Mapping points on a sphere: many to one

Question

Is there a finite sequence of rotations that maps any point on a sphere's equator to the north pole (or at least very close to the pole, it would be sufficient for the purposes of my question to reach a latitude of > 80 degrees)? If not, what is the reason why?

Answer 1

Take two points on the equator that are opposite to each other. Then they will stay opposite no matter how much you rotate, so if you get one of these close to the north pole, then the other one will be far away.

In general, rotations leave the distance of points invariant, so you will never be able to get points closer together.

Answer 2

It's not clear what order your question means to express. If it's

"Is there a rotation sequence that sends all points of the equator to somewhere near the NP?"

then @DirkLiebhold's answer is correct: there is not.

If it's

"For an arbitrary point $P$ on the equator, is there a rotation sequence that sends it to somewhere near the NP?"

the answer is "yes, a single rotation suffices." Take your point $P$ and walk 90 degrees away from it on the equator (in either direction) to reach a point $Q$. Draw a line from $Q$ to the center, $C$ of the sphere. Rotating the sphere about the line $QC$ by either $+90$ or $-90$ degrees will send $P$ to the north pole.

If it's

"For an arbitrary point $P$ on the equator, is there a sequence of rotations about the x-, y-, and z-axes that sends it to somewhere near the NP?"

then the answer is "yes", but it's a little more complicated. Here's the sequence

1. Rotate about the north-south pole axis (which I'll call the $z$ axis, because I personally like $z$ to point up; others might call this the $y$ axis) enough to move $P$ to lie on the $x$-axis at a point I'll call $P'$. Assuming that your $x$-axis goes through longitude $0$, you're rotating by the negative of the longitude of $P$.

2. Now rotate about the $y$-axis by either $90$ or $-90$ degrees to take $P'$ to the north (or south) pole.

In short: two axis-aligned rotations suffice to solve this third problem.