Anybody looking at this answer should also check the Wikipedia link provided therein - Sub-orbital spaceflight. The section Tourist flights specifically answers some of the points raised in this question

Yours is a daring approach which I couldn't gather the courage for! I've just one worry - if the ball is smaller than the cross section of the tube, and tube happens to be of quite smaller mass than that of the ball, then somewhere during the ball's stay inside the tube, it may happen that tube attains more $v_x$ than ball's $v_x$. In that case, the ball will continuously collide back and forth with the walls of the tube. And that will leave the exit part very uncertain if the arc is not ending exactly vertically upward (in which case the collisions may die out). Not sure.

Yes, many thanks, I was fixing this mistake in the meanwhile. Just posted with correction.

This means that at this point the ball and the tube will be moving horizontally like a single body just before the ball leaves the tube.

Radial direction is the direction along the radius of curvature of the tube. The ball will have no freedom of movement in that direction if its diameter is equal to that of the cross-section of the tube. As far as that direction is concerned, the ball and the tube are just one body for as long as the ball is inside the tube. Now, at the topmost part of the tube, this radial direction is the horizontal direction, i.e. the line from the center of curvature of the tube to this point is horizontal.

In the question you made up, you added circular motion, so you need to bring in the force say, $F$, that is helping gravity in generating the circular motion. Once you have it, $F + F_{gravity} = \frac{mv^2}{r}$, and there is no discrepancy.