The earth’s rotation axis, which is tilted 23.5­ from the plane of the earth’s orbit, today points to Polaris, the north star. But Polaris has not always been the north star because the earth, like a spinning gyroscope, precesses. That is, a line extending along the earth’s rotation axis traces out a 23.5­ cone as the earth precesses with a period of 26,000 years. This occurs because the earth is not a perfect sphere. It has an equatorial bulge, which allows both the moon and the sun to exert a gravitational torque on the earth. Our expression for the precession frequency of a gyroscope can be written Ω=𝜏/ω. Although we derived this equation for a specific situation, it’s a valid result, differing by at most a constant close to 1, for the precession of any rotating object. What is the average gravitational torque on the earth due to the moon and the sun?

Step by step solution

01

Given information

The tilt of the earth's rotation axis = 23.5^o

Period of earth, T=26000 years
Mass of earth, M=5.9 x10²⁴ kg
The radius of the earth, R=6400 km
The period of the earth on its own axis is t=24 hr

02

Explanation

Calculate moment of inertia of earth assuming it is sphere

$$\begin{gathered} I=\frac{2}{5}M{R}^2 \\ =\frac{2}{5}\times (5.9\times {10}^{24}kg)\times {\left(6.4\times {10}^{6}m\right)}^2 \\ =9.67\times {10}^{37}{\mathrm{kgm}}^2 \end{gathered}$$

From the period calculate precision frequency

$\Omega =\frac{2\pi }{T}$

Substitute values

$$\begin{gathered} =\frac{2\pi rad}{(26000\times 365\times 24\times 60\times 60sec)} \\ =7.67\times {10}^{-12}\mathrm{rad}/\mathrm{s} \end{gathered}$$

And natural frequency

$$\begin{gathered} \omega =\frac{2\pi }{t}=\frac{2\pi rad}{24\times 60\times 60sec} \\ =7.2\times {10}^{-5}\mathrm{rad}/\mathrm{s} \end{gathered}$$

we can calculate torque as
Torque = precision frequency x moment of inertia x natural frequency

$$\begin{gathered} =\left(9.67\times {10}^{37}\mathrm{kg}{\mathrm{m}}^2\right)\times \left(7.67\times {10}^{-12}\mathrm{rad}/\sec \right)\times \left(7.2\times {10}^{-5}\mathrm{rad}/\sec \right) \\ =5.34\times {10}^{22}\mathrm{N}.\mathrm{m} \end{gathered}$$

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