The little prince looked around his asteroid. Everything seemed to be in its place; his flower was watered, the baobabs had been weeded, and he had cleaned out both the active volcanoes and the extinct one (one never knows). And yet, he couldn’t help but think it could be improved somehow.

The little prince on his spherical planet
The little prince on his spherical planet.

“Little flower, what can I do to make you even more beautiful?” asked the prince.

“Well,” replied the flower, looking bashful at the request, “there is one thing. This is such a little planet, the gravity is a bit weak. Maybe you could increase the gravity, so the air is thicker and my roots hold more firmly?”

This seemed like a difficult request, and the prince thought hard about it. He couldn’t increase the mass of his planet (importing elephants didn’t occur to him), but he could redistribute the rocks of his small planet so that gravity by the flower would be stronger.

His first idea was to pile all the rocks of his planet below the flower, making a long cylinder.

The cylindrical planet

As the prince piled more and more of the planet below the flower, to his surprise he seemed to feel lighter rather than heavier! “Oh no,” the prince said to himself, “making my planet into a cylinder is just moving the flower further away from the center!”

The two active volcanoes also seemed to be unhappy with the new arrangement. One near the middle of the planet was spewing out an alarming amount of fire, while the other at the end of the planet spluttered and seemed about to go out. Acting quickly, the prince began to search for a better configuration.

Rather than rearranging stones all day, the prince decided to calculate the best shape for his planet. He had always found this type of problem fun, unlike the boring spreadsheets grownups were always calling “math.”

First, the prince worked out the gravity the flower would feel for an arbitrary axial-symmetric planet shape. He wasn’t sure what the density \rho of his planet was, but he hoped it would just cancel out in the end. By imagining dividing the planet into thin circles and adding the gravity from each circle, he got

g = 2 \pi G \rho \int_0^{x_{\max} } \left( 1 - \frac{x}{x^2+y(x)^2} \right) dx

Of course, not just any function y(x) would work! The planet had only limited mass to work with, so he would have to restrict to shapes that kept the total mass fixed:

m=\pi \rho \int_0^{x_{\max}} y(x)^2 dx

First he tried applying these to a sphere of radius R.

\begin{aligned}
y_{sphere}(x)^2 &= R^2 - (x-R)^2 \\
&= 2xR - x^2
\end{aligned}
\begin{aligned}
m_{sphere} &= \pi \rho \int_0^{2R} (2xR-x^2) dx\\
&= \left. \pi \rho \left( Rx^2 - \frac{x^3}{3} \right) \right\vert_0^{2R} \\
&= \frac{4}{3} \pi R^3 \rho
\end{aligned}
\begin{aligned}
g_{sphere} &= 2\pi G \rho \int_0^{2R}\left( 1 - \frac{x}{ \sqrt{x^2+2xR-x^2}} \right) dx \\
&= 2\pi G \rho \int_0^{2R}\left( 1 - \sqrt{\frac{x}{2R}} \right) dx \\
&= \left. 2\pi G \rho \left( x - \frac{2}{3}\sqrt{\frac{x^2}{2R}} \right) \right \vert_0^{2R} \\
&= \frac{4}{3}\pi G \rho R \\
&= \frac{G m}{R^2}
\end{aligned}

These equations matched those the little prince remembered for the volume of a sphere and the gravity on a planet’s surface, so the prince felt confident continuing with the ungainly cylinder on which he found himself.

\begin{aligned}
m_{cyl} &= \pi \rho \int_0^h r^2 dx \\
&= \pi \rho r^2 h
\end{aligned}

He got up to measure the height and radius of the current cylinder, but then decided to just include this as a variable a = r/h.

r = \left( \frac{m}{\pi \rho} \right)^\frac{1}{3} a^\frac{1}{3}
h = \left( \frac{m}{\pi \rho} \right)^\frac{1}{3} a^{-\frac{2}{3}}
\begin{aligned}
g_{cyl} &= 2\pi G \rho \int_0^h \left( 1-\frac{x}{\sqrt{x^2+r^2}} \right) dx \\
&= \left. 2\pi G \rho \left( x-\sqrt{x^2+r^2} \right) \right\vert_0^h \\
&= 2\pi G \rho \left( h + r -\sqrt{h^2+r^2} \right) \\
&= 2\pi G \rho  \left( \frac{m}{\pi \rho} \right)^\frac{1}{3}
   \left(  a^{-\frac{2}{3}} + a^{\frac{1}{3}} -\sqrt{ a^{-\frac{4}{3}} -  a^{\frac{2}{3}} }\right) \\
\end{aligned}

Was that stronger or weaker than the original sphere gravity?

\begin{aligned}
\frac{g_{cyl}}{g_{sphere}} 
&= \frac{
    2\pi G \rho  \left( \frac{m}{\pi \rho} \right)^\frac{1}{3}
    \left(  a^{-\frac{2}{3}} + a^{\frac{1}{3}} -
        \sqrt{ a^{-\frac{4}{3}} - a^{\frac{2}{3}} }
    \right)
}{
    \frac{4}{3}\pi G \rho  \left( \frac{3 m}{4 \pi \rho} \right)^{\frac{1}{3}}
} \\
&= \left( \frac{9}{2} \right)^{\frac{1}{3}}
    \left(  a^{-\frac{2}{3}} + a^{\frac{1}{3}} -
        \sqrt{ a^{-\frac{4}{3}} - a^{\frac{2}{3}} }
    \right)
\end{aligned}

He graphed it, and maximized for a.

At a=\frac{9-\sqrt{17}}{8} \approx 0.61 the flower’s gravity is 0.7% higher than the sphere case! “I was on the right track,” the prince said to himself, “except I made my cylinder too long and skinny!

“But is a squat cylinder the best shape, or can I do better?” the prince wondered. He though and thought. He realized that the cylinder couldn’t be the best because the flower would feel less pull from rocks at the distant corners than it would from rocks at other parts of the surface, so he could improve the gravity by moving some rocks. “The best shape would be one where the flower would experience equal gravity from all rocks on the surface,” the prince realized.

\frac{x}{\left( x^2+y^2 \right) ^ {\frac{3}{2}}} = C
y_{opt}(x)^2 = \left( \frac{x}{C} \right)^{\frac{2}{3}} - x^2

He wasn’t sure exactly what that constant would be, but he had a feeling it must relate to the size of his asteroid.

\begin{aligned}
\frac{m}{\pi \rho} &= \int_0^{C^{-\frac{1}{2}}} y^2 dx \\
&=  \int_0^{C^{-\frac{1}{2}}} \left( \left( \frac{x}{C} \right)^{\frac{2}{3}} - x^2 \right) dx \\
&= \left. 
    \frac{3}{5} \frac{x^{\frac{5}{3}}}{C^{\frac{2}{3}}} 
    - \frac{x^3}{3}
\right \vert_0^{C^{-\frac{1}{2}}} \\
&= 
    \frac{3}{5} \frac{C^{-\frac{5}{6}}}{C^{\frac{2}{3}}} 
    - \frac{C^{\frac{3}{2}}}{3} \\
&= \frac{4}{15C^{\frac{3}{2}}} \\
C &= \left( \frac{4\pi\rho}{15m} \right)^{\frac{3}{2}} \\
y_{opt}^2 &= \left( \frac{15m}{4\pi\rho} \right)^{\frac{4}{9}} x^{\frac{2}{3}} - x^2
\end{aligned}

Finally, he had found the shape with the highest gravity for his flower!

\begin{aligned}
g_{opt} &= 2\pi G \rho \int_0^{C^{-\frac{1}{2}}} 
    \left( 1 - \frac{x}{\sqrt{x^2+y^2}} \right) dx
\\
&= 2\pi G \rho \int_0^{C^{-\frac{1}{2}}} 
    \left( 1 - \frac{x}{\left(\frac{x}{C}\right)^{-\frac{1}{3}}} \right) dx
\\
&= \left. 2\pi G \rho
    \left( x - \frac{3}{5}{C^{\frac{1}{3}} x^{\frac{5}{3}}} \right)
\right\vert_0^{C^{-\frac{1}{2}}}
\\
&= 2\pi G \rho
    \left( C^{-\frac{1}{2}} - \frac{3}{5}C^{-\frac{1}{2}} \right)
\\
&=\frac{4}{5} \pi G \rho
    \left(\frac{15m}{4\pi \rho} \right)^{\frac{1}{3}} 
\end{aligned}

“How much stronger will gravity be with this new shape?” asked the prince.

\begin{aligned}
\frac{g_{opt}}{g_{sphere}} &= \frac{
    \frac{4}{5} \pi G \rho
    \left(\frac{15m}{4\pi \rho} \right)^{\frac{1}{3}}
}{
    \frac{4}{3}\pi G \rho  \left( \frac{3 m}{4 \pi \rho} \right)^{\frac{1}{3}}
} \\
&= \frac{3}{5^{\frac{2}{3}}} \approx 1.026
\end{aligned}

So the prince set about rearranging his planet once again. Soon he had it configured to the optimal shape. And the flower flourished under the additional 2.6% gravity.

The ideal planet for the flower

Many thanks to

  1. Antoine de Saint-Exupéry. (1943) Le Petit Prince. Gallimard
  2. I Datsenko, O Lozovenko and Yu Minaiev. “Comment on ‘The optimal shape of an object for generating maximum gravity field at a given point in space.'” European Journal of Physics. 37:5. doi: 10.1088/0143-0807/37/5/058003
  3. Xiao-Wei Wang and Yue Su. (2015) “The optimal shape of an object for generating maximum gravity field at a given point in space.” European Journal of Physics. 36(5). doi:10.1088/0143-0807/36/5/055010. arXiv:1412.5541

Images from this post are ©Spencer Bliven 2020. They may be shared with attribution under the CC-BY-SA 4.0 license.

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