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?

  • 3
    $\begingroup$ 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... $\endgroup$ Jan 25, 2020 at 22:51
  • $\begingroup$ @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. $\endgroup$
    – Schwern
    Jan 25, 2020 at 22:56
  • $\begingroup$ 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. $\endgroup$
    – notovny
    Jan 25, 2020 at 23:16
  • 1
    $\begingroup$ 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. $\endgroup$
    – HopDavid
    Jan 27, 2020 at 12:07
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 27, 2020 at 16:42

1 Answer 1


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


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.

  • $\begingroup$ "This exceeds the current supply." Then he'd best get busy! Thank you for going all in on this absurd question. $\endgroup$
    – Schwern
    Jan 27 at 21:00

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