# Escape Velocity Question: Why is the final kinetic energy = 0?

JoeyBob
So Ekf-Eki+Epf-Epi=0. I understand that the final potential energy is 0 (distance away approaches infinity), but don't get why the final kinetic energy becomes 0. If the final kinetic energy was 0, wouldn't that mean the object no longer has any velocity and would start being effected by the gravity again, even if it was really far away from the planet?

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If the final kinetic energy was 0, wouldn't that mean the object no longer has any velocity and would start being effected by the gravity again, even if it was really far away from the planet?
The final kinetic energy is zero only at infinity, which it never reaches. At any finite distance the object is moving away, so it will never return no matter how long you wait. With any lower initial velocity it would come to a stop at finite distance in finite time and then return. But with escape velocity it never quite stops.

JoeyBob and sophiecentaur
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If the final kinetic energy was 0, wouldn't that mean the object no longer has any velocity and would start being effected by the gravity again, even if it was really far away from the planet?
You are correct. When the kinetic energy drops to zero, the mass will fall back or orbit around the Earth. Gravity is a long range force that acts on masses even if they are in a galaxy far, far away.

You might think of the escape velocity as the threshold value that separates bound from unbound systems. If the initial kinetic energy of the mass is greater that ##\frac{1}{2}m v_{\text{esc}}^2##, then the mass will always have some kinetic energy left no matter how far it is from the Earth and the mass is considered no longer bound to the Earth. In another thread you have indicated that you are in a chemistry-related field. Think of the quantity ##\frac{1}{2}m v_{\text{esc}}^2## as the gravitational analogue of the ionization energy.

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JoeyBob
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So Ekf-Eki+Epf-Epi=0. I understand that the final potential energy is 0 (distance away approaches infinity), but don't get why the final kinetic energy becomes 0.
Approaches and becomes are not synonyms.

JoeyBob, nasu, phinds and 1 other person
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Summary:: Why final kinetic energy = 0?

So Ekf-Eki+Epf-Epi=0. I understand that the final potential energy is 0 (distance away approaches infinity), but don't get why the final kinetic energy becomes 0. If the final kinetic energy was 0, wouldn't that mean the object no longer has any velocity and would start being effected by the gravity again, even if it was really far away from the planet?
What non-zero velocity "at infinity" does the object need in order not to fall back?

Staff Emeritus
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Summary:: Why final kinetic energy = 0?

So Ekf-Eki+Epf-Epi=0. I understand that the final potential energy is 0 (distance away approaches infinity), but don't get why the final kinetic energy becomes 0. If the final kinetic energy was 0, wouldn't that mean the object no longer has any velocity and would start being effected by the gravity again, even if it was really far away from the planet?
As already pointed out the final PE and KE, approach 0 as the distance approaches infinity. But the gravitational force also approaches 0 as the distance approaches infinity ( in the way that 1/x^2 approaches 0 as x approaches infinity.)
So, if you are going to say that KE and PE become 0 at infinity, you'd also have to say that the gravitational force acting on the object becomes 0 also.

PeroK, Ibix and jbriggs444
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Escape velocity is a good measure of the minimum rocket energy you need but it has its limitations because, for a non-orbital journey, you will mostly want to be going a lot faster than that - in order to get somewhere in a useful period of time. If you leave Earth with just a solar escape velocity, by the time you get to Neptune's orbit, you would be going pretty slowly (and it would take you a long time). Leaving the Solar System is always done using a slingshot orbit so you can't really compare the theoretical case of a rocket escaping from Earth with just its own fuel AND making it out of the Solar System too.
If you look at the orbital speeds of the planets, you find that Neptune's speed is about 1/6 of Earth's speed and that gives you an idea of the relative speeds of a simple projectile at those radii.
If you consider that gained Gravitational Potential Energy will be lost Kinetic Energy on the way out, going four times as far out as Neptune would involve halving the speed.
The speed of Voyager craft is about three times the orbital speed of Neptune and, as the graph in that link shows, it's hardly slowing down at all because it started off (with a little slingshot help) with a much higher speed than escape velocity from Earth orbit. They went past Neptune in the late 1980s iirc and are passing the heliopause now.

hmmm27
JoeyBob
Thanks guys, makes sense to me now.

Ibix and PeroK
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What non-zero velocity "at infinity" does the object need in order not to fall back?
The answer is, of course, 0 ;-).

The solutions for the Kepler problem for ##E=0## are parabolae, i.e., unbound trajectories. For the zero angular-velocity case the bodies move radially away from each other approaching 0 velocity at infinite distance.

Proof: The solution of the radial equation of motion for ##L=0## and ##E=0##,
$$\frac{\mu}{2} \dot{r}^2-\frac{\alpha}{r}=0 \quad \text{with} \quad \alpha=\gamma m_1 m_2, \quad \mu=\frac{m_1 m_2}{m_1+m_2}$$
is (using Mathematica ;-))
$$r(t)=\left (r_0^{3/2} +\sqrt{\frac{9 \alpha}{2m}} t \right)^{2/3}.$$

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Delta2
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The answer is, of course, 0 ;-).
Yeah, but zero does not satisfy the condition to be non-zero. Maybe just approximately.

sophiecentaur
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I understand that the final potential energy is 0
As already pointed out the final PE and KE approach 0
In contrast to the OP, I get why KE is approaches zero at infinity, but why isn't PE equal to approaching ##mv_{init}^2/2## ? ##PE=mgd## is too simplistic, assuming ##g## as a constant.
If you consider that gained Gravitational Potential Energy will be lost Kinetic Energy
like that.

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Mentor
In contrast to the OP, I get why KE is approaches zero at infinity, but why isn't PE equal to approaching ##mv_{init}^2/2## ?
The difference between the initial potential energy and the final potential energy approaches ##mv_{init}^2/2## (when ##v_{init}## is exactly equal to the escape velocity from the initial height) as the kinetic energy approaches zero. We have made the arbitrary but convenient (when grinding through the algebra) choice to set the potential energy at infinity to zero; thus the initial potential energy is negative.
##PE=mgd## is too simplistic, and assumes that ##g## as a constant
True, but this just means that to calculate the potential we have to integrate the non-constant force over the distance instead of simply multiplying a constant force by the distance. (Actually, that latter calculation is an integral too, it just so happens to immediately simplify down to a multiplication, which is why it's taught in intro classes where the students may not have already learned calculus)

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PE=mgd is too simplistic, assuming g as a constant.
It not just "too simplistic"; it totally wrong in the context of Escape Velocity. Gravitational Potential (Energy per unit of mass) is:
E = MG/r
where M is the mass of the central attractor and G is the universal gravitational constant. You can only use g when you are very near a planet and where you can also treat the ground as flat etc. etc.. To understand orbits and trajectories in space you would never start with mgh.

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The total energy, of course, reads
$$E=\frac{m}{2} \vec{v}^2-\frac{\gamma m M}{r}.$$

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Yeah, but zero does not satisfy the condition to be non-zero. Maybe just approximately.
The question was what's the minimal total energy that the particle escapes when shot radially away (angular momentum 0). The answer to that question is exactly 0 (not only approximately). It's also easy to see by drawing the potential and some constant total energy lines in the same diagram. Then you'd not even need to solve the equation of motion to get this result. You just read it off from the diagram, considering that the physically allowed region is given by ##V(r) = E-T \leq E## since ##T=m v^2/2 \geq 0##.

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The question was what's the minimal total energy that the particle escapes when shot radially away (angular momentum 0).
The question was:
What non-zero velocity "at infinity" does the object need in order not to fall back?

nasu
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Then an infinitesimal velocity is sufficient. If it's directed away from the other body both bodies just fly appart forever. If it's directed towards the other body the bodies crash together at some time ;-)).

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The mention about zero being just approximately non-zero was intended as a joke. Sorry.😀

sophiecentaur and vanhees71
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You are correct. When the kinetic energy drops to zero, the mass will fall back or orbit around the Earth. Gravity is a long range force that acts on masses even if they are in a galaxy far, far away.

You might think of the escape velocity as the threshold value that separates bound from unbound systems. If the initial kinetic energy of the mass is greater that ##\frac{1}{2}m v_{\text{esc}}^2##, then the mass will always have some kinetic energy left no matter how far it is from the Earth and the mass is considered no longer bound to the Earth. In another thread you have indicated that you are in a chemistry-related field. Think of the quantity ##\frac{1}{2}m v_{\text{esc}}^2## as the gravitational analogue of the ionization energy.
so actually we are considering satellite and Earth as a system but when satellite breaks the limit to infinty its no longer the part of the system?

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so actually we are considering satellite and Earth as a system but when satellite breaks the limit to infinty its no longer the part of the system?
I would not use those words. We are free to draw system boundaries where we please. The distinction is between a pair of objects that continue separating further and further forever (an unbound system) and a pair of objects that never move more than a particular distance apart (a bound system).

vanhees71