Daria Ivanova - Affording a planet with geometry

For my first video recommendation, it seemed obvious to me to spotlight Daria Ivanova’s SoME3 submission. The topic is light-hearted, the explanations are clear, and of course the presentation style is so amazing and original.

Bonus: gravity at the other pole #

Now, we know that the gravity at one pole is equal to \(g\), and that the volume is approximately 92.6% of a sphere with the same gravity. But what’s the gravity at the other pole?

Using the law of cosines, \(\cos(\varphi’) = 1 + r(\varphi)^2 -2r(\varphi) \cos(\varphi)\), and since by definition, \(r(\varphi)\cos(\varphi) + \cos(\varphi’)^{3/2} = 1\), we find \[ 1 - r(\varphi)\cos(\varphi) = \bigl( 1 + r(\varphi)^2 - 2r(\varphi)\cos(\varphi) \bigr)^{3/2} \] Therefore, at fixed angle \(\varphi\), we’re looking for the smallest positive root of the parametric function, for \(\varphi \in (0, \pi/2)\), \[ \rho \mapsto f(\rho, \varphi) := \bigl(1 - \rho\cos(\varphi) \bigr)^2 - \bigl( 1 + \rho^2 - 2\rho\cos(\varphi) \bigr)^3 \]

At first, I was surprised to see that for small angles, the root was greater than 1, but that actually holds up with the drawing. Also, it seems that \(\rho = 1\) would be a good initial condition for root finding.

To compute the force, we use the formula from the video, $$ F = \int_{\theta = 0}^{2\pi} \int_{\varphi=0}^{\pi/2} \int_{\rho=0}^{r(\varphi)} \underbrace{\frac{Gm\delta}{\rho^2}}_{\text{force}} \underbrace{ \cos(\varphi)}_{\substack{\text{project on}\\ \text{vertical}\\ \text{component}} } \underbrace{ \rho^2 \sin(\varphi) {\rm d}\rho {\rm d}\varphi {\rm d}\theta }_{ \text{infinitesimal volume} } $$ This simplifies into \[ F = \underbrace{2\pi G m \delta}_{\frac{5}{2}mg} \int _0^{\pi/2} r(\varphi)\cos(\varphi)\sin(\varphi) , {\rm d}\varphi . \] A numerical computation code yields \[ \frac{F}{mg} \approx \frac{5}{2} \cdot 0.3949 \approx 98.7%. \] Therefore, if this is indeed the minimal spot, the gravity is off by at most 1.3% on this planet. That seems very manageable.