0

The question is as follows:

What is the expected distance of any point on Earth and the north pole? Take Earth radius 1. (Clarification: Shortest distance cuts through the sphere, instead of lying on surface.)

This (poorly written but official) solution states:

Another approach is to imagine a horizontal ring of $dy$ thickness at distance y from N (north pole).

Area of ring = $2 \times \pi \times dy$ Probability of choosing point on this ring = $\frac{dy}{2}$ Distance of N & a point on ring = $\sqrt{2y}$

Exp length = $\int_0^2\sqrt{2y}\,\frac{dy}{2} = \frac{4}{3}$

Source: https://brainstellar.com/puzzles/203

I understand the overall logic, ( we divide the total area of the ring by the total surface area to get the probability, we find the distance with pythagoren theorem, and do an expected value LOTE kind of thing) but the one thing I am struggling with is why the area of the ring = $2 \times \pi \times dy$. In my eyes, it should be something like $\sqrt{2y-y^2}$ from some pythagorean theorem-ing, but that doesn't give me the right answer (instead it gives something like 16/15). I may be just rusty on my calculus and there might be a formula for such a "ring''s surface area, but intuitively it just makes no sense because $2\times\pi\times dy$ is like assuming every ring occurs at the equator (with radius 1), which can't be the case, right?

Claire
  • 159
  • 3
    I couldn't make sense out of that solution either. I would use the fact (a bit surprising but explained on our site many times) that the z-coordinate of a point on the surface is uniformly distributed in $[-1,1]$. And the distance to the North pole is $d=\sqrt{2-2z}$ by Pythagoras. Hence the expected distance is $$E(d)=\frac12\int_{-1}^1\sqrt{2-2z},dz=\frac43.$$ – Jyrki Lahtonen Sep 01 '24 at 03:31
  • About the main question and intuition on $2\pi\times dy$: Can the Surface Area of a Sphere be found without using Integration? – peterwhy Sep 01 '24 at 04:45
  • We require you to credit the original source of all material originally written by others. See https://math.stackexchange.com/help/referencing. Please [edit] your post to provide appropriate attribution, following our policy on this. – D.W. Sep 01 '24 at 06:41
  • Euclidean or geodesic distance? –  Sep 01 '24 at 10:10

1 Answers1

0

By the euclidean volume measure $$dV = dr\wedge r\ d\theta\wedge\ r\ \sin\theta \ d\phi = dr\wedge r\ d\cos\theta\ \wedge r d\phi $$ the area between to circles of latitude is $2\ \pi\ r^2\ dz$ with $z \ = \cos\theta$ . $$\frac {1} {4\pi r^2}\ \int_ {-1}^1 2\pi r^2\, dz = 1 $$

It follows that the mean distance is

$$\frac {1} {4\pi r^2}\int_ {-1}^1 2\pi r^2\ r | (0, 1) - (\sqrt {1 - z^2}, 1-z) | \ dz\ = \ \frac {1} {4\pi r^2}\ \int_ {-1}^1 2\pi \ r^3\sqrt {2 (1 - z)}\, dz\ = \ \frac {4 r} {3} $$

Roland F
  • 5,122