Question

One of saturn's moon is named as Mimas. The orbital radius of Mimas is $1.87\times$ ${10}^{8}m$ from saturn. The mean orbital period of Mimas is approximately 23 hours. Calculate the speed of Mimas with which it revolves around saturn. Assume that the path taken is circular.

Answer 1

Given,
Orbital radius, $R=1.87\times {10}^{8}m$
Time period:
$T=23hours$
$T=23\times 3600=82,800s$

From definition of speed,
$\text{Speed}=\frac{\text{Circumference}}{\text{Time period}}$
$v=\frac{2\pi R}{T}$
$v=\frac{2\times 3.14\times 1.87\times {10}^8}{82,800}$
$v=14183m{s}^{-1}$
$\therefore$Speed of Mimas with which it revolve around saturn is $14183m{s}^{-1}$.