Name: Circular Motion and Centripetal Force
Uploaded: 2019-07-19
Duration: 6 min 24 s
Description: Here is an interesting video on the uniform circular motion which explains the circular motion of the Ferris wheel, merry go round and roller coaster. Get in!

Question

A satellite is in a circular orbit around a planet of mass $M$ , as shown in the diagram.

$d .$ Titan and Enceladus are two of Saturn’s moons. Data about these moons are given in the table.

Moon	Orbit radius / $m$	Angular speed / $rad s^{- 1}$
Titan	$1.22 \times 10^{9}$
Enceladus	$2.38 \times 10^{8}$	$5.31 \times 10^{- 5}$

$i .$ Determine the mass of Saturn.

Accepted Answer

The relation between angular speed and gravitational field strength is: $ω^{2} r^{3} = G M$

Rearranging the above values, we get: $M = \frac{ω^{2} r^{3}}{G}$

Using values corresponding to Enceladus, $ω = 5.31 \times 10^{- 5} rad s^{- 1}$ and $r = 2.38 \times 10^{8} m$ .

Since, universal gravitational constant is $G = 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}$ .

Substituting these values in above equation. We get,

$M = \frac{{(5.31 \times 10^{- 5})}^{2} \times {(2.38 \times 10^{8})}^{3}}{(6.67 \times 10^{- 11})} = 5.70 \times 10^{26} kg$

Hence, the mass of the Saturn is $5.70 \times 10^{26} kg$ .

Important Questions on Circular Motion and Gravitation

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