<blockquote>
<p>Why is final kinetic energy zero?</p>
</blockquote>
<p>From a comment by the original poster, &quot;this is regarding escape velocity&quot;.</p>
<p>The answer is simple: That's how escape velocity is defined.</p>
<p>Gravitational potential energy is always non-positive and is only zero at an infinite distance. Kinetic energy is always non-negative and is only zero when the relative velocity is zero. This means that the total energy, gravitational potential energy plus kinetic energy, of a pair of objects orbiting one another can be negative, positive, or zero. This total energy is a constant in Newtonian mechanics.</p>
<p>If the total energy is negative the objects remain forever bound to one another. They don't escape. They instead follow elliptical or circular orbits about one another.</p>
<p>If the total energy is positive the objects do escape from one another and eventually (in an infinite amount of time) become separated by an infinite distance. The objects follow hyperbolic arcs as they separate. Gravitational potential energy drops to zero at an infinite distance, but since total energy is conserved, the kinetic energy must be positive even at this infinite separation.</p>
<p>Finally, if the total energy of the system is exactly zero, the objects do escape from one another following parabolic arcs. Eventually (in an infinite amount of time at an infinite distance), the gravitational potential becomes zero, just as it does when the total energy is positive. Since the total energy is constantly zero, the kinetic energy must also become zero at an infinite distance in this case.</p>
<p>This special case of zero total energy is how escape velocity is defined. It's a very useful but unachievable concept. Useless concepts in science tend to be discarded. Escape velocity is unachievable in the sense that the probability of obtaining exactly zero total energy is exactly zero. However, the concept is very useful and is easy to calculate. Equal or exceed this easily calculable velocity and you escape. Don't exceed it and you don't escape.</p>

CC BY-SA 4.0
2023-07-24T20:57:00.047Z
I don't quite follow the last paragraph about unachievability. A smoothly accelerating body moving from sub- to super-escape velocity must at some point be moving at *exactly* escape velocity, at which point, the negative PE is exactly balanced by the positive KE, for a total energy of exactly zero. What isn't achievable is actual *escape*, since that requires traveling an infinite distance.
Nuclear Hoagie
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2023-07-25T02:15:02.590Z
@NuclearHoagie Hitting **exactly** zero is a probability space  of measure zero.
David Hammen
CC BY-SA 4.0
2023-07-25T12:54:11.620Z
I still don't follow. You can't explicitly set your velocity to *exactly* escape velocity as that would require infinite precision. But since velocity and acceleration are continuous values, you *cannot avoid* moving at exactly escape velocity when accelerating past it. Every object that has ever escaped the earth has achieved an instantaneous velocity of exactly the escape velocity value - in what sense is escape velocity unachievable? I can't tell if this is just a general comment about physically realizable values having limited precision.
Nuclear Hoagie
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2023-07-26T13:41:02.157Z
@NuclearHoagie Yes, your accelerating vehicle will pass through exactly the point where total energy (potential plus kinetic) is identically zero. Stopping acceleration exactly at that point is impossible. Or more precisely, it is a [probability space of measure zero](https://math.stackexchange.com/questions/3625725).
David Hammen
CC BY-SA 4.0
2023-07-26T14:51:58.223Z
Measure zero doesn't imply "impossible" or "unachievable", though. When you stop accelerating you are clearly traveling at *some* velocity, despite all velocities having a probability space of measure zero. Zero probability doesn't imply impossible, things that occur "almost never" are possible yet have zero probability. By this logic, *all* velocities are "unachievable".
Nuclear Hoagie
CC BY-SA 4.0
2023-07-24T10:43:50.190Z
2023-07-24T10:49:03.987Z
2023-07-24T10:49:03.987Z
David Hammen
<p>Regarding escape velocity, you are ending your trip at the highest point, at maximum <span class="math-container">$E_p$</span>. At this point, you will have a minimum <span class="math-container">$E_k$</span>, and the lowest kinetic energy is zero.  The principle that <span class="math-container">$E_p+E_k=\text{constant}$</span> is &quot;conservation of Energy (not of momentum) and assumes no friction losses, or rocket power.</p>
<p>If your kinetic energy is not zero at that point you would have energy to go higher. So if you have energy left over at the end of your trip you must have had excess energy at the start: ie you were not going at minimum speed at the start.</p>
<p>That is if the final <span class="math-container">$E_k \neq 0$</span> then the initial speed was not minimum.  Equivalently, if the initial speed is minimum, the final <span class="math-container">$E_k = 0$</span></p>
<p>This is not true if, for example, your trajectory does not end at a maximum <span class="math-container">$E_p$</span>, for example, the minimum speed to project a stone so it goes around the moon and lands where it started does not have <span class="math-container">$E_k=0$</span> at the end of the trajectory.</p>

CC BY-SA 4.0
2023-07-23T09:17:49.763Z
2023-07-23T10:50:48.450Z
2023-07-23T10:50:48.450Z
James K
<p>If your final kinetic energy is positive, then you have escaped from the gravity well and will never return -- when time is at infinity, when the gravity well has pulled you as much as it can ever pull, you still have some kinetic energy left over to keep going. That's what 'escape' means. (Don't think too hard about what &quot;keep going <em>after</em> infinite time&quot; could possibly mean, it's just a limit equation.)</p>
<p>If your final kinetic energy is negative (which is physically impossible), then you didn't escape; the negative value is indicating that you never reached that 'final' infinitely-far point. Instead, sometime before infinity, you fell back down, so that means you're still in orbit (even if that orbit is billions of years long).</p>
<p>Now, if you can have a positive value that means 'escape' and a negative value that means 'orbit', there's a line between those two, and that line is where the final kinetic energy is exactly zero: in infinite time, you come to a complete stop and never return, but you don't keep going either.</p>
<p>So that zero outcome is what we use to calculate escape velocity. It's drawing a line where if your speed goes past that line, you go up and never come down again.</p>

CC BY-SA 4.0
2023-07-24T20:43:31.773Z
2023-07-25T17:55:09.287Z
2023-07-25T17:55:09.287Z
Darth Pseudonym
<p>When the question is about the minimum speed to reach a certain point, we would usually use conservation of momentum, i.e. <span class="math-container">$\mathrm{E_k + E_p}$</span>(initial) = <span class="math-container">$\mathrm{E_k(0) + E_p}$</span>(final). I am curious in why we take the final <span class="math-container">$\mathrm{E_k}$</span> to be 0.</p>
<p>Senior high level.</p>

CC BY-SA 4.0
2023-07-23T05:57:40.133Z
Hello Xuan,  can you explain the relevance of this question to astronomy, as posed it looks like a physics homework question. . .
James K
CC BY-SA 4.0
2023-07-23T08:44:06.037Z
this is regarding escape velocity
Just want to know more
<astrophysics>
Why is final kinetic energy zero for escape energy calculations?
Just want to know more
# Why is final kinetic energy zero for escape energy calculations?

(Asked by: username_0 on 2023-07-23)

When the question is about the minimum speed to reach a certain point, we would usually use conservation of momentum, i.e. $\mathrm{E\_k + E\_p}$(initial) = $\mathrm{E\_k(0) + E\_p}$(final). I am curious in why we take the final $\mathrm{E\_k}$ to be 0.

Senior high level.

**Tags:** astrophysics

> **Comments:**
> **username_1**: Hello Xuan,  can you explain the relevance of this question to astronomy, as posed it looks like a physics homework question. . .  
> **username_0**: this is regarding escape velocity  

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## 3 answers

### Answer by username_2 on 2023-07-24. Score: 5

> Why is final kinetic energy zero?

From a comment by the original poster, "this is regarding escape velocity".

The answer is simple: That's how escape velocity is defined.

Gravitational potential energy is always non-positive and is only zero at an infinite distance. Kinetic energy is always non-negative and is only zero when the relative velocity is zero. This means that the total energy, gravitational potential energy plus kinetic energy, of a pair of objects orbiting one another can be negative, positive, or zero. This total energy is a constant in Newtonian mechanics.

If the total energy is negative the objects remain forever bound to one another. They don't escape. They instead follow elliptical or circular orbits about one another.

If the total energy is positive the objects do escape from one another and eventually (in an infinite amount of time) become separated by an infinite distance. The objects follow hyperbolic arcs as they separate. Gravitational potential energy drops to zero at an infinite distance, but since total energy is conserved, the kinetic energy must be positive even at this infinite separation.

Finally, if the total energy of the system is exactly zero, the objects do escape from one another following parabolic arcs. Eventually (in an infinite amount of time at an infinite distance), the gravitational potential becomes zero, username_0 as it does when the total energy is positive. Since the total energy is constantly zero, the kinetic energy must also become zero at an infinite distance in this case.

This special case of zero total energy is how escape velocity is defined. It's a very useful but unachievable concept. Useless concepts in science tend to be discarded. Escape velocity is unachievable in the sense that the probability of obtaining exactly zero total energy is exactly zero. However, the concept is very useful and is easy to calculate. Equal or exceed this easily calculable velocity and you escape. Don't exceed it and you don't escape.

> **Comments:**
> **username_3**: I don't quite follow the last paragraph about unachievability. A smoothly accelerating body moving from sub- to super-escape velocity must at some point be moving at *exactly* escape velocity, at which point, the negative PE is exactly balanced by the positive KE, for a total energy of exactly zero. What isn't achievable is actual *escape*, since that requires traveling an infinite distance.  
> **username_2**: @username_3 Hitting **exactly** zero is a probability space  of measure zero.  
> **username_3**: I still don't follow. You can't explicitly set your velocity to *exactly* escape velocity as that would require infinite precision. But since velocity and acceleration are continuous values, you *cannot avoid* moving at exactly escape velocity when accelerating past it. Every object that has ever escaped the earth has achieved an instantaneous velocity of exactly the escape velocity value - in what sense is escape velocity unachievable? I can't tell if this is username_0 a general comment about physically realizable values having limited precision.  
> **username_2**: @username_3 Yes, your accelerating vehicle will pass through exactly the point where total energy (potential plus kinetic) is identically zero. Stopping acceleration exactly at that point is impossible. Or username_0 precisely, it is a [probability space of measure zero](https://math.stackexchange.com/questions/3625725).  
> **username_3**: Measure zero doesn't imply "impossible" or "unachievable", though. When you stop accelerating you are clearly traveling at *some* velocity, despite all velocities having a probability space of measure zero. Zero probability doesn't imply impossible, things that occur "almost never" are possible yet have zero probability. By this logic, *all* velocities are "unachievable".  

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### Answer by username_1 on 2023-07-23. Score: 3

Regarding escape velocity, you are ending your trip at the highest point, at maximum $E\_p$. At this point, you will have a minimum $E\_k$, and the lowest kinetic energy is zero. The principle that $E\_p+E\_k=\text{constant}$ is "conservation of Energy (not of momentum) and assumes no friction losses, or rocket power.

If your kinetic energy is not zero at that point you would have energy to go higher. So if you have energy left over at the end of your trip you must have had excess energy at the start: ie you were not going at minimum speed at the start.

That is if the final $E\_k \neq 0$ then the initial speed was not minimum. Equivalently, if the initial speed is minimum, the final $E\_k = 0$

This is not true if, for example, your trajectory does not end at a maximum $E\_p$, for example, the minimum speed to project a stone so it goes around the moon and lands where it started does not have $E\_k=0$ at the end of the trajectory.


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### Answer by username_4 on 2023-07-24. Score: 0

If your final kinetic energy is positive, then you have escaped from the gravity well and will never return -- when time is at infinity, when the gravity well has pulled you as much as it can ever pull, you still have some kinetic energy left over to keep going. That's what 'escape' means. (Don't think too hard about what "keep going *after* infinite time" could possibly mean, it's username_0 a limit equation.)

If your final kinetic energy is negative (which is physically impossible), then you didn't escape; the negative value is indicating that you never reached that 'final' infinitely-far point. Instead, sometime before infinity, you fell back down, so that means you're still in orbit (even if that orbit is billions of years long).

Now, if you can have a positive value that means 'escape' and a negative value that means 'orbit', there's a line between those two, and that line is where the final kinetic energy is exactly zero: in infinite time, you come to a complete stop and never return, but you don't keep going either.

So that zero outcome is what we use to calculate escape velocity. It's drawing a line where if your speed goes past that line, you go up and never come down again.


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