Over The Pole
Two places are diametrically opposite each other on the same line of latitude. Compare the distances between them travelling along the line of latitude and travelling over the nearest pole.
Age
16 to 18
Challenge level
Problem
Compare the distance between two points diametrically opposite each other on a line of latitude by travelling around the line of latitude in the first case and by travelling directly over the nearest pole along a line of longitude in the other case.
Make this comparison for different lines of latitude and show that the ratio of the distances tends to a limit as the lines of latitude get nearer to the pole.
The Earth Animation requires Flash Player version 9. Click on the Earth to draw the lines of latitude and longitude. Move the mouse pointer to the window edges to navigate.Getting Started
Knowing the angle of latitude you can work out the radius of the circle of latitude and half its circumference. You can also work out the arc length of the route over the pole.
It would be a good idea to draw a graph for the ratio of these distances for different angles of latitude.
Can you explain why this ratio tends to a limit?
Student Solutions
Simon from Elizabeth College, Guernsey and Andrei from Tudor Vianu National College, Romania have both solved this problem and both solutions are used below.
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To solve this problem we shall consider that the point is
situated at latitude $\alpha$. To travel around the line of
latitude, the distance from P to Q would be half the circumference
of the circle at latitude $\alpha$. This circle has a radius $R
\cos \alpha$, where $R$ is the radius of Earth.
So, the distance traveled from P to Q on the line of latitude
is $d_{lat} = \pi R \cos \alpha$.
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Traveling over the line of longitude, the circle on which we
have to calculate the distance is a great circle of the sphere, and
the angle of displacement is $2(\pi /2 - \alpha)$ radians. The
distance is therefore $4\pi R(90-L)/360$ where $L$ is the angle of
latitude in degrees or equivalently $d_{long} = 2R (\pi/2 -
\alpha)$.
It is clear that the path on a great circle is always
shorter.
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A graph of the ratio of these distances shows that this ratio
seems to tend to $\pi /2$ as the line of latitude approaches the
pole, that is the ratio $${d_{lat}\over d_{long}}= {\pi R \cos
\alpha \over 2R (\pi/2 - \alpha)}$$ tends to a limit as $\alpha \to
\pi/2$.
This ratio can also be written as $${d_{lat}\over d_{long}}=
{\pi R \sin (\pi/2- \alpha) \over 2R (\pi/2 - \alpha)}.$$ As we
know ${\theta\over \sin \theta} \to 1 $ as $\theta \to 0$ we can
take $\theta = \pi/2 - \alpha $ and we see that this limit is $\pi
/2$.
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Andrei calculated this limit using L'Hopital's Rule (case
$0/0$): $$\lim_{\alpha\to \pi/2}{\pi R\cos \alpha \over R(\pi
-2\alpha)} = \lim_{\alpha\to \pi/2}{\pi \cos \alpha \over (\pi
-2\alpha)} = \lim_{\alpha\to \pi/2}{-\pi \sin \alpha \over -2} =
{\pi \over 2} \approx 1.57$$
Teachers' Resources
This problem affirms the idea that great circle distances on a sphere are the shortest distances. It prepares the way for more complicated calculations of distances between any two places on the globe.
See Flight Path.