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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.. enter image description here

Joebevo
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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 07:20
  • 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 07:41
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    Huh? $E_2$ is already defined, it's $0$. Now $V'$ looks the way it is. – Chris Gerig Aug 16 '12 at 07:45
  • 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 07:58
  • Another question by OP from Chapter 3 in Goldstein: http://physics.stackexchange.com/q/33713/2451 – Qmechanic Nov 16 '12 at 22:44

2 Answers2

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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

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

See also this Phys.SE question.

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.

Community
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Qmechanic
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The effective potential energy is not the same as the orbital energy. The orbital energy is a constant of the orbit, while the effective potential energy varies throughout the orbit. For a parabolic orbit the orbital energy is always zero, but the effective potential energy varies between positive and negative depending on the phase of the orbit.

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

Nick
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