands; stage 2 are pushing down against 5LJ, but trying to push up 10LJ so the net upward thrust is less than that developed by 1LJ. Even at stage 4, the wasteage is 50%; that's a poor efficiency! – Starfish Prime Jan 26 '20 at 12:20
  • @notovny that was what I was reaching towards, I think. You probably want an exponential decay in LJs per stage, but assuming perfect synchronisation and efficiency you could get a 2x velocity increase per stage, right? Might be able to manage it with under 8 stages, then... – Starfish Prime Jan 26 '20 at 12:25
  • @StarfishPrime Marketing stunts don't have to be efficient. – Schwern Jan 26 '20 at 17:40
  • @Schwern depends on your production rate of LeBrons, right? – Starfish Prime Jan 26 '20 at 19:04
  • @StarfishPrime They're using Professor Farnsworth's Larry Bird cloner. – Schwern Jan 26 '20 at 19:56
  • 1
    You don't need escape velocity to escape Titan. Go past the Saturn-Titan L1 or L2 points and Saturn can wrest the basketball player from Titan's capture. These points are about 332 kilometers above Titan's surface. – HopDavid Jan 27 '20 at 12:07
  • @StarfishPrime You can't do exponential increases per stage because they have to stand somewhere. Realistically the best you can do is 1, 4, 9, 16, 25... However, you can get a lot of stages in--if you could get even half a meter per second out of a stage it would work. – Loren Pechtel Jan 27 '20 at 15:43
  • @LorenPechtel half a metre per second seems like it should be doable, and low-balling things a little. But is exponential really that impractical? $2^n$ seems like it should be fine, given that you probably don't need even as many as 8 stages. – Starfish Prime Jan 27 '20 at 15:52
  • @StarfishPrime Where do you put them? Look at the tower in the question--it's only 2D so it goes 1, 2, 3, 4, 5. In 2D it goes 1, 4, 9, 16, 25. I don't see how you can fit more people in unless you can somehow find 4D versions of LeBrons. – Loren Pechtel Jan 27 '20 at 16:02
  • 1
    @LorenPechtel from the OP, "Each "stage" is holding onto a platform for the next stage to stand on". You don't need direct LJ-to-LJ contact. And if the gain per stage is high enough, it won't be far from $x^2$ anyway. – Starfish Prime Jan 27 '20 at 16:42
  • @StarfishPrime That helps but your lower stages are going to get pretty massive themselves. – Loren Pechtel Jan 27 '20 at 22:44
  • @LorenPechtel sure, but the total weight of stages 1 through n-1 should not exceed the lifting capacity of stage n (on the assumption that LJ can lift his weight) whereas for $x^2$ staging once you get past 4 stages the weight of the upper stages starts impairing the thrust of the lower ones. – Starfish Prime Jan 28 '20 at 09:56
  • @StarfishPrime I'm questioning the weight of the platform itself. – Loren Pechtel Feb 01 '20 at 05:24
  • @LorenPechtel I just put it in there so you can stack however you like, ignore its mass. What I'm interested in is if each stage jumping off the stage below works at all, and a ballpark efficiency. – Schwern Feb 01 '20 at 07:09
  • @LorenPechtel a lightweight carbon fibre framework big enough to support the weight of 256 people on Mimas is not going to be problematic. – Starfish Prime Feb 01 '20 at 10:55

1 Answers1

3

At least $3.8 \cdot 10^{15}$ LeBrons are needed. This exceeds the current supply.

Biomechanics

Modelling a human jumper as being able to push with a constant force over some fixed distance, we find how much velocity the jumper with mass $m$ gains when pushing away from some base mass $m_{base}$

$$\Delta v = \sqrt{\frac{\nu}{m^2 \left(\frac{1}{m} + \frac{1}{m_{base}}\right)}}$$

Working backwards from the stated capability of a 110kg mass being capable of jumping 1 meter up under Earth gravity ($4.43m/s$), with $m_{base} \approx \infty$, we can find the applicable $\nu = 2150 kg m^2/s^2$ for our "rocket parts".

Discrete rockets

Ignoring the crushing forces at the bottom of a LeBronian stack for a moment, and the exceedingly poor thrust-to-weight ratio, having a single man in each stage is the optimum as far as total $\Delta v$ is concerned.

For a 100 stage stack (not counting for the initial push against the ground), we get:

$$\sum_{i=1}^{99} \sqrt{\frac{\nu}{(i \cdot m_{LJ})^2 \left(\frac{1}{(i \cdot m_{LJ}} + \frac{1}{m_{LJ}}\right)}} \approx 20.44 m/s$$

At this point, the expended mass units are small enough compared to the rocket that the continuous approximation of reglar rockets would be quite precise, yielding the total mass of LeBrons:

$$M_T \approx 100M_{JL} \cdot e^{(159m/s - 20.44m/s) / 4.43m/s}$$

Arriving at the aforementioned $3.8 \cdot 10^{15}$ figure.

I would like to note that this is already 1% of the mass of Mimas.

Wider stages

Of course, even the mass of just a couple of hundred men are going to crush the bottom stages of this flesh-rocket (flocket?) under Mimas gravity. The problem of not being able to accelerate faster than the gravity of the Saturn moon is also a problem.

The usual solution to this is to make lower stages wider with more thrust, in an exponentially growing pyramid.

(Alternate tricks, like each expended LeBron turning around and jumping off the underside of each stage, using the stack as a "spring" with spinal force transfer, re-using landed stages to jump up again and push off the stack, and so on will be ignored)

An easy to calculate (but not necessarily optimal) growth ratio is to double the amount of LeBrons in each stage. This gives each stage a$\Delta v$ of $2.21 m/s$, exactly half of a single mono-LeBron (insert twice the base mass, twise the load mass into the first equation).

This gives 72 stages in total, with a total LeBron number just shy of Avogadro's number.

The optimal taper ratio is certainly less than 2, but an approximate region of $10^{15} - 10^{22}$ should be precise enough given that this amassment should collapse into a sphere in hydrostatic equilibrium under its own gravity.

SE - stop firing the good guys
  • 43,384
  • 3
  • 143
  • 244
  • "This exceeds the current supply." Then he'd best get busy! Thank you for going all in on this absurd question. – Schwern Jan 27 '23 at 21:00