Question regarding Kleppner Kolenkow Example 5.5 "Escape Velocity—the General Case"

Question

In example 5.5 from Kleppner Kolenkow (2nd edition), the general case of finding the escape velocity of a mass $m$ projected from the earth at an angle $\alpha$ with the vertical, neglecting air resistance and Earth's rotation, is presented.

Their analysis goes as follows:

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The force on $m$, neglecting air resistance, is

$$\mathbf{F} = -mg\dfrac{R_e^2}{r^2}\mathbf{\hat{r}}$$

where $\mathbf{\hat{r}}$ is a unit vector directed radially outward from earth's center, $R_e$ is the radius of the Earth and $r$ is the distance of $m$ from the center of the Earth. We don't know the trajectory of the particle without solving the problem in detail, but for any element of the path the displacement $d\mathbf{r}$ can be written as
$$d\mathbf{r} = dr \mathbf{\hat{r}} + rd\theta \boldsymbol{\hat{\theta}}$$ where $\boldsymbol{\hat{\theta}}$ is a unit vector perpendicular to $\mathbf{\hat{r}}$, see the picture below for a sketch of the images they present.

Because $\mathbf{\hat{r}} \cdot \boldsymbol{\hat{\theta}} = 0$ we have
$$\mathbf{F} \cdot d\mathbf{r} = -mg \dfrac{R_e^2}{r^2}\mathbf{\hat{r}} \cdot (dr \mathbf{\hat{r}} + rd\theta \boldsymbol{\hat{\theta}}) = \\ -mg\dfrac{R_e^2}{r^2}dr.$$ The work-energy theorem becomes
$$\dfrac{m}{2}(v^2-v_0^2) = -mgR_e^2 \int_{R_e}^r \dfrac{dr}{r^2} = \\ -mg R_e^2(\dfrac{1}{r} - \dfrac{1}{R_e})$$ Here is where I get lost: They say the escape velocity is the minimum value of $v_0$ for which $v=0$ when $r \to \infty$. We find
$$v_{\text{escape}} = \sqrt{2gR_e} = 1.1 \times 10^4 \text{m/s}$$
which is the same result as in the example when $\alpha=0$, that is when the mass is projected straight up (presented earlier in the book). They write, "In the absence of air friction, the escape velocity is independent of the launch direction, a result that may not be intuitively obvious".

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Indeed I find this very unintuitive and have one gripe with the analysis presented. I don't understand how we can assume that $r$ even will go to infinity in the first place. As far as I can see, what have been shown is that if $r \to \infty$, then $v_0 = \sqrt{2gR_e}$ will make $v=0$. What I don't see have been shown, however, is that setting $v_0 = \sqrt{2gR_e}$ will make $r \to \infty$, which I think really is what escape velocity should be. (This is what was done in the case of a pure vertical projection of the mass, it was shown that if $v_0 = \sqrt{2gR_e}$ then $r_{\text{max}} \to \infty$ but this has not been done in the more general case presented here I think.)

Question
Based on this similar question the curvature of the Earth seems to make it so that even if you fire the mass horizontally it won't crash down or simply stay at a constant height above the earth, but this would have to be shown in the analysis right, it is not merely enough to assume that $r$ can go to infinity?

Answer 1

You are right. In the case of more than 1 dimension, having enough energy to escape to infinity does not necessarily mean that the body will do so. For example, if we had force of the form $$ \mathbf{F} = -mge^{-10(\frac{r}{R_e}-1)}\mathbf{\hat{r}}, $$ a similar argument would show that $$ \dfrac{m}{2}(v^2-v_0^2) = -mg \frac{R_e}{10}\left(e^{-10(\frac{r}{R_e}-1)} -1\right), $$ so one may be tempted to say that any body launched with speed greater or equal that $\sqrt{\frac{gR_e}{5}}$ will escape to infinity. However, it's not true: a body launched horizontally with speed $\sqrt{gR_e}$ will move in a circular orbit. One can show that even small deviations from the horizontal launch angle will not make the body to escape to infinity.

This sort of thing does not happen for the force $\mathbf{F} = -mg\dfrac{R_e^2}{r^2}\mathbf{\hat{r}}$, though, so for this force, the magnitude of escape velocity is indeen independent of direction, but some additional analysis is required to show it. (I don't know if the book in question does this sort of analysis.)

Answer 2

however, is that setting $v_0 = \sqrt{2gR_e}$ will make $r \to \infty$, which I think really is what escape velocity should be? (This is what was done in the case of a pure vertical projection of the mass, it was shown that if $v_0 = \sqrt{2gR_e}$ then $r_{\text{max}} \to \infty$ but this has not been done in the more general case presented here I think.).

I think the problem is you are focusing too much on the mathematics and forgetting the physical scenario. I don't blame you, it's tricky to switch between the two lines of thinking's even for me.

Here is an analogy to help grasp this: Imagine you are a car with a sufficiently sized fuel tank for the purposes of this discussion and you want to travel to some point X along a line(You just have to reach there and you win, no need to stay there). Now, there is a strong wind blowing retarding you at all times with a constant deacceleration and the car has an issue that it accelerates to some $y$ speed at the beginning of the trip instantaneously but can't deaccelerate again. Finally for the sake of discussion, let's say all the fuel is converted to an equivalent amount of kinetic energy (speed in direction to reach X) at the beginning of the trip.

Question: What is the minimum fuel that you need to complete the trip? The answer: Exactly opposite the energy depleted by the wind pushing back on you for the whole travel.

What about if you wanted to go a bit more? Then you would need some non zero kinetic energy when you reach X to push further and hence you must begin with an energy greater than the amount the wind depletes.

Hope this helps.