Q72
General SciencePhysics
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?
- a.zero
- b.$F \times 2\pi r$
- c.$F \times \pi r$
- d.$F \times 2r$
Answer: (A) zero
Work done is $W = Fs\cos\theta$, where $\theta$ is the angle between the force and the displacement. In a circular orbit the gravitational pull acts along the radius towards the Sun while the planet moves along the tangent, so $\theta = 90^\circ$ at every instant and the work done is zero over any part of the path, including half the circumference. That is also why the planet's speed and kinetic energy stay constant in a circular orbit. Options B, C and D wrongly multiply the force by the path length or the diameter. General rule: a centripetal force never does work. (In an elliptical orbit gravity does work over parts of the path, but the total over one full revolution is still zero.)