A planet of mass $m$ is revolving around the sun in a circle of radius $r$. What is the work done by the gravitational force $F$ in moving the planet over half the circumference of the circle?

Work done ($W$) by a force is given by the dot product of the force vector ($\vec{F}$) and the displacement vector ($\vec{d}$):
$$W=\vec{F}\cdot \vec{d}=Fd\cos \theta$$

For a planet revolving in a uniform circular orbit, the gravitational force acting on the planet acts as a centripetal force directed toward the center of the circle. At any point along the path, this force is perpendicular to the direction of motion (velocity and instantaneous displacement), meaning the angle $\theta =90^{\circ }$. Since $\cos 90^{\circ }=0$, the work done by the gravitational force is always zero, regardless of the distance covered.