the angle between two great circles on the sphere

1.4k Views Asked by At

I am trying to learn spherical geometry, but I have difficulty resolving a simple issue.

enter image description here

Let's define a sphere's equator and it's poles N, S. if we create a great circle by tilting the equator circle by degree of $\alpha$. In that great circle point P is the closest to N (the point in the sphere where the latitude is highest). Now, let's look at the great circle that is created by the points N, P on the sphere. Those great circles intersect in point P on the sphere.

My question is about the angle between those great circles at the point P. It seems to me that the angle is indeed $90$. but according the definition of this source (point 9):

"By the angle between two great circles is meant the angle of inclination of the planes of the circles."

It seems that the planes of the great circles are not perpendicular - meaning the angle between them is not 90.

I would appreciate if someone would instruct me what I'm missing here, and how to show, formally, that the angle is 90?

2

There are 2 best solutions below

4
On BEST ANSWER

You are right. The angle between the two plane is ${90}°$. Here is two ways to see it.

First, the point $P$ is the one closest to the North pole. Since great circle are the straigh line of spherical geometry, the shortest distance between a point and a line is obtain with a perpendicular.

Second, the angle between the plane is given by the angle between their normal vector.

Let consider the sphere centered at the origin and the poles on the $z$ axis. Rotate the sphere so the point $P$ is on the $yz$-plane. Note: the point $Q$ is also on the $yz$-plane since $P$ and $Q$ are the ends of the same diameter.

The normal of the plane passing thru $P$, $Q$ and the two poles is on the $x$ axis. E.g. the normal could be the vector starting at $O$ pointing toward were the equtorial plane meet the oblique plane.

The normal vector of the oblique plane passing thru $P$ and $Q$ is in the $yz$-plane. The normal could be the vector starting at $O$ pointing toward the mid point between $P$ and $Q$ on the great circle passing by the poles.

The two vectors are perpendicular.

Hope it helps

0
On

think coordinates for a unit sphere centered at the origin. take the plane of the paper (for your diagram) to be the $xz$-plane. this contains the circle PNQS whose equation is: $$ x^2 + z^2 = 1 $$ the normal to this plane lies along the $y$-axis.

this is also the axis about which the horizontal circle has been rotated around to bring it to the position shown. thus a normal to the rotated circle lies in the $xz$-plane, and is thus always perpendicular to the y-axis, for any angle of rotation.

NB two planes are perpendicular if and only if their normals are perpendicular.