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In the book Le Petit Prince by "French aristocrat, writer, poet, and pioneering aviator Antoine de Saint-Exupéry" the main character lives on an extremely tiny asteroid with the name B612. It is this "asteroid" that the B612 Foundation was named after, founded by retired astronaut and entrepreneur Dr. Ed Lu and Drs. Clark Chapman and Piet Hut.

What "escapes" this XKCD-based answer (pardon the pun) is that it is the $\frac{mass}{radius}$ ratio that is key to the escape velocity, not just the surface gravity.

$$v_{esc} = \sqrt{\left(\frac{2 GM}{r_0} \right)}$$

Question: Using numbers from my answer there, what would be the largest radius sphere (and corresponding mass) that had Earth's surface gravity of about 9.8 m/s^2 that you could jump at escape velocity?

"Bonus points" for an approximate scale height of an atmopshere on the unusual theoretical body in your answer (not B612 of course)

Le Petit Prince

Source

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This is quite a nice calculation. Suppose someone can jump (more exactly raise their centre of mass) $h$ meters on Earth. Then $$mgh = 1/2 m v^2$$. So, if $v > v_{esc}$ then

$$mgh > \frac{m}{2} \frac{2GM}{r}$$

so $$gh > \frac{GM}{r}$$

However we know that our planetoid has surface gravity $g$ so

$$g = \frac{GM}{r^2}$$

Combining these, $$\frac{GMh}{r^2} > \frac{GM}{r}$$ or

$$h > r$$

An Olympic high jumper probably lifts their centre of mass about $1m$ so $r$ is certainly going to be of this order. Now you have to worry about your definitions. Is your centre of mass on the surface, or your feet at the start, for instance, but a small number of meters is the general range.

Taking $r = 1$ we get $9.8 = GM$ so $M = 1.4 \times 10^{11}$ and the density is about $10^{10} kg/m^3$ a bit denser than white dwarf material, but nowhere near as dense as neutronium.

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    $\begingroup$ OK, so the largest asteroid Le Petit Prince can escape, assuming he can jump 1 meter, is of circumference $2\pi$ meters. Rather a short walk! I bet his asteroids have a surface gravity of more like 0.1g :-) $\endgroup$ – Carl Witthoft Oct 31 '18 at 15:00
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    $\begingroup$ The illustration is pretty much spot on, then :-) $\endgroup$ – Simon Oct 31 '18 at 16:03
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    $\begingroup$ I wonder, since the length of a human is on the same scale as the size the body, would the gravitational gradient make a big difference? $\endgroup$ – Avi Cherry Oct 31 '18 at 20:09
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    $\begingroup$ @AviCherry absolutely. Your head would experience about 1/9 of the gravity that your feet do and the question of jumping off the asteroid depends crucially on what position you start in (how far from the centre of the asteroid your centre of mass is). $\endgroup$ – Steve Linton Oct 31 '18 at 20:53
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    $\begingroup$ @sǝɯɐſ The sphere masses about 100 million tons. It's not going anywhere much. $\endgroup$ – Steve Linton Oct 31 '18 at 20:54

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