Question

– Box 7: Question During its circular orbit, in a very small amount of time dt, Venus moves an infinitesimal amount ds. What is the work done on Venus by the gravitational force during that time? Recall that the work done by gravity is calculated as dWgrav = Fgrav · ds. Hint: You might want to draw a diagram of Fgray and ds. Will Venus' kinetic energy change as it moves along its circular orbit?

Answer 1

Fgravity = GMm/r^2

ds= $\widehat{}$ dr$r\widehat{^}$+r$\Theta$$\Theta \widehat{^}$ ( resolving into polar co ordinates)

Workdone dw = Fg. ds ( r. $\Theta$ = 0 and r. r = 1 )

dw = Fdr

Integrating both sides = W= $\int$ Fdr = F$\int$dr = F×r = GMm/r

According to work energy theorem work done = change in kinetic energy

Here work done not equal to zero hence kinetic energy changes