can I travel in the same direction along the surface of an ellipsoid without ever returning home? Starting from an arbitrary point on an ellipsoid, moving straight at a random direction along the surface, are you always guaranteed to come back to the starting point eventually?
 A: I'd say the anser is no, for the reasons outlined below.
I interpret your question as choosing a random initial direction from the arbitrary point, then moving in said direction along a geodesic path.
Searching the web I found http://geographiclib.sourceforge.net/1.29/triaxial.html which states:

There are two classes of simple closed geodesics (i.e., geodesics which close on themselves without intersection): the equator and all the meridians. All other geodesics oscillate between two equal and opposite circles of latitude; but after completing a full oscillation in latitude these fall slightly short (for an oblate ellipsoid) of completing a full circuit in longitude.

The amount by which these “fall short” of closing after a single turn around can be tuned by choosing the dimensions of the ellipsoid. So it should be possible to make this distance incommensurable with the full circuit. The distance measure used to express this incomesurability would have to be some which corrects for the elliptic form, so that you get the same offset for every iteration around the ellipsoid. But the general idea of some form of irrational ratio between full circuit and geodesic revolution should always be possible.
The page quoted above also mentions other classes of geodesics. In that sense, the formulation of “all other geodesics” in the quotation above isn't strictly true, but for us here it is enough to know that geodesics of this kind do exists. Here is a more interesting class for your question:

If the starting point is $β_1=90°$, $ω_1=0°$ (an umbilical point), and $α_1=45°$ (the geodesic leaves the ellipse $y=0$ at right angles), then the geodesic repeatedly intersects the opposite umbilical point and returns to its starting point. However on each circuit the angle at which intersects $y=0$ becomes closer to $0°$ or $180°$ so that asymptotically the geodesic lies on the ellipse $y=0$.

So you can choose a point which lies on such an umbilical geodesic, but not opn $y=0$, and the direction of the umbilical geodesic that goes with it. Since the geodesic will converge towards the $y=0$ plane, you can be certain that you'll never return to your starting point if you didn't return there after a few iterations. The illustration on that page should be helpful in seeing that the latter won't happen. Of course, the probability for randomly choosing the correct angle for such an umbilical geodesic for an arbitrary starting point is strictly speaking zero. So it will “almost never” happen, but it is still possible.
The interesting aspect about these umbilical geodesics is that you won't even come close to your starting point. This is in contrast to the incommensurable situation above, where you won't return to your exact starting location but will come arbitrary close to it.
