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I would like some help in understanding the derivation of the formula for escape velocity. Why does $t$ approach infinity? And why is $v(t)$ approaching zero? I know that $r(t)$ is infinite because the gravitational field is infinite. But I don't understand why $t$ (time) and velocity would be zero.

I don't think I understand what $v(t)$ is. Is it velocity at a given point in time? And when time increases, you are farther away from the body producing the gravitational field and thus velocity needed for escape is lower? ....

Here is a photo of the derivation:

From the wikipedia page: http://en.wikipedia.org/wiki/Escape_velocity

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2 Answers 2

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You are right in saying that v(t) is the velocity at a given time.

1) The first equation is the equation for conservation of energy for an object in a gravitational field. The left side is the change in kinetic energy and the right side is the change in gravitational potential energy.

2) For escape velocity, we consider what happens a looooong time after the rocket is launched, so we let t go to infinity. For a rocket launched exactly at escape velocity, it has just enough energy to reach a distance infinitely far away from the planet, so r goes to infinity. If the rocket has "just enough energy" to go infinitely far, that means that once it is at r = infinity, its velocity must be 0, since there would be no more leftover energy. Hence v(t) also approaches 0.

3) The article substitutes these values in to the energy equation and solves for the initial velocity, which is the escape velocity.

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Thank you for the explanation! @Qmechanic thank you for adding the photo. I can't add photos because I don't have 10 rep points. –  EngieOP Feb 8 '13 at 23:31

Your interpretation of $v(t)$ is correct. It is indeed the instantaneous velocity at time $t$. Now the idea is the following. The gravitational potential goes to zero for $r \rightarrow \infty$. Since the velocity is finite at all times, the average velocity $$\langle v \rangle = \frac{r}{t}$$ must also remain finite (where $t$ is the time passed since the launch of the projectile). Now, since $r\rightarrow\infty$, $\langle v \rangle$ can only remain finite if also $t\rightarrow\infty$. So that's a different way of showing why we let $t$ go to infinity.

Of course, you can look at it the other way around as well. Let $t$ approach infinity. Since we have a finite velocity at all times, the distance travelled $r$ will also approach infinity because the projectile has been travelling at a finite velocity for an infinite period of time.

Your reasoning for why $v \rightarrow 0$ is also correct. If the gravitational potential is zero, there is no need for any velocity. In fact, this defines the escape velocity. It is the velocity needed to come to an exact stop at an infinite distance away from the body you want to escape from (or, equivalently, at an infinite time after launch). The rest of the calculation is very straightforward and I think you didn't look for any help on that part so I'll leave it at this.

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Quick comment - I'm not sure that your statement that $v < c$ is accurate for all $v_0$ in this scenario, since the given energy conservation equation is non-relativistic. –  Draksis Feb 8 '13 at 23:39
    
Well I didn't specify the equations to be used, but perhaps it is an unnecessary addition to the conceptual idea that the velocity is finite at all times. I'll remove it. –  Wouter Feb 8 '13 at 23:43

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