What is geodesic curvature of the sphere/great circles on the sphere? What is the geodesic curvature of the sphere? The great circle on the sphere is curved. Why is the geodesic curvature of the great circle equal to zero? I don't understand.
 A: Part of the confusion is that "curvature" is a highly overloaded term. Some kinds of curvature (such as mean or Gaussian curvature) are properties of the surface and measure, in different ways, how curved a piece of the surface is "on average." Other kinds of curvature, like geodesic curvature, are properties of curves. So it doesn't make sense to ask what the "geodesic curvature of the sphere" is.
Suppose you have a curve on the sphere. It has to bend, as viewed by a 3D observer, because the underlying surface is bending. But you can ask whether the curve is "as straight as possible" given that it must bend to stay on the sphere; the failure to be as straight as possible is geodesic curvature. You can quantify the amount of in-plane bending by looking at the acceleration vector $\gamma''(s)$ of a curve $\gamma(s)$; this vector will have some component along the normal of the sphere and some component that is tangent. The magnitude of the part that's tangent (also often written as the covariant derivative of the curve in its own direction, $\nabla_{\gamma'} \gamma'$) is the geodesic curvature. Notice that for a great arc, the derivative of the tangent $\gamma'$ is purely in the direction normal to the sphere, and so the geodesic curvature is zero. A great arc is "as straight as possible" with no bending in the tangent plane of the sphere.

Note for nitpickers: I'm assuming here that the sphere is equipped with the metric pulled back from ambient space. You can of course equip the sphere with other metrics or connections so that the geodesics are no longer great arcs.
