It just has more horizontal speed, so it travels sideways a decent amount during its fall.
Anyways, yes, the ISS is scooting sideways so fast that, in the time it would have taken to fall straight down, the ground beneath it has dropped away by an amount equal to its initial height, so the thing stays at the same altitude.
[1]: *This hurt my brain when I first heard it, but I promise it’s true: it’s just that cannons are normally fired at an upward angle, giving them upward velocity and thus “hang time.”
I think this is assuming the ground is flat, which isn't a good assumption when talking about orbits.
I discuss orbits in the next paragraph, where I hope it’s evident I don’t assume a flat plane :)
Do you have a citation for this? I'd expect, as with all physics questions, there's a caveat.
In this case, "... over sufficiently small distances."
If you fire, say, the main guns of a battleship "over the horizon" (from your initial vantage point), I'd strongly suspect this doesn't hold.
In order for it to, the gravitational force vector, integrated over flight path, would have to perfectly counter the curvature of the Earth... which doesn't seem like it would line up so neatly.
In reality, air friction and maximum muzzle velocities probably render most of these concerns moot for practical purposes.
This is Newtonian physics.
Obviously, air, other things, friction mess it up.
But essentially that's it.
Horizontal motion of a body won't affect gravitations vertical pull.
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.
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.
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.
...then you are not firing horizontally, but upwards. If you would insist on orienting those guns perpendicular to the gravity vector you would get a very big splash not too far away.
I'm not an artillerist, but as far as I know (and supported by a quick glance at Wikipedia), the range advantage of those big guns over smaller ones doesn't come from higher muzzle velocity (which is limited by the physical property of the propellant independent of gun size), but from the much higher kg/CdA value of their very big projectiles.
Air.
No, it doesn't, because the Earth is curved, and air resistance. With a relatively dense, aerodynamic shell and energy sufficient for only a short flight time, both of these effects are minimal, so it's approximately true, but lose any of those and it stops being a good approximation.