A satellite while revolving around the earth | vedclass

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Question

(A) The time taken for one revolution is $t = 1 	ext{ hour } 30 	ext{ minutes} = 90 	ext{ minutes}$.Converting this into seconds: $t = 90 	imes 60

Vedclass · Accepted Answer

$1.16 	imes 10^{-3} 	ext{ rad/s}$

Vedclass · Answer

(A) The time taken for one revolution is $t = 1 	ext{ hour } 30 	ext{ minutes} = 90 	ext{ minutes}$.Converting this into seconds: $t = 90 	imes 60 = 5400 	ext{ s}$.The angle covered in one complete revolution is $	heta = 2\pi 	ext{ radians}$.The formula for angular speed is $\omega = \frac{	heta}{t}$.Substituting the values: $\omega = \frac{2\pi}{5400} = \frac{\pi}{2700} 	ext{ rad/s}$.Calculating the numerical value: $\omega \approx \frac{3.14159}{2700} \approx 1.1635 	imes 10^{-3} 	ext{ rad/s}$.Thus,the angular speed is approximately $1.16 	imes 10^{-3} 	ext{ rad/s}$.

$A$ satellite while revolving around the earth completes one revolution in $1$ hour and $30$ minutes. What is the angular speed of the satellite?

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