Discussion
For a particle moving under the action of a central force, the correct relation between the energy (E) of the particle, eccentricity (e) of the conic section describing the orbit, reduced mass (μ), angular momentum (\(\bar{l}\)), and the force constant (k) is given by :
Answer: A
The relation is \(E = \frac{\mu k^{2}}{2l^{2}}(e^{2} - 1)\).
For the inverse-square force with potential \(-k/r\), the orbit equation gives the eccentricity in terms of energy and angular momentum:
\[ \begin{aligned} e &= \sqrt{1 + \frac{2El^{2}}{\mu k^{2}}} \\ e^{2} - 1 &= \frac{2El^{2}}{\mu k^{2}} \end{aligned} \]
Rearranging gives the first option. It also has the right physics: bound orbits (e < 1) have negative energy, the parabola (e = 1) has zero energy and hyperbolas (e > 1) positive energy. The options with \(1 + e^{2}\) or \(e^{2} + 1\) are always positive, and \((1 - e^{2})\) without the factor 1/2 has the wrong sign and size.
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