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Q6.1-4PE

Expert-verifiedFound in: Page 221

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) What is the period of rotation of Earth in seconds?**

**(b) What is the angular velocity of Earth?**

**(c) Given that Earth has a radius of **${\mathbf{6}}{\mathbf{.}}{\mathbf{4}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{6}}}{\mathbf{}}{\mathbf{m}}$** at its equator, what is the linear velocity at Earth’s surface?**

a) The period of rotation of the earth is 86400 s.

b) The angular velocity of the earth is $7.291159\times {10}^{-5}$ rad/s.

c) The linear velocity at the earth’s surface is $470\text{\hspace{0.33em}}\text{m/s}$.

**The actual value of the period of rotation of the earth is 23 hours, 56 minutes, and 4 seconds****.**

**This happens because a solar day is longer than a sidereal day. While the earth rotates, it also moves around the sun with an interval of one day.**

The period of rotation of the earth can be calculated as:

** **

** $\begin{array}{c}=1\times 24\times 60\times 60\text{\hspace{0.33em}s}\\ =86400\text{\hspace{0.33em}s}\end{array}$**

** **

So, the period of the rotation of the earth is 86400 s**.**

The angular velocity can be calculated as:

$\begin{array}{c}=\frac{2\pi}{86400}\\ =7.275\times {10}^{-5}\text{\hspace{0.33em}rad/s}\end{array}$

The earth rotates at a moderate angular velocity of $7.275\times {10}^{-5}\text{\hspace{0.33em}rad/s}$.

Earth’s radius$=6.4\times {10}^{6}\text{\hspace{0.33em}}\mathrm{m}$.

Thus, the linear speed of any point on the earth’s surface at the equator due to the earth’s rotation is

$\begin{array}{c}V=r\omega \\ =6.4\times {10}^{6}\times 7.3\times {10}^{-5}\\ =470\text{\hspace{0.33em}}\text{m/s}\end{array}$

Therefore, the linear velocity at the earth’s surface is $470\text{\hspace{0.33em}}\text{m/s}$.

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