Aka orthogonal vectors won't affect each other.

Does this hold true as an object's ballistic trajectory approaches significant fractions of a planet's diameter?

Granted, the object is constantly being accelerated towards the center of the planet.

But that force vector's direction changes with respect to the initial "horizontal" launch vector as the object continues on a straight path, until they're longer orthogonal.

The force of gravity is always orthogonal to the direction of motion when in orbit. So never loses momentum. Which is why the moon is still going around the earth a billion years later and has not 'fallen' into it.

This is only true for circular orbits. It’s pretty transparently obvious that elliptical orbits move closer to and farther from the center of mass (and lose and gain momentum accordingly).

We're talking suborbital velocities here.

If I'm wrong, I'd love to hear exactly why, but regurgitating basic physics doesn't resolve the difficulties in modelling a straight flight path around a curved surface, in relation to a dropped object.

in a completely Newtonian universe (ignoring relativity), if you shoot the ball at high enough speed it will never hit the earth. horizontal motion doesn't affect the vertical pull, but the direction of the vertical pull changes by the time the ball's traveled well over the horizon.

> in a completely Newtonian universe (ignoring relativity), if you shoot the ball at high enough speed it will never hit the earth.

Right. Assuming a perfectly spherical earth, at below orbital velocity, it hits the ground somewhere; at orbital velocity up to (but excluding) escape velocity it (assuming the cannon gets out of the way) orbits with the low point at (and opposite) the firing position, and beyond escape velocity it takes a curving path getting ever farther away.