Let the elliptical orbit be bisected through the semi-minor axis... Ellipses are symetrical and both sides here are mirror images of each other, true? If so, then they share matching curvature at opposite ends of the semi-major axis (and opposite ends of the semi-minor axis, and reflections across both the semi-major and semi-minor axses...) Comparing the parts of the orbit at each end of the semi-major axis, both should have the same curvature... But one is close to the focus with the central mass and the other is further away. So at the close end the path is in a stronger gravitational acceleration, the other end in a weaker one. This difference appears in the orbiter at the close end moving faster than when at the far end. The question is: If the distances from the central mass are different, and the speeds along the orbital path are different, why do both ends of the orbit describe the identical curvature?