ISS constantly falls down to earth. It is also moving so fast that it constantly misses earth.
I'm gonna have to roll up some herb while you explain this comment further.
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.
Anything in orbit it literally the exact same. You get something moving fast enough horizontally that even though the Earth's gravity is still pulling on it, its trajectory towards earth is perfectly parallel (actually often not perfectly, but in principle) with the curvature of the planet. But by doing it high enough in the sky, you can escape the atmosphere of the earth, to where there is no wind resistance, meaning once you get up to a high horizontal speed, you can turn off the engines and coast perpetually without slowing down.
So everything in orbit, from the ISS to satellites, to the moon, are all in Earth's gravity well and are falling towards the planet, but they have a horizontal speed as well that keeps them from colliding with the surface. The Earth and the rest of the planets in the solar system do the same thing in terms of their orbit with the Sun.
It's also why sometimes satellites or space junk that have been in the sky for years will come crashing down to Earth. Sometimes the calculations for how fast you need to be going are off, or something will throw off it's horizontal momentum, and that will cause it's trajectory to dip just enough that it is no longer orbiting the earth perfectly, but instead is spiraling ever so slightly towards the surface, and will over the course of months or years or decades dip closer and closer until it enters the atmosphere, at which time air resistance becomes a factor again and it breaks up and really plummets.
Not quite. Ignoring air resistance, any throw you could possibly produce will have a parabolic trajectory, which means it could be parallel with the mountain slope only if the slope itself is parabolic.
A uniform gravitational field is such a good approximation for the situation described that I didn't think of including that proviso. Ballistic trajectory calculations close to the Earth's surface assume constant g.
The force always pointing towards a single point is what causes an ellipse to form. If the force always pointed down it would be a parabola, and over short distances on the surface this is a really good approximation.
Actually that's not quite correct, I believe it is not quite out of the atmosphere completely and as such there is a (very) small amount of air resistance and they have to fire engines to boost back into the appropriate orbit every once in a while.