Question 705023

SCQHARD

A body of mass $m$ is acted upon by a central force $\vec{f}(\vec{r}) = -k \vec{r}$ , where $k$ is a positive constant. If the magnitude of the angular momentum is $l$ , then the total energy for a circular orbit is

(A)

$2 \sqrt{\frac{k l^2}{m}}$

(B)

$\frac{1}{2} \sqrt{\frac{k l^2}{m}}$

(C)

$\frac{3}{2} \sqrt{\frac{k l^2}{m}}$

(D)

$\sqrt{\frac{k l^2}{m}}$

Detailed Solution

The effective potential is $V_{eff} = \frac{l^2}{2mr^2} + \frac{1}{2}kr^2$ . For a circular orbit, $\frac{d V_{eff}}{d r} = 0$ , so $-\frac{l^2}{mr^3} + kr = 0$ , which gives $r_0^4 = \frac{l^2}{mk}$ or $r_0^2 = \sqrt{\frac{l^2}{mk}} = \frac{l}{\sqrt{mk}}$ . The total energy $E = V_{eff}(r_0) = \frac{l^2}{2m r_0^2} + \frac{1}{2} k r_0^2 = \frac{l^2}{2m} (\frac{\sqrt{mk}}{l}) + \frac{1}{2} k (\frac{l}{\sqrt{mk}}) = \frac{l}{2} \sqrt{\frac{k}{m}} + \frac{l}{2} \sqrt{\frac{k}{m}} = l \sqrt{\frac{k}{m}} = \sqrt{\frac{k l^2}{m}}$ .

Question 705023

Detailed Solution

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