Question

question 12 of 13 a day on a distant planet observed orbiting a nearby star is 41.9 hr, and a year on the planet lasts 38.7 earth days. calculate the average angular speed ωₚ of the planet about its own axis of rotation in radians per second. ωₚ = rad/s calculate the average angular speed ωₒ of the planet as it travels around its neighboring star. ωₒ = rad/s © macmillan learning

Explanation:

Part 1: Angular speed about its own axis ($\omega_p$)

Step1: Recall angular speed formula

The angular speed $\omega$ for a full rotation (angle $\Delta\theta = 2\pi$ radians) in time $\Delta t$ is given by $\omega=\frac{\Delta\theta}{\Delta t}$. For the planet's rotation about its own axis, one day is the time period. First, convert the time from hours to seconds.
Given a day is $41.9$ hours. Since $1$ hour $ = 3600$ seconds, $\Delta t_p=41.9\times3600$ seconds.

Step2: Calculate $\omega_p$

$\Delta\theta = 2\pi$ radians (for one full rotation). So,
$\omega_p=\frac{2\pi}{\Delta t_p}=\frac{2\pi}{41.9\times3600}$
Calculate the denominator: $41.9\times3600 = 41.9\times3.6\times10^3=150.84\times10^3 = 150840$
Then $\omega_p=\frac{2\pi}{150840}\approx\frac{6.2832}{150840}\approx4.165\times 10^{-5}\ \text{rad/s}$ (approximate value, more precise calculation: $2\times3.1416\div150840\approx6.2832\div150840\approx4.165\times 10^{-5}$)

Part 2: Angular speed about the star ($\omega_o$)

Step1: Convert year to seconds

A year on the planet is $38.7$ Earth days. First, convert days to hours: $38.7$ days $\times24$ hours/day $ = 928.8$ hours. Then convert hours to seconds: $928.8\times3600$ seconds.
$\Delta t_o=38.7\times24\times3600$
Calculate $38.7\times24 = 928.8$; $928.8\times3600=928.8\times3.6\times10^3 = 3343.68\times10^3=3343680$ seconds.

Step2: Calculate $\omega_o$

For orbital motion, one full orbit (angle $\Delta\theta = 2\pi$ radians) takes time $\Delta t_o$. So,
$\omega_o=\frac{2\pi}{\Delta t_o}=\frac{2\pi}{3343680}$
Calculate this: $2\times3.1416\div3343680\approx6.2832\div3343680\approx1.879\times 10^{-6}\ \text{rad/s}$

Answer:

For $\omega_p$: $\approx\boldsymbol{4.17\times 10^{-5}\ \text{rad/s}}$ (or more precisely $4.165\times 10^{-5}\ \text{rad/s}$)
For $\omega_o$: $\approx\boldsymbol{1.88\times 10^{-6}\ \text{rad/s}}$ (or more precisely $1.879\times 10^{-6}\ \text{rad/s}$)

(Note: Depending on the precision of $\pi$ used, the values may vary slightly. Using $\pi = 3.14159265$ will give more accurate results.)