0
$\begingroup$

Does the gravitation of Earth have a limit? when a body projected vertically with $v=11km/s$ (escape velocity) from Earth's surface does this means that it does not return back to Earth?

$\endgroup$
1
$\begingroup$

I assume this question is about classical mechanics. Therefore gravitational force doesn't have a limit.

According to conservation law of energy (P.S. Gravitational energy $E_p = -\frac{GMm}{r}$):

$\frac{1}{2}mv^2-\frac{GMm}{r}=0,\ when\ r\rightarrow\infty$

Therefore, the escape velocity is $v=\sqrt{\frac{2GM}{r}}$.

$\endgroup$
  • $\begingroup$ By the way, in general relativity, gravitational force transmits at the space of light. Therefore, it has a range of about 4.6 billion light years (since the Earth is 4.6 billion years old). $\endgroup$ – Qianyi Guo Apr 12 '14 at 14:56
  • $\begingroup$ Really, there's a point when gravity from Earth was non-existent and then suddenly it is there? What about the mass from the gas nebula that the solar system came from? Doesn't that count? At 4.6 billion light years out, you can't determine that the gravity is from a planetary body, or a gas cloud; it would appear to be a point. $\endgroup$ – LDC3 Apr 12 '14 at 16:03
0
$\begingroup$

Yes, this is exactly the definition of the escape velocity.

I cannot remember the number for it, but I'll trust you that it is 11.

Either way, you derive this by doing an integral, where you take the final position at infinity. That is, you assume that the particle that you shoot up into the air/ space goes off forever and never comes back.

It follows from Classical Mechanics and Newtonian gravity, so

$$ {F} = m {a} = G\frac{Mm}{r^2} $$

Then we can write

$$ a = \frac{d^2x}{dt^2} = v \frac{dv}{dx} $$

So we can set up an integral over displacement to solve for velocity.

As regards you're first question, 'Does the gravitation of earth have a limit?', the answer would be no. By General Relativity, all of the 'stuff' (matter and energy) in the universe causes gravitational interactions with everything else. This gravitational interaction would propagate at the limiting speed to infinite distance (for a massless graviton), and so even when the particle is really really really ... really far away, the Earth can still be exchanging this gravitational interaction with the particle that we 'shot' out into space.

But we must make some assumptions to 'solve' the problem in the Classical case and come up with an escape velocity.

Of course, we can also say that since we observe the accelerating cosmic expansion of the universe, if that particle went so far away that it traveled hundreds of Mpc to a distant galaxy cluster, then the expansion would mean that it's distance to us would indeed keep getting larger.

But then we would need to know what the fate of the universe is, which we do not!

If, say, there will be a 'Big Crunch', as some people say, then everything in the universe would eventually come back to the same infinitesmial small point, and it would be correct to say that 'the particle are the Earth (or at least what would be left of it) are to be found at the same place'. I don't know if that's good enough for you to say that the particle 'came back' to the Earth.

But if the Cyclic Model of the universe is correct, such as in String Cosmology and Brane World Scenarios of cosmology, the universe does not actually contract, rather some extra dimensions do, so then there would not be the particle 'comming back'.

So in short, for your second question, yes.

For your first question, we don't know for sure what will happen!

$\endgroup$
0
$\begingroup$

In classical mechanics the gravitational force does not have a spacial limit, i.e. any mass no matter how far it is from the earth will experience its gravitational pull (and vice versa).

However, this does not mean that any object thrown into space will eventually return to earth. This is due to the fact, that the gravitational force on an object becomes smaller (proportionally to the square of the distance) as it moves further away from earth. So, if you shoot a mass away at a high enough velocity (for earth this is about 11 km/s as you mentioned), the decelerating gravitational force will decrease faster than the velocity of the mass. The mass will never stop decelerating but also never halt and return. It has thus "escaped" earth's gravitation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.