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.

• A. $21,274m{s}^{-1}$
• B. $7,091m{s}^{-1}$
• C. $28,366m{s}^{-1}$
• D. $14,183m{s}^{-1}$

Answer 1

The correct option is D $14,183m{s}^{-1}$

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 $14183m{s}^{-1}$.