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This topic really screwed with my head in school and my teacher could never answer my questions so I thought I would ask here. I will break it down with mathematics and then logic.

Let's assume we only have 1 body which is earth and nothing else exist.

We can work out the escape work done using this formula where $m = 1$:

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Then transfer work done into velocity:

enter image description here

And you get the answer 11182.36 meters per second, which is correct for earth escape velocity. So the mathematics check out. The reason I'm including this is because I'm going to use the same variables(mainly ∞).

The definition for gravity is that it's always attractive and is infinite. So the change in velocity(Because we're escaping it is presumed it's moving away from the mass) is always decreasing, or $\dfrac{dy}{dx}$ is $\downarrow$ and we can assume when $R = \infty$, $\dfrac{dy}{dx}$ is $0$. So going back to the equations above to reach infinite distance your going to need infinite time(that's just the nature of how infinite works), but if we have infinite time and it's multiplied by this negative $\dfrac{dy}{dx}$ surely it's impossible to escape a gravity field. Because no matter how small, any number times $\infty$ is $\infty$ and will result in an ∞ work done.

Hopefully you can understand that, and there is 2 separate mathematics ways of looking at it and both give different answers.

That was the mathematics part and that brings me onto the logic part. If gravity is infinite how can you ever escape?

I really hope someone can answer this because I literally can't sleep because I think about this, maybe my understanding of $\infty$ is just wrong. Also don't try and use a calculator, it will get to a point where R is too large and the calculator doesn't have enough bits so just rounds to $0$.

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  • $\begingroup$ What's wrong with escaping in infinite time? $\endgroup$
    – user258881
    Aug 6, 2020 at 16:53
  • $\begingroup$ There's no mystique here. Yes, gravity acts even over infinitely far distances (but then infinitely small) but the integral for work done (for $o \to \infty$) is still finite. This is common with many decay functions $\endgroup$
    – Gert
    Aug 6, 2020 at 16:55
  • $\begingroup$ @FakeMod But in that infinite time you're only taking away energy not giving it. Because a rocket only get all it's energy at the start over that infinite time you will all ways losing energy due to gravity. If it's over an infinite time that means it will be given infinite work done and there for it hasn't escaped the gravitational field as we can't supply a rocket with infinite energy and there is no where for it to gain more energy. $\endgroup$ Aug 6, 2020 at 17:27
  • $\begingroup$ @Gert gravitational force is only 0 when R = ∞ $\endgroup$ Aug 6, 2020 at 17:31
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    $\begingroup$ That one-before-last comment was pure gobbledigook. Incomprehensible wordsalad. $\endgroup$
    – Gert
    Aug 6, 2020 at 17:34

1 Answer 1

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“Escape velocity” is mis-named in two respects.

First, it a speed not a velocity (it is a scalar quantity, not a vector).

Secondly, it is not the minimum speed needed to (ballistically) escape from a planet’s gravitational field since, as you point out, this is impossible. It is the minimum speed needed to get as far away from the planet as you like (again, on a ballistic trajectory) given an unlimited amount of time.

But “minimum speed for an unbounded ballistic trajectory” does not sound as snappy as “escape velocity”.

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  • $\begingroup$ @Gert Indeed. Which is why I emphasised ballistic i.e. unpowered trajectories. $\endgroup$
    – gandalf61
    Aug 6, 2020 at 17:41

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