Answer:
[tex]1.99\cdot 10^{-7} rad/s[/tex]
Explanation:
The average angular speed of the Earth about the Sun is given by:
[tex]\omega = \frac{2 \pi}{T}[/tex]
where
[tex]2 \pi[/tex] rad is the total angle corresponding to one revolution of the Earth around the Sun
T is the orbital period of the Earth
The orbital period of the Earth is 365.25 d. We must convert it into seconds first:
[tex]T=365.25 d \cdot (24 h/d) \cdot (60 min/h) \cdot (60 s/min)=3.16\cdot 10^7 s[/tex]
And by substituting into the equation above, we find the average angular speed:
[tex]\omega=\frac{2 \pi rad}{3.16\cdot 10^7 m/s}=1.99\cdot 10^{-7} rad/s[/tex]
The average angular speed of Earth about the sun is [tex]1.99 \times 10^{-7} \;\rm rad/s[/tex].
Given data:
The orbiting period of Earth is, T = 365.25 days
The rate of change of angular displacement of an object with respect to time is known as angular speed.
We know that,
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds
Therefore,
[tex]1\;\rm day = 365.25 \times24 \times 60 \times 60 \;\rm s\\1 \;\rm day =3.16 \times 10^{7} \;\rm s[/tex]
Now, the expression for the angular speed of the Earth is,
[tex]\omega = \dfrac{2 \pi}{T}[/tex]
Solving as,
[tex]\omega = \dfrac{2 \pi}{T}\\\\\omega = \dfrac{2 \pi}{3.16 \times 10^{7} \;\rm s}\\\\\omega = 1.99 \times 10^{-7} \;\rm rad/s[/tex]
Thus, we can conclude that the average angular speed of Earth about the sun is [tex]1.99 \times 10^{-7} \;\rm rad/s[/tex].
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