A particle P of mass m moves under the action of the force F~ (~r) = −βm/r^3 * ~ r, where ~r is the position vector of the particle with respect to the origin O, r = |~r|, β is a positive constant.
Assume that the orbit of P is a circle of radius R. Find the magnitude of the
centripetal acceleration and the speed v of the particle P.
Best Answer
The acceleration is, of course, Force divided by mass. Here force is $\frac{-\beta m}{r^3}\vec{r}$ so the acceleration is $\frac{-\beta}{r^3}\vec{r}$ so has magnitude $\frac{-\beta}{R^2}$.
If an object is moving in a circle of radius R then we can write its position, relative to the center of the circle, as $Rcos(vt)\vec{i}+ Rsin(vt)\vec{j}$. Its velocity vector is the derivative of that, $-Rvsin(vt)\vec{i}+ Rvcos(vt)\vec{j}$ and its acceleration vector is $-Rv^2cos(vt)\vec{i}-Rv^2sin(vt)\vec{j}$. The absolute value is $Rv^2= \frac{\beta}{R^2}$ so $v^2= \frac{\beta}{R^3}$ and $v= \frac{\beta^\frac{1}{2}}{R^\frac{3}{2}}$