[Math] Finding the shortest distance to the north of the sphere

spherical trigonometryspherical-geometry

I found these problems in Alan F Beardon's Algebra and Geometry:

  1. Verify that any point with latitude α is a spherical distance R(π/2−α) from the north pole.

  2. Suppose that an aircraft flies on the shortest route from London (latitude 51◦ north, longitude 0◦) to Los Angeles (latitude 34◦ north, longitude 151◦ east). How close does the aircraft get to the north pole?

Could you give me any hints as to how to solve them? All I know is spherical distance and spherical trigonometry. It's pretty clear to me that if I can prove the first one the second should become easy, but I've no clue as to how to proceed.

Thanks for your help.

Best Answer

Hint:

The shortest distance between two points on the surface of a sphere is the distance measured on the great-circle between them.

For 1) use the the definition of latitude $\alpha$ as the shortest distance between a point a the equator and note that the distance from the pole and the equator is $\pi/2$.

For 2) find the great circle that pass between the two points.

The two cities and the north pole define a spherical triangle $ANB$ and you know: $$ A=(\phi_A,\theta_A)\qquad B=(\phi_B,\theta_B) \qquad N=(0,\pi/2) $$ You can find:

the arcs $AN=b$ and $BN=a$ (as in 1))

the angle in $N$: $\nu=\angle ANB=\theta_B-\theta_A$

the arc distance $AB=n$ between the two points $A$ $B$ given by:

$$ \cos n= \sin \theta_A\sin \theta_B+\cos \theta_A\cos \theta_B\cos(\phi_A-\phi_B) $$

Now, using the sine rule $$ \dfrac{\sin a}{\sin \alpha}=\dfrac{\sin b}{\sin \beta}=\dfrac{\sin n}{\sin \nu} $$

you can solve the triangle finding the angles $\alpha=\angle NAB$ and $\beta=\angle NBA$

The minimum arc distance of the arc $AB$ from $N$ is the arc height $h$ of the triangle, given by: $$ \dfrac{\sin h}{\sin \alpha}=\dfrac{\sin b}{\sin \pi/2} $$

Related Question