Will a point moving on a sphere always at an angle x (0 deg. < x < 90 deg.) to the "equator" reach a "pole"? Formulating my question seems to have given me the answer: that the point will continue   getting closer to the pole but never reach it. Am I correct?  
Edit in response to Martin Argerami:
I see your point. So let's make the point always move at an angle x to the latitude line that includes the point. Then will the point reach a pole? 
 A: Spherical coordinates $(a,b)$ of a point on the unit sphere are related to its Cartesian coordinates $(x,y,z)$ through the relations
$$
(x,y,z)=(\cos a\cos b,\sin a\cos b,\sin b).
$$
In the Northern hemisphere, the longitude $b$ is an angle in $[0,\frac\pi2]$ and the latitude $a$ is an angle in $\mathbb R/2\pi\mathbb Z$. Assume that the point is at $(a(0),b(0))=(0,0)$ at time $0$, is at $(a(t),b(t))$ at time $t\geqslant0$, and moves at constant speed, crossing parallels at a constant angle. 
Then, writing down the direction of the parallel at point $(x,y,z)$ as $(-y,x,0)$, that is, the direction of the parallel at point $(a,b)$ as $(-\sin a,\cos a,0)$, and the speed vector at time $t$ as
$$
(-\sin a,\cos a,0)\cos b\cdot a'+(-\cos a\sin b,-\sin a\sin b,\cos b)b',
$$
one gets the conditions that 
$$
(\cos b)^2(a')^2+(b')^2,\qquad\cos b\cdot a',
$$ 
should both be constant. That is, $b(t)=\beta t$ for some $\beta\gt0$ and $a'(t)=\alpha/\cos(\beta t)$ for some positive $\alpha$, the ratio $\alpha/\beta$ characterizing the angle the particle crosses parallels at. 
Thus, the point is at the North pole at time $t_N=\pi/(2\beta)$, which is finite. Since the point moves at constant speed, the total distance it moved when it reaches the North pole is finite. But the number of turns around the North-South axis is described by
$$
a(t_N)=\int_0^{t_N}\frac\alpha{\cos(\beta t)}\mathrm dt=\frac\alpha\beta\int_0^{\pi/2}\frac{\mathrm dt}{\cos t},
$$
which diverges, hence it is infinite.
