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Suppose we have an object with mass $m$ placed (in two dimensions) at an initial position $r_{0}$ (disregarding $\theta$) from the origin and having an initial velocity $\textbf{v}_{0}=v_{r0}\textbf{e}_{r} +v_{\theta0} \textbf{e}_{\theta}$.

Let us also assume the object is in a gravitational field generated by a mass $M$ many times greater than $m$, the potential energy of the particle thus being $U(r)=-GMm/{r}$.

Assuming the gravitational pull generated by $M$ is the only force present, we have

$$\frac{1}{2}m[v_{r}^2+v_{\theta}^2]+U(r)=cst=E_{0},$$

where

$$E_{0}=\frac{1}{2}mv_{0}^2+U(r_{0})=\frac{1}{2}m[v_{r0}^2+v_{\theta0}^2]+U(r_{0}).$$

We also have conservation of angular momentum $$rmv_{\theta}=cst=l_{0},$$

where $$l_{0}=r_{0}mv_{\theta0}.$$

We can specify $v_{r0}$ so as to make $E_{0} =0$; the only requirement for this $v_{r{0}}$ to exist is

$$-U(r_0)-\frac{1}{2}mv_{\theta 0}^2 >0$$

or $$r_{0}<\frac{2GM}{v_{\theta 0}^2}.$$

So we first specify $v_{\theta 0}$, then $r_{0}$, and finally $v_{r0}$.

Thus, we have $E=E_{0}=0$ and $mrv_{\theta}=l_{0}$.


There is an $r_{1}$ below which $U(r)+\frac{1}{2}mv_{\theta}^2 >0$; it is given by $$r<\frac{2l_{0}^2}{GMm^2}=\frac{2r_{0}^2v_{\theta {0}}^2}{GM}=r_{1}$$

and it is readily determined given $r_{0}$ and $v_{\theta 0}$.

$E=0$ is always true regardless of $r$. But in the region $r<r_{1}$, we have $U(r)+\frac{1}{2}mv_{\theta}^2 >0$; thus, the only way to have $E=0$,$\,$ i.e. $\frac{1}{2}mv_{\theta}^2+U(r)+\frac{1}{2}mv_{r}^2=0$, is to have $$\frac{1}{2}mv_{r}^2<0,$$ which is impossible. Where did this illogical result come from?

The following diagram is the reason I made this post, it was excerpted from A.P. French's Newtonian Mechanics, page $577$. enter image description here

It depicts $U'(r)=U(r)+\frac{1}{2}mv_{\theta}^2$ as a function of $r$ for a given $l$. The straight white line above the x-axis represents $E_{0}$ (the total energy of the object). Is the diagram in the red zone, where $U'(r) >E_{0}$, physically correct? ? We know that $E_{0}=U'(r)+\frac{1}{2}mv_{r}^2$, so the contribution from $\frac{1}{2}mv_{r}^2$ will be no less than zero; thus, we can't have $E_{0}=U'(r)+\frac{1}{2}mv_{r}^2$ in the red zone. So is the diagram accurate in this region?

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FWIW, the total potential energy is $U^{\prime}(r)=U(r)+U_{\rm cf}(r)$, where $U_{\rm cf}(r)=\frac{\ell_0^2}{2mr^2}$ is the centrifugal potential. The region with $E<U^{\prime}(r)$ is classically forbidden as it would require the radial kinetic energy to be negative, which is classically impossible.

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By setting $E=0$ you are saying that the object's speed is equal to escape velocity $v_e(r)=\sqrt{\frac{2GM}{r}}$. By fixing angular momemtum $l_0$ you are saying that the angular speed of the object is $v_{\theta}(r)=\frac{l_0}{mr}$.

But if $r \lt \frac{l_0^2}{2GMm^2}$ then $\frac{l_0^2}{m^2r^2} > \frac{2GM}{r}$ so $v_\theta(r) > v_e(r)$. All this is saying is that there is a region within which the constraints speed = escape velocity and angular momentum = $l_0$ cannot both be satisfied because the angular speed alone is greater than the escape velocity.

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