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Comment by JumpCrisscross | Hacker News Reader

> what's the science behind this?

Earth's gravity in low-earth orbit is almost the same as on the surface [1]. What makes things on the ISS float isn't their distance from earth. It's horizontal (or more accurately, tangential) speed. Objects in orbit are in free fall.

To replicate that, parabolic flights free fall. (Parabolic paths are similar to the elliptical paths orbiting objects take [2].) They then stop free falling (to avoid hitting the ground) before climbing back up and doing it again and again. And again.

[1] https://en.wikipedia.org/wiki/Low_Earth_orbit#Orbital_charac...

[2] https://history.nasa.gov/conghand/traject.htm

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This always reminds of the Hitchhikers Guide: "There is an art, it says, or rather, a knack to flying. The knack lies in learning how to throw yourself at the ground and miss."

ISS constantly falls down to earth. It is also moving so fast that it constantly misses earth.

>> 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.

Think of firing a canon sideways: it takes a while for the cannon ball to drop and hit the earth. Now imagine firing it faster and faster, till eventually it's dropping at the same rate as the earth curves. That is essentially what the ISS is doing at a height of about 400km above the surface.

It’s worth noting that if you shoot a cannonball dead horizontally, it hits the ground at the same time as it would if dropped from the same height [1].

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.”

> It’s worth noting that if you shoot a cannonball dead horizontally, it hits the ground at the same time as it would if dropped from the same height

I think this is assuming the ground is flat, which isn't a good assumption when talking about orbits.

It’s a perfectly fine assumption when talking about cannons, which are typically earthbound and have low projectile speeds.

I discuss orbits in the next paragraph, where I hope it’s evident I don’t assume a flat plane :)

> It’s worth noting that if you shoot a cannonball dead horizontally, it hits the ground at the same time as it would if dropped from the same height

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.

"Do you have a citation for this?"

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.

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.

It doesn't take into account the curvature of the earth. The commenter was assuming a simpler flat plane.

> If you fire, say, the main guns of a battleship "over the horizon"

...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.

> there's a caveat

Air.

> It’s worth noting that if you shoot a cannonball dead horizontally, it hits the ground at the same time as it would if dropped from the same height

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.

And the important part about orbits is that wind resistance won't slow it down significantly. At a low orbit like the ISS it touches the atmosphere ever so slightly and needs a boost every now and then.

What they mean is by "moving so fast" is, it's moving so fast horizontally. Imagine being at the top of a steep mountain and throwing a rock directly out from you. The rock will continue to move horizontally, due to the force behind your throw, as well as vertically, as it drops due to gravity. If you don't throw it hard enough, then the trajectory of the rock won't clear the mountain slope. If you throw it too hard, it will clear the base of the mountain by a wide margin. But if you throw it just right, the rock could hypothetically have a trajectory that is perfectly parallel to the mountain slope (in reality, due to things like air resistance, this isn't all that feasible).

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.

> But if you throw it just right, the rock could hypothetically have a trajectory that is perfectly parallel to the mountain slope (in reality, due to things like air resistance, this isn't all that feasible).

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.

This is correct, and I was making a simplification. A better example would be a curved hill.

Nitpicking here, but the trajectory would be approximately elliptical, not parabolic. A parabolic trajectory would require a uniform gravitational field.

> A parabolic trajectory would require a uniform gravitational field.

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.

To clarify for anyone reading, the difference is not constant vs variable magnitude of g, it's constant vs variable direction of g. Gravity is a vector pointing at the (weighted by inverse of distance squared) centre of mass, so once you are operating on a global scale you have to account for the direction of g changing as you move around.

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.

Only now when re-reading your comment do I realize that you said "any throw you could possibly produce", in which case you are correct. I was thinking that you were talking more about the "fire a stupendously powerful cannon to get an orbital (or closeish) trajectory" situation, in which case the velocities would be high enough to not use a uniform gravitational field. My apologies for the misunderstanding.

How do they maintain horizontal velocity?

The ISS? It's above the atmosphere so there is no air resistance to slow it down. The velocity is constant after the initial boost into orbit.

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.

It is always accelerating straight toward the center of the planet at ~1G.

https://xkcd.com/2011/ - so ignoring the sarcasm in the comic, the two red lines are "throwing a ball" like you are used to - and the black line is the ball that never quite reaches the ground - i.e. the ISS.

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