[r-bryan]

Cool. Now solve it for a circular fence on the surface of a sphere. In fact, solve all four cases {exterior, interior of sphere} X {exterior, interior of fence}. Spherical trig can only make the solution(s) even more heroic, right?

[scatters]

Surface of a sphere depends on how large the fence is compared to the sphere. If the fence is small, the answer is the same as the plane version. If the fence is as large as possible (an equator) then the rope needs to be precisely one quadrant in length, equalling the "radius" of the fence. If the fence encloses more than half the sphere... well, if it encloses all of the sphere (that is, it is a small circle with the "inside" declared to be the outside) then the rope is again one quadrant, so half the "radius" of the fence.

More interesting is a space-goat tethered to the interior of a hollow sphere, hypersphere etc.; no closed-form solutions for higher-dimensional cases, but the answer tends to sqrt(2) as the number of dimensions approaches infinity.

[Akronymus]

Why not try hyperbolic space too?

[antiquark]

You forgot about the hypergoats.