How are you able to achieve 0G while still within the escape velocity of earth, 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...
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.”
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.
Air.
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.
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.
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.
...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.
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.
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 :)
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.
All of those orbits represents trajectories in which you are in free fall.
The primary reason these orbits don't "exist" is that they impact the surface of the Earth [1]. However, if you "ride" one of these orbits briefly, you still get the free fall effect, just as if you were in orbit! It's just that you can't stay on the orbit freely because of the minor issue of surface impact.
When you juggle or something, the balls are actually, very briefly, in orbit. Minus air resistance, which is a big deal for a ball. (I initially wrote "throw a baseball", but air resistance becomes a factor very quickly even then.) With a powered projectile, like, say, an airplane, you can deliberately overcome the air resistance for a while and stay in the orbit for longer, even through the atmosphere.
It may seem like a bizarre idea of "powering" through an orbit, but it's perfectly reasonable. It's still a gravitational orbit, it's just that electromagnetic forces (contact forces) are getting involved and mucking up the purity of the orbit, so we cancel them out briefly with other ones via the jet engines. If you had a device that could continue powering through the Earth, you could stay at free-fall even through the rock.
[1]: The secondary reason is that the Earth isn't the aforementioned point mass. A neutrino could hypothetically "orbit" through the Earth for a while before being captured, but its orbit will be complicated by the fact that as it goes below the surface, the "mass" of the Earth starts dropping. Which makes me wonder if there ever are any such neutrinos. The only thing stopping such a thing from existing is the need for a process to create neutrinos that have less than escape velocity for the Earth, and I'm not sure if there is such a process.
You owe it to yourself to see these amazing videos filmed on planes that simulate zero G:
OK Go - https://www.youtube.com/watch?v=LWGJA9i18Co
Physics Girl - https://www.youtube.com/watch?v=q1_AJWZajEk
Veritasium - https://www.youtube.com/watch?v=SAQ-iIJkLzA
"Contrary to popular misconception, the 0 g freefall phase of flight begins as the aircraft climbs, and does not occur solely as the aircraft descends. Although the aircraft has upward velocity during the initial 0 g phase, its acceleration is downward: the upward velocity is decreasing" [1].
Basically, gravity can accelerate you all that it wants, but if there is no force resisting that acceleration (air drag, your feet against the ground, the seat pushing on your back in an airplane) then you don't feel that acceleration.
https://en.wikipedia.org/wiki/G-force
"Gravitation acting alone does not produce a g-force, even though g-forces are expressed in multiples of the acceleration of a standard gravity. Thus, the standard gravitational acceleration at the Earth's surface produces g-force only indirectly, as a result of resistance to it by mechanical forces. These mechanical forces actually produce the g-force acceleration on a mass. For example, the 1 g force on an object sitting on the Earth's surface is caused by mechanical force exerted in the upward direction by the ground, keeping the object from going into free fall."
You can choose your frame of reference when considering interacting masses. I choose mine, which gives the interpretation above.
[1] ignoring buoyancy and drag from air, which are small in this scenario.
If you're in an elevator, and the elevator cable snaps, you'll be in free fall (at least initially), although you are nowhere near the escape velocity.
("Zero G" is admittedly a bad term, since there's no such thing as "zero G" literally anywhere in the universe. It's all freefall.)
The reason is that the air itself is imparting substantial acceleration upon you from you hitting it, which leads to terminal velocity. Once you reach terminal velocity the air pressure is imparting exactly the same kind of force against you to repel gravity as the floor would be at ground level, so you experience one full g. You only really feel weightless the moment you jump off a surface, before you've accumulated any velocity and thus before you're being accelerated by air resistance.
And what's interesting, is that what feels "down" is actually "up". Which is one of the gotchas in flying.
> You only really feel weightless the moment you jump off a surface, before you've accumulated any velocity and thus before you're being accelerated by air resistance.
This is a very useful thing in ice skating, skiing, martial arts, etc. Especially combined with changing moment of inertia.
G is a unit. 9.81 Newtons. So 3G is 3*9.81 newtons.
The same: in both cases the vessel is falling toward earth accelerating at 1g, so for the passenger that are also experiencing 1g of gravity it feels like they're floating
So why horizontally? Imagine there was no air resistance on Earth. If you shot a bullet that bullet would keep going until the force of gravity pulled it down and it hit the Earth. But now imagine that you shoot it fast enough that the vertical distance gravity is pulling it down is less than the vertical distance it gains due to the curvature of the Earth. That equilibrium is exactly what orbit is. It also leads to the highly counter intuitive fact that the height of a given circular orbit is determined exclusively by how fast an object is moving relative to the body it's orbiting. Mass doesn't matter.
Okay so back to the plane. If it's not intuitive yet imagine throwing a ball. It works exactly the same as our bullet, but we can visualize one important part easier. The ball's trajectory will be a parabola. And at the highest point of that parabola the net vertical force on the ball is zero. It's where the force you exerted on it to send it up, and the force of gravity pulling it down eventually reach an equilibrium. Something inside of that ball would experience 0g at the moment when it was at its parabolic peak. And that's exactly what these planes do. They simply 'throw' the planes into a parabolic path, and the passengers experience near 0 g while traveling through the parabolic peak.
However the earth is a sphere and the direction of the free-fall constantly adjusts itself to point towards the center of the sphere creating an orbit.
It is literally throwing oneself at the ground and missing.
This is not correct. If we count gravity as a force, then it is pulling on the ball just as much at the peak of the parabola as anywhere else, and once the ball leaves your hand gravity is the only force on the ball (leaving out air resistance); your hand doesn't magically exert force on the ball once it's thrown.
If we do not count gravity as a force (which is the approach taken in General Relativity), then there is no force on the ball at all (leaving out air resistance) once it leaves your hand.
There is no vernacular in my post. When I use the word force, I am stating it in a purely colloquial sense. And in this regard everything is completely cogent and clear description of the forces (har har) in play. By contrast look at the top post. It provides a couple of sentences along with a link to Wiki for further elaboration that immediately jumps into orbital mechanics, assuming an understanding of delta v, etc. There's nothing wrong with the comment in and of itself, but it's an absolutely awful comment in regards to the audience it's talking to.
And I think this pedantry a big part of the reason that so many individuals are completely scientifically illiterate. Most of all science is relatively simple, but one of the biggest issues is vernacular. And indeed within a field there is extremely good reason for this vernacular. It is not only vastly more concise than trying to obtusely explain every single concept from the ground up, but it is also more precise. Do I mean force? Do I mean momentum? Speed? Velocity? Every concept is entirely different, but in the world outside outside of the field -- none of this matters. Theories are just ideas, speed and velocity are same thing, and so on.
The point of this is, do you think my post would be clear and accurate in what it is understood to mean from the demographic that the message was directed at? I think the answer is absolutely yes. And the casual use of terms that have more precise meanings within a vernacular is in no way going to mislead them as to the meaning of what is said. Far from it, in my opinion - using more appropriate terminology is likely to lead to a less elucidating post!
Yes, and did you notice that I didn't object at all to the part of your post that corrected the "escaping Earth's gravity" misconception? That's because there was nothing wrong with it. I only objected to the part of your post that was incorrect.
> he knows nothing about orbital mechanics and, most likely, next to nothing about physics in general.
In which case the last thing you should want to do is to tell him things about physics that are wrong. Which is why I corrected the wrong thing you told him.
> There is no vernacular in my post.
My objection had nothing whatever to do with your choice of words. You made a factually incorrect statement and I corrected it. That's all there is to it.
It’s factually wrong to say that the net force is zero at the peak, when what is zero, is the vertical speed.
What was questioned was the feeling of weightlessness. You imply that weightlessness is only felt at the peak of the parabolic path. When the ball leaves the throwers hand, the only force, disregarding air resistance, acting on the ball is the force of gravity, until it hits the ground. Someone inside the ball will feel weightlessness all the way from leaving the throwers hand to hitting the ground. There is no special feeling at or around the peak.
Not everything you said, no. That's why I objected. If you want more detail, you said:
> The ball's trajectory will be a parabola. And at the highest point of that parabola the net vertical force on the ball is zero. It's where the force you exerted on it to send it up, and the force of gravity pulling it down eventually reach an equilibrium. Something inside of that ball would experience 0g at the moment when it was at its parabolic peak.
Actually, the net force on the ball (disregarding air resistance) is the same throughout the entire trajectory once it leaves your hand. (Here I'm taking the Newtonian view that considers gravity to be a force.) That's why the ball continuously accelerates downward by the same amount throughout the entire trajectory once it leaves your hand--which it has to in order for the trajectory to be a parabola. The force you exert on it to throw it upward stops as soon as it leaves your hand, so the only force thereafter is gravity. And since the force of gravity is not felt, the ball is in free fall, feeling 0 g, for the entire parabola.
So the part of your post that I quoted was not a "completely cogent and clear description of the forces"; it was a wrong description of the forces. That's why I corrected it.