An artificial Earth satellite moves in an orbit between 60^°…

An artificial Earth satellite moves in an orbit between $60^\circ$ north and $60^\circ$ south latitude. Assuming that the satellite can fall on the Earth's surface with equal probability at any point between these parallels, find the probability that it will fall north of $30^\circ$ north latitude.

Step 1: Area of a zone

At latitude $\varphi$, the height of the parallel's plane above the equator is $R\sin\varphi$. By Archimedes' theorem the zone between two parallels has area $2\pi R\cdot(\text{height difference})$:

\[ S(\varphi_1,\varphi_2) = 2\pi R^2(\sin\varphi_2-\sin\varphi_1). \]

Step 2: The whole region

From $-60^\circ$ to $60^\circ$:

\[ S = 2\pi R^2\big(\sin60^\circ-\sin(-60^\circ)\big) = 2\pi R^2\cdot\sqrt3. \]

Step 3: The favourable region

From $30^\circ$ N to $60^\circ$ N (the satellite cannot go further north):

\[ S_1 = 2\pi R^2(\sin60^\circ-\sin30^\circ) = 2\pi R^2\cdot\frac{\sqrt3-1}{2}. \]

Step 4: The probability

\[ \boxed{P = \frac{S_1}{S} = \frac{\sqrt3-1}{2\sqrt3} = \frac{3-\sqrt3}{6}\approx0.211.} \]

Step 5: A common mistake

Comparing latitudes linearly gives $\frac{60-30}{120}=0.25$. That is wrong: a band of latitude near the equator is wider (has more area) than a band of the same width near $60^\circ$, where the parallels are shorter.