Question

A particle of mass 'm' moves in a circular orbit of radius 'r' under action of an attractive force $\frac{-k^2}{r^2}$ which always acts towards the centre of the orbit. What is the total energy in the particle?

• A. $\frac{k}{r^2}$
• B. $\frac{2k}{r}$
• C. $\frac{k^2}{r}$
• D. $\frac{k^2}{2r}$

Answer 1

The correct option is D $\frac{k^2}{2r}$

As the particle is rotating in a circular orbit the net force on it should be equal to the centripetal force .
therefore, $(m\times V^2)/r=k^2/r^2$
$(m\times V^2)=k^2/r$
$KE=(m\times V^2)/2=k^2/(2\times r)$