A planet is moving in a circular orbit. It co | vedclass

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(C) Given: The planet completes $2$ revolutions in $360$ days.One complete revolution corresponds to an angle of $2\pi$ radians.Therefore, the total a

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$3.5 	imes 10^{-2} 	ext{ rad day}^{-1}$

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(C) Given: The planet completes $2$ revolutions in $360$ days.One complete revolution corresponds to an angle of $2\pi$ radians.Therefore, the total angle $	heta$ covered in $2$ revolutions is $	heta = 2 	imes 2\pi = 4\pi$ radians.The time taken $t$ is $360$ days.The angular frequency $\omega$ is defined as $\omega = \frac{	heta}{t}$.Substituting the values: $\omega = \frac{4\pi}{360} = \frac{\pi}{90} 	ext{ rad day}^{-1}$.Using $\pi \approx 3.14159$, we get $\omega \approx \frac{3.14159}{90} \approx 0.0349 	ext{ rad day}^{-1}$.This can be written in scientific notation as $3.49 	imes 10^{-2} 	ext{ rad day}^{-1}$, which is approximately $3.5 	imes 10^{-2} 	ext{ rad day}^{-1}$.

$A$ planet is moving in a circular orbit. It completes $2$ revolutions in $360$ days. What is its angular frequency?

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