Great circle on $S^2$ Suppose that $\alpha$ is a great circle on $S^2$. I want to write mathematically, a proof if posible, the reason why $\alpha''$ is pointing towards the interior of this great circle.
I perfectly understand the intuition, in any other case the trajectory would not be a circle.
Do you know how to prove, mathematically, that $\alpha''$ points towards the interior of this great circle?
 A: All this assumes that $\alpha$ is the set which is the great circle; if it's merely a parameterization of that set, then its second derivative might be zero at various points, so the claim would not be true. So starting from the set, I'll parameterize it, and then show that this constant-speed parameterization actually has an acceleration vector that points towards the circle center.  (That's not very deep --- every constant speed parameterization of any circle has an acceleration vector that points towards the center of the circle! My only contribution here is to make it very explicit for this particular case.)
Suppose that $\alpha$ is the great circle in the plane orthogonal to $\pmatrix{a\\b\\c}$, but not lying in either the $x = 0$ or the $z = 0$ plane. (I'll let you deal with those two circles on your own). Then letting $$
s = \pmatrix{-b\\a\\0}, t = \pmatrix{0\\ -c\\ b}, 
\\ u = \frac{1}{\|s\|} s,\\
w = t - (t \cdot u) u,\\
v = \frac{1}{\|w\|}w\\
$$
we have that $u$ and $t$ are both perpendicular to the plane-normal (check this!), hence so is $w$, and therefore $v$,  and $u$ and $v$ are orthogonal (check this!), and are unit vectors.
So we can write
$$
h(r) = \cos(r) u + \sin(r) v
$$
as a parameterization of $\alpha$. Since $h''(r) = -h(r)$ (I'll let you take the derivatives, too!), we can see that $h(r) + h''(r) = (0,0,0)$, which is inside the circle (being the center of the sphere!)
