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The rotation rate of the Earth (Ω = 7.2921 × 10 −5 rad/s) can be calculated as 2 π / T radians per second, where T is the rotation period of the Earth which is one sidereal day (23 h 56 min 4.1 s). [2] In the midlatitudes, the typical value for is about 10 −4 rad/s.
The Earth's moment of inertia is roughly constant, so as the Earth's rate of rotation \(\omega\) decreases, its angular momentum \(L=I \omega\) must also decrease. But angular momentum is conserved; where does the angular momentum go?
4 Σεπ 2016 · Linear speed in circular motion depends on angular velocity (here: 1 rotation per day, for all latitudes) and radius of the circle - distance from the axis of rotation. $ v = \omega r$ If $\alpha$ is your latitude, the radius of the circle you make with each revolution of Earth, is $r$.
Rotational frequency can be obtained dividing angular frequency, ω, by a full turn (2 π radians): ν =ω/ (2π rad). It can also be formulated as the instantaneous rate of change of the number of rotations, N, with respect to time, t: n =d N /d t (as per International System of Quantities). [4] .
The centrifugal force at the equator is \(\Omega^{2} R=\left(7.27 \times 10^{-5} \mathrm{s}^{-1}\right)^{2}\left(6.378 \times 10^{6} \mathrm{m}\right)=0.033 \mathrm{m} \mathrm{s}^{-2}\), and hence accounts for more almost 2/3 of the difference in g between the equator and the poles.
This makes the problem simple because the rotation rate (omega) is constant, so the difference between your head and feet is r1/r2. For a person standing in a 224 m radius structure, that's 2/224 = 0.9%.
1 Ιαν 2014 · Changes in the Earth rotation speed are expressed as deviations of Universal Time 1 (UT1) from the uniform atomic time (Universal Time Coordinated, UTC) dUT1 = UT1–UTC or as variations in the length of day (LOD). The subgroup of polar motion and dUT1 or LOD is called Earth rotation parameters (ERP).
