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enter image description here Assuming that the earth is a perfect sphere with radius 6378 kilometers, what is the expected straight line distance through the earth (in km) between 2 points that are chosen uniformly on the surface of the earth?

SNEHIL SINHA
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    The first thing to notice is that you can fix one of the points. Then, you just have to compute $$\cfrac{\int\limits_{S(0,6378)}d(x,t)dt}{\operatorname{area}(S(0,6378))}$$ – xavierm02 Aug 31 '14 at 11:28
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    I do not understand if this question is asking how far to walk from A to B or how far to dig from A to B? – Nick Aug 31 '14 at 12:33
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    @Nick: "through the earth" means you have to dig. – TonyK Aug 31 '14 at 12:56
  • @xavierm02 It's not obvious to me that fixing one of the points does not change the uniform distribution of the other point, or perhaps I'm just not seeing it. Can you explain? – John Smith Aug 31 '14 at 17:24
  • @Nick : Well. It seems obvious to me because rotations are isometries so moving an arbitrary point to a fixed one doesn't change the length of the straight line. But I wouldn't be able to justify that much further. – xavierm02 Aug 31 '14 at 18:30

2 Answers2

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You may assume that the first point is the north pole $N$ ($\theta=0)$ of the earth sphere $S$. All points in an "infinitesimal lampshade" at latitude $\theta$ and of latitude width ${\rm d}\theta$ have the same distance from $N$, namely $$\rho(\theta):=2r\sin{\theta\over2}\ .$$ The area of this "infinitesimal lampshade" amounts to $${\rm d}A=2\pi r\sin\theta\cdot r{\rm d}\theta\ .$$ In this way the mean distance $\bar\rho$ becomes $$\bar\rho={1\over{\rm area}(S)}\int_S\rho(\theta)\>{\rm d}A={1\over 4\pi\>r^2}\int_0^\pi 2r\sin{\theta\over2}\ 2\pi r^2 \sin\theta\>d\theta\ .$$ As $$\int_0^\pi \sin{\theta\over2}\sin\theta\ d\theta=2\int_0^\pi \sin^2{\theta\over2}\cos{\theta\over2}\>d\theta={4\over3}\sin^3{\theta\over2}\Biggr|_0^\pi={4\over3}$$ we finally obtain $$\bar\rho={4\over3}\>r\ .$$

Christian Blatter
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View the sphere as the surface of revolution of the circle $$x^2+y^2=r^2 \tag 1$$ about the $x$-axis. Differentiating both sides of $(1)$ we get $$ 2x\,dx+2y\,dy=0\quad\text{or}\quad x\,dx+y\,dy=0\quad\text{or}\quad\frac{dy}{dx}=\frac{-x}{y}. $$

The distance from $(r,0)$ to $(x,y)$ is $\sqrt{(x-r)^2+y^2}$, by the Pythagorean theorem.

The element of arc length is \begin{align} & \sqrt{(dx)^2+(dy)^2} \quad\text{(by the Pythagorean theorem)} \\[8pt] = {} & \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ \ dx \\[8pt] = {} & \sqrt{1+\frac{x^2}{y^2}}\ \ dx = \frac{\sqrt{y^2+x^2}}{y} \, dx = \frac r y \, dx. \end{align} When the graph is revolved about the $x$-axis, the point $(x,y)$ traverses a circle of circumference $2\pi y$. So we get an infinitely narrow strip of length $2\pi y$ and infinitely small width $\dfrac r y\, dx$, hence with area $2\pi y\dfrac r y\,dx=2\pi r\,dx$, and every point on that strip is at the same distance from $(r,0)$, namely $$ \sqrt{(x-r)^2+y^2} = \sqrt{(x-r)^2+(r^2-x^2)}= \sqrt{2r^2-2rx\ {}}. $$ So summing all these infinitely small quantities we get \begin{align} & \int_{-r}^r\ \overbrace{{}\ \sqrt{2r^2-2rx\ {}}\ {}}^{\text{distance}} \ \overbrace{{}\ 2\pi r\,dx\ {}}^{\text{element of area}} \\[8pt] = {} & (2r)^{3/2}\pi \int_{-r}^r \sqrt{r-x\,{}}\ dx \\[8pt] = {} & (2r)^{3/2}\pi \int_{2r}^0 \sqrt{u}\,(-du) \\[8pt] = {} & (2r)^{3/2}\pi \frac 2 3 (2r)^{3/2} = \frac {16} 3 \pi r^3. \end{align} Dividing this by the whole surface area $4\pi r^2$ to get the average, we have $$ \text{average} = \frac{(16/3)\pi r^3}{4\pi r^2} = \frac{4r}3. $$

Where we saw $2\pi y\dfrac r y\, dx$, the cancelation of the $y$ seems a bit startling. It was in effect discovered by Archimedes in around 250 BC.

Michael Hardy
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