A particle moves in a circular orbit under the action of a central attractive force inversely proportional to

Question

A particle moves in a circular orbit under the action of a central attractive force inversely proportional to the distance $r$. The speed of the particle is:

Accepted Answer

1. **Identify Forces:** For a particle moving in a circular orbit, the centripetal force required is provided by the given central attractive force.
2. **Centripetal Force Formula:** The centripetal force acting on a particle of mass $m$ moving with speed $v$ in a circle of radius $r$ is $F_c = \frac{mv^2}{r}$ [Source 61].
3. **Given Force Law:** The problem states the attractive force $F$ is inversely proportional to distance $r$. So, $F \propto \frac{1}{r}$ or $F = \frac{k}{r}$ where $k$ is a constant.
4. **Equate Forces:** 
   $$ \frac{mv^2}{r} = \frac{k}{r} $$
5. **Solve for Speed ($v$):** Multiplying both sides by $r$ eliminates it from the equation.
   $$ mv^2 = k $$
   $$ v = \sqrt{\frac{k}{m}} $$
6. **Conclusion:** Since $k$ and $m$ are constants, the speed $v$ is constant and **independent of $r$**.

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