# Does effective potential for a gravitational force have a maximum below $E=0$?

The relevant figure is below (taken from Goldstein's Classical Mechanics). This figure plots the effective potential for a gravitational force. Does the effective potential $V'$ go flat below $E_2=0$? After finding $r_{flat}$, the point where the effective force $f'=0$ (or equivalently, where $V'$ goes flat), I got $$V'(r_{flat})=-\frac{mk^2}{2l^2}$$ Now, it looks like this is expression is negative since $m$,$k$, and $l$ are all positive. So my question is: Doesn't this result in a possible parabola that could have an energy less than $E_2=0$? I know it isn't possible for a parabola to have negative energy, so where am I going wrong in my reasoning?

Thanks..

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The picture even demonstrates that $V'$ is flat below $E_2$... why is this a problem? Your phrase "possible for a parabola to have negative energy" doesn't make any sense, can you please clarify? –  Chris Gerig Aug 16 '12 at 7:20
Well, if you look at the picture, one can lower the position of $E_2$ well below where it currently is and still not have the orbit be bounded. I think the minimum-energy unbounded orbit is a parabola, in which case the figure implies that the energy associated with a parabolic path is less than zero. According to Goldstein, $E_{parabola}=0$. –  Joebevo Aug 16 '12 at 7:41
Huh? $E_2$ is already defined, it's $0$. Now $V'$ looks the way it is. –  Chris Gerig Aug 16 '12 at 7:45
What I mean is, it $\it looks$ like you could have a parabolic path of energy $<0$, based on this figure. Just imagine an $E_{parabola}<E_2$ but above the $V'$ curve(for large $r$). –  Joebevo Aug 16 '12 at 7:58
Another question by OP from Chapter 3 in Goldstein: physics.stackexchange.com/q/33713/2451 –  Qmechanic Nov 16 '12 at 22:44

I) The fictitious potential

$$\tag{1} V^{\prime}~=~V+\frac{\ell^2}{2mr^2}$$

is a sum of a Newtonian gravitational potential

$$\tag{2} V~=~-\frac{k}{r},$$

and a centrifugal potential. The mechanical energy is a constant of motion and given by

$$\tag{3} E~=~\frac{1}{2}m\dot{r}^2+V^{\prime}.$$

II) OP correctly calculates that the minimum point is

$$\tag{4} r_0~=~\frac{\ell^2}{mk},$$

and that the minimum value is

$$\tag{5} E_4~:=~V^{\prime}(r_0)~=~-\frac{mk^2}{2\ell^2}.$$

III) Despite what the Figure 3.3 may suggest, there is no gap between the limit $\lim_{r\to\infty} V^{\prime}(r)$ and $E_2:=0$. The potential $V^{\prime}$ is a monotonically growing function in the whole interval $r\in[r_0,\infty[$, with the limit

$$\tag{6} \lim_{r\to\infty} V^{\prime}(r)~=~0,$$

as can be easily deduced from the first two formulas (1) and (2). The energy $E=E_2:=0$ corresponds to a parabolic orbit, while the energy $E<E_2:=0$ corresponds to an elliptic orbit.

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The effective force is zero when the effective potential equals $E_4$. $E_4$ is always negative regardless of the eccentricity of the orbit.