I am a little confused about escape velocity. Does the escape velocity always approach 0 as we go to an infinitely far distance even if there isn't any friction?
If so, why does it approach 0? Shouldn't it be moving with constant speed?
$\begingroup$
$\endgroup$
3
-
11$\begingroup$ "Shouldn't it be moving with constant speed" - note that zero is a constant speed, there is no contradiction here. $\endgroup$– Nuclear HoagieDec 8, 2021 at 21:50
-
$\begingroup$ Aside: should we redefine escape velocity to account for the expansion rate of the universe? I.e., should it be defined not by the infinite distance concept, but the velocity required to distance from the reference body to the point where gravitational acceleration is exactly equal to the expansion of the intermediate space? $\endgroup$– Cristobol PolychronopolisDec 9, 2021 at 16:33
-
1$\begingroup$ If I'm infinitely far away, haven't I already escaped, so don't need any motion to "escape more"? $\endgroup$– Eric TowersDec 9, 2021 at 17:03
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
Escape velocity is a minimum velocity a body at a given point must have to escape the gravitational field of some other body.
Escape velocity depends on initial location of the body. If the initial position is very far from this other body you only need to push it a little bit and it will fly away and never return. The escape velocity is very small, almost zero.
If initial velocity of a body is higher than escape velocity it will fly away, never return, and after a long time far away from the source of gravitational field the velocity of our body will be constant and not zero. But this velocity is not an escape velocity (not sure if there is some special name for it)
-
16$\begingroup$ The remnant constant velocity you mention is known as the hyperbolic excess velocity, or $v_\infty$. $\endgroup$ Dec 8, 2021 at 21:09
-
1$\begingroup$ Note that initial velocity higher than escape velocity, isn't enough of a criterion. For example if its headed on collision course with the other body $\endgroup$– StilezDec 9, 2021 at 12:02
-
2$\begingroup$ @Stilez Agree. Interesting fact - escape velocity does not depend on direction. Yes, you have to choose initial velocity direction in such a way that objects would not collide, but except for that the direction does not matter. F.e. if both bodies are mathematically small you only need not to direct the escaping body exactly into the other one. It does not matter if you direct the velocity exactly away or almost into or prependicular - the minimum required to escape will be the same. $\endgroup$– lesnikDec 9, 2021 at 12:33
$\begingroup$
$\endgroup$
I think there is some confusion in the phrasing of your question - there are 2 interpretations:
If you are at an infinite distance from a body then you have already 'escaped' - so 0 velocity is required to achieve that.
However - I THINK you mean why does the escaping body HAVE zero velocity when it has reached 'infinite' distance. This is because, by definition, the escape velocity is such that the kinetic energy of the body is equal to the total potential energy required to reach infinite distance' i.e. once it HAS reached that position then its kinetic energy is zero.
If the body started at MORE than escape velocity then it would have 'excess' KE and hence would have a remaining velocity once it reaches 'infinity'
$\begingroup$
$\endgroup$
Time reversal : Free fall
I find it easier to understand the concept of escape velocity by reversing the time. It's okay to do so because we ignore any friction (e.g. atmospheric drag).
We consider one primary body (e.g. the Earth) and another object (e.g. a rock), in an otherwise empty universe.
At $t=0$, the rock is extremely far away from the Earth. It doesn't matter how far away, it will still feel some attraction from Earth. The force will be extremely small, but since there's no other force applied to the rock, it will very slowly begin to move towards the Earth. The closer it gets, the more it will be attracted, and the higher its acceleration.
Finally, after a long free fall, it will crash against the Earth at a speed of $\approx 11\frac{\mathrm{km}}{\mathrm{s}}$.
Escape velocity
We ignored any friction, which means that in theory, the process would work just as well in reverse:
we can shoot the rock away from Earth with a powerful cannon, at $\approx11\frac{\mathrm{km}}{\mathrm{s}}$, and it will land back to the starting point, very far away from Earth, with no velocity.
In order to see what happens at infinity, you can simply repeat the experiment, increasing the rock-Earth distance each time. The impact speed will slowly increase and tend to $\approx 11.186\frac{\mathrm{km}}{\mathrm{s}}$, the escape velocity!
