# On a succession of steps that leads to kinetic energy being negative (Newtonian Mechanics)

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 $$r, we have $$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 $$577$$. 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?

## 2 Answers

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 is classically forbidden as it would require the radial kinetic energy to be negative, which is classically impossible.

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.