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JoeyBob

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- Thread starter JoeyBob
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JoeyBob

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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.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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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.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 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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- #4

jbriggs444

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Approaches and becomes are not synonyms.So Ekf-Eki+Epf-Epi=0. I understand that the final potential energy is 0 (distance awayapproachesinfinity), but don't get why the final kinetic energybecomes0.

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What non-zero velocity "at infinity" does the object need in order not to fall back?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?

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As already pointed out the final PE and KE,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?

So, if you are going to say that KE and PE

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sophiecentaur

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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.

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JoeyBob

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Thanks guys, makes sense to me now.

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

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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- #10

nasu

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

- #11

hmmm27

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I understand that the final potential energy is 0

In contrast to the OP, I get why KEAs already pointed out the final PE and KEapproach0

like that.If you consider that gained Gravitational Potential Energy will be lost Kinetic Energy

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- #12

Nugatory

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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.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## ?

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)##PE=mgd## is too simplistic, and assumes that ##g## as a constant

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sophiecentaur

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It not just "too simplistic"; it totally wrong in the context of Escape Velocity. Gravitational Potential (Energy per unit of mass) is:PE=mgd is too simplistic, assuming g as a constant.

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

$$E=\frac{m}{2} \vec{v}^2-\frac{\gamma m M}{r}.$$

$$E=\frac{m}{2} \vec{v}^2-\frac{\gamma m M}{r}.$$

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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 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##.Yeah, but zero does not satisfy the condition to be non-zero. Maybe just approximately.

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jbriggs444

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The question was:The question was what's the minimal total energy that the particle escapes when shot radially away (angular momentum 0).

What non-zero velocity "at infinity" does the object need in order not to fall back?

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nasu

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

- #19

hxlfbloodpxince

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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?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.

- #20

jbriggs444

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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).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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To elaborate on @jbriggs444's reply: We can always consider an Earth and a satellite as a system. If the satellite orbits the Earth, it is aso 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 think it is fair to say that Halley's comet forms a bound system with the Sun but not with the Earth. However, Earth, Sun and Halley's Comet, and the rest of the planets and asteroids too, form a bound system.

- #22

sophiecentaur

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That is unless some small extra amount of energy is supplied to modify the trajectory into an elliptical orbit. You always need a second burst in a tangential direction at some stage.If it's directed towards the other body the bodies crash together at some time ;-)).

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