Answer:
The average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²
Explanation:
Given data,
The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec
= 3.1536 x 10⁷ +0.840
= 31536000.84 s
The period of 365 rotation of Earth in 2006, T₀ = 365 days
= 31536000 s
Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365
= 86400.0023 s
The time period of rotation is given by the formula,
<em>Tₐ = 2π /ωₐ</em>
ωₐ = 2π / Tₐ
Substituting the values,
ωₐ = 2π / 365.046306
= 7.272205023 x 10⁻⁵ rad/s
Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365
= 86400 s
Time period of rotation,
Tₓ = 2π /ωₓ
ωₓ = 2π / T
= 2π /86400
= 7.272205217 x 10⁻⁵ rad/s
The average angular acceleration
α = (ωₓ - ωₐ) / T₁
= (7.272205217 x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84
α = 6.152 X 10⁻²⁰ rad/s²
Hence the average angular acceleration of the Earth, α = 6.152 X 10⁻²⁰ rad/s²
