Answer to Question #204153 in Classical Mechanics for Mile

Question

The moon Titan orbits the planet Saturn with a period of 1.9 × 106 s. The average radius of

this orbit is 1.6 × 109 m.

a) What is Titan’s centripetal acceleration?

b) What is Titans Velocity?

Answer 1

The information given is

The period of a complete orbit is $T=1.9\times 10^{6}\;\text{s}$

The average radius of the orbit is. $r=1.6\times 10^{9}\;\text{s}$

Considering the circular orbit.

The angular speed is given by

$w=\dfrac{2\;\pi\;\text{rad}}{T}$

Where.

$2\;\pi\;\text{rad}$ is the angular displacement of a complete orbit

$T$ is the period of a complete orbit.

Evaluating numerically.

$w=\dfrac{2\;\pi\;\text{rad}}{T}\\ w=\dfrac{2\;\pi\;\text{rad}}{1.9\times 10^{6}\;\text{s} }\\ w= 3.3\times 10^{-6} \text{rad}/\text{s}$

The centripetal acceleration is given by.

$a_{c}=w^{2}\;r$

Where.

$w$ is the angular speed

$r$ is the radius.

Evaluating numerically.

$a_{c}=w^{2}\;r\\ a_{c}=( 3.3\times 10^{-6} \text{rad}/\text{s} )^{2}\times 1.6\times 10^{9}\;\text{s} \\ a_{c}=1.7\times 10^{-2}\;\text{m}/\text{s}^{2}$

Answer A

The centripetal acceleration is $\displaystyle \color{red}{\boxed{a_{c}=1.7\times 10^{-2}\;\text{m}/\text{s}^{2}}}$

Part b

The linear velocity is given by

$V=w\cdot r$

Where

$w$ is the angular speed

$r$ is the radio

Evaluating numerically

$V=w\cdot r\\ V= 3.3\times 10^{-6} \text{rad}/\text{s}\times 1.6\times 10^{9}\;\text{s}\\ V=5.3\times 10^{3}\;\text{m}/\text{s}$

Answer B

The velocity is $\displaystyle \color{red}{\boxed{V=5.3\times 10^{3}\;\text{m}/\text{s}}}$