# Why isn't the movement of Newtons cannonball superposed with a constant initial velocity?

According to my understanding of movements, if you throw a ball (or shoot a cannonball) with a certain initial velocity $v_0$ in horizontal direction, its movement can be understood as a superposition of two independent movements: a free fall in vertical direction (e.g. vertical in relation to the surface of the earth) with a constant acceleration $g$ and a horizontal movement with constant speed $v_0$:

But if that is correct, shouldn't the movement of Newton's cannonball also be superposed with a one-directional movement with constant velocity $v_0$, as depicted below?

But as far as I can tell the second picture has to be incorrect, because the constant one-directional velocity would render the orbit unstable, wouldn't it?

• Where does $v_0$ come from? When a satellite was placed in orbit it was given a tangential speed as exact as possible – Steeven Jan 16 '18 at 7:41
• @Steeven : Let's assume that the satellite was placed in orbit with that initial velocity $v_0$ exactly tangential. – Wamseln Jan 16 '18 at 8:04
• Then in the next instant gravity has turned the direction of that velocity. Same magnitude, since gravity doesn't pull against the direction, but new direction. – Steeven Jan 16 '18 at 10:16

The first picture is a simplification of the situation that works at scales where the distances (height, range) are much less than the diameter of the Earth. At a scale of a few kilometres, we neglect the curvature of the Earth and assume that the gravitational field is uniform. In this case, the horizontal motion is always orthogonal to the gravitational force, so it (the horizontal motion) is unaffected by the gravitational force and $v_0$ remains constant.