How is escape velocity possible? No matter how far away an object is from the Earth...

How is escape velocity possible? No matter how far away an object is from the Earth, it will still experience a small gravitational force. Is this force and acceleration towards the Earth just so negligible we ignore it? Can anything truly escape the Earth's pull?

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All objects have their own gravity, so at some point the object's gravity will be stronger than the effect of gravity from the Earth. Why would it then fall back to Earth?

This isn't the reason why. Even if Earth was the only object in there universe there exists a velocity such that you'll escape and never come back.

Wrong. An object can't attract itself. Anyone who's taken calc based physics should know that.

en.wikipedia.org/wiki/Escape_velocity#Deriving_escape_velocity_using_calculus

If you don't know calculus, it's basically because the gravitational potential energy approaches zero as the object gets farther and farther from the Earth

It's not attracting itself (well it kinda is, all the particles that make it up are attracted to each other). If you think of it like that bedsheet spacetime remember that the craft will make an indent in the sheet too.
At some point distant from the Earth the bedsheet will be sufficiently horizontal that the craft's indent will cause it not to have any attraction to the Earth.

As you say, the gravitational potential approaches 0 as you get further away from the Earth, but it would never reach 0 if your craft didn't have mass and thus it's own gravity.

That velocity is just one that'll take your craft far enough that it's own gravitational potential overcomes that of the Earth.

It's a matter of potential energy. PE= GMm/R
G is the gravitational constant, M the mass of the Earth, m the mass of some object, and R the distance between the mass-centers of the Earth and the object.

When R is infinite, PE is zero. As an object falls, PE decreases (becomes negative) and the loss shows up as kinetic energy. The object accelerates until it hits the Earth's surface. So an object can only gain so much speed no matter how far it falls. That's the Escape Velocity.

Conversely, an object moves slower and slower if projected upwards from the Earth. But the Earth's field can only subtract that finite amount of energy from it. If the initial velocity is greater (even a tiny bit) than the Escape Velocity, it will still have some kinetic energy (upwards velocity) even after receding an infinite distance. In other words, it'll never fall back.

are gibberish. An object's self-gravitation (left side pulling on right-side and vice-versa) has zero effect on its motion.

and are right.

>An object's self-gravitation (left side pulling on right-side and vice-versa) has zero effect on its motion.
That's not even the point you dufus.

But that's what is claiming.
The object's own gravity will be stronger than the gravity from Earth. That would make escape velocity dependent on the mass and density of the object.
Wrong! Wrong! Wrong!

An object's mass and density control the Roche Limit but have zero connection to Earth's escape velocity.

Gravity is an inverse square law (although it might not be at intergalactic distances according to entropic gravity theory) meaning it falls off VERY fast

Gravity propagates at the speed of light so basically no you will never truly escape earth’s pull but there’s no point in calculating it past a certain distance as you can see from the picture you posted

As no object can reach infinity in practice, does that not mean there is always a minuscule amount of PE?

You can never escape it if the Earth and You are the only two massive objects in the universe.

That is not the case, even though you always experience a force from every massive object (therefore the Earth) at a certain point the attraction from one body would be greater than that of the Earth's. In that sense, you "escape" earth's gravity in favor of another body's.

It decreases asymptotically

Correct.
No matter how far you go, you'll always have a little potential energy which could be lost. And you'll always feel the retarding pull of Earth, even light years away. I mean, your velocity relative to Earth will decrease forever.

You're forgetting inertia.

Your question has been answered already but its also worth considering how absurdly weak gravity is compared to every other force.

... No?

If kinetic energy > potential energy it's not gravitationally bound.

Your velocity will decrease but the rate at which your moving away from Earth will be such that the gravitational pull is getting weaker faster than you're slowing down. If you take the limit as t-> infinity you're asymptotically free.

how so?

two objects with a relative velocity to each other of 0, separated from each other in an empty, non expanding universe with the same laws of gravity would attract, no matter the distance
gravity is exponential, but the effect isn't non-zero

space-time and shiet is why it's actually zero at certain distance

If Earth and the object are the only masses in the system then there comes a point where the gravitational influence is so low that it cannot overcome inertia to attract the object back to it.

Nice trips, also I see the confusion, I was making a point about the net force exerted by gravity, as if you were the only pair of objects you'd constantly be accelerating towards each other, even if indeed your inertia was so high you'd never actually reverse direction. Using different senses of the word escape

Leads into my next point where the direction of the net gravitational force can point away from earth if there are more bodies

What about eliminating gravitational pull.

>gravitational potential energy approaches zero
mega-brainlet pants-wetting retard checking in

it never becomes zero, right?

Any number divided by infinity is so close to 0 that it's actually considered exactly 0. This is used all over calculus and physics concepts.

so can the adage about all of the bodies always exerting some tug on all the other be said to be correct? Or does a moon on the on the other side of the universe exert nothing on me?

It's called a limit. It approaches 0 but never actually reaches it. It just gets closer each time you do it. If you divide one by two, you get a half. That by two, a quarter. An eighth. A sixteenth. A thirty-second. Same thing is going on here. It's how inverse functions work.

Has a point because you can approximate it as zero, like when integrating a function like this (it's called an improper integral).

>It's called a limit. It approaches 0 but never actually reaches it
So, I take it can always be said to be "something," but what about , how can I reconcile it being both zero and something.

I need maths, don't I?

Yes.
The the effect of the gravity travels at the speed of light, so there are parts of the universe it doesn't reach anymore (parts outside the observable universe) and maybe parts it never reached.
But anywhere within the observable universe within the number of light years equal or less than the age of the Moon has part of the Moon's gravitational influence in it.

It never becomes zero, but as I noted inertia is an issue.

>so there are parts of the universe it doesn't reach anymore (parts outside the observable universe)
What about inflation models? Couldn't gravity had been flung further out of the horizon quickly at the beginning, before light even began its march? Shouldn't I be able to "feel" that?

The magnitude of gravitational force on the object decreases fast enough so that the Earth's gravity will never perform enough work on the object to bring it to stop it's outward movement.

>You can never escape it if the Earth and You are the only two massive objects in the universe.
Yes you can.
See

Yes there is always a pull but .0000000000001 N is comparatively 0 to something even as small as 1N so we write it as 0

Neat, I remember asking my physics teacher in high school the same question.

The force of gravity between two masses is inversely proportional to the distance between them. That means the farther and farther a rocket gets from Earth, the force with which gravity pulls it back to Earth also decreases. At escape velocity, it just means you're moving fast enough that the force of gravity accelerating you towards Earth would take infinitely long to actually bring your speed down to "zero" and then begin pulling you back.

You do slow down, but the rate at which you slow down is also diminishing as you get farther away.

Square of the distance.

It doesn't escape the earth's pull, it just moves away despite being pulled

brainlet here again, farting in the bath and cracking myself up

So, doesn't the fact that you use zero render it computationally a non-factor?

See

Yes.
At a point the gravitational force acting on the object is less than the inertia of the object so the gravitational force can no longer alter its path or velocity. Even if the gravitational force is not 0 it is functionally 0 for all it can do.

>the absolute state of Veeky Forums

>it just moves away despite being pulled
the definition of escaping.

fucking brainlets

Please don't post about things you have less than a high school education on.

if something is moving away from you, despite any of your efforts to keep up with it, or get a hold on it, it is in fact escaping

Okay, last question and I'll get back to eating my play-dough.

If the effect is zero for all intents and purposes, why do we even bother saying the influence exists at all? If it's not important to the mathematics or a "non-factor," couldn't we just as easily say the effect "drops off," or am I dancing around something fundamental because I don't calculus?

Ooga booga nig nog terrier

I swear to god this board is either filled with the most incorrigible retards or people who are masterfully talented at trolling

Because we can technically still measure it (see LIGO).
Also, technically, if another object was exerting gravitational influence on that object and the two gravitational influences were above the object's inertia, then their additive influences would affect the object.

So it is still there, but when it comes to affecting the object's motion it's not worth factoring into an equation.

It's like dropping a down feather on a stable bowling ball. It would need to be on the cusp of being able to move for that down feather to have any effect, but the feather exists regardless.

So, there's an equation for it, (GMm)/r^2 and when r approaches infinity, or the mass/distance ratio is too small, we don't consider it at all. (unless told to for a very specific reason i.e. to show how weak gravity is)

Like, when we calculate the acceleration due to gravity of earth to be 9.81, we don't say "Well, let's include Jupiter, Saturn, etc because they exist and are relatively close compared to infinity" It's just too small of a number

As long as a number is close to 0, then it most certainly is a "non-factor", as you say.

I just did the calculation, and Jupiter exerts 0.0000034 N on me. That's incredibly small for something pretty close and very large.

>mfw nobody got that it's bait

Actually, it wasn't bait. The mistake I made was referring to the object's gravity instead of just its mass, or gravitational mass.

I mean, you could just go mgR = 0.5mv^2 => v = sqrt(2gR) at that rate, as mgR here would be a valid expression for GPE.

Imagine it as an integral.
When you are over the escape velocity, you escape earth fast enough that the (decreasing) gravitational pull can't slow you down fast enough. And you still have residual velocity at infinite distance. Otherwise you can't escape fast enough and the earth slows you down and flips your speed vector around.