Finding the shortest distance to the north of the sphere

Question

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.

Answer 1

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.

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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: $$ \dfrac{\sin h}{\sin \alpha}=\dfrac{\sin b}{\sin \pi/2} $$

Answer 2

Spherical geometry uses radians rather than degrees simply to make the formulas look more natural by excluding factors of conversion.

So a full turn, which is a rotation of 360 degrees, in radians is $2\pi$; and half a turn, or 180 degrees is $\pi$ radians.

Now consider a circle of radius$R$; and draw two lines from the centre seperates by an angle $\theta$ to the circumference; the arc-length cut out by these two lines is simply $R\theta$ (remembering that the angle needs to be expressed in radians).

This ought to be sufficient to solve the first question remembering that altitude is not measured from the North Pole but from the Equator; and that a right-angle (that is an angle of 90 degrees) expressed in radians is $\pi/2$.