Question 300: Work, Energy and Power & Circular Motion

Question

A particle moves in a circular orbit of radius '$$r$$' under a central attractive force, $$F=-\frac{k}{r}$$, where $$\mathrm{k}$$ is a constant. The periodic time of its motion is proportional to

Accepted Answer

$$\begin{aligned} & \mathrm{mr} \omega^2=\frac{\mathrm{k}}{\mathrm{r}} \ & 	herefore \omega^2=\frac{\mathrm{k}}{\mathrm{mr}^2} \quad 	herefore \omega^2 \propto \frac{1}{\mathrm{r}^2} \ & 	herefore \quad \omega \propto \frac{1}{\mathrm{r}}, \mathrm{T}=\frac{2 \pi}{\omega} \quad 	herefore \mathrm{T} \propto \mathrm{r}\end{aligned}$$

Answer

$$r^{\frac{1}{2}}$$

Answer

$$\mathrm{r}^{\frac{2}{3}}$$

Answer

$$r$$

Answer

$$r^{\frac{3}{2}}$$

Work, Energy and Power & Circular Motion

A particle moves in a circular orbit of radius '$r' under a central attractive force, F=-\frac{k}{r}, where \mathrm{k}$ is a constant. The periodic time of its motion is proportional to

Choose the correct answer: