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Expert-verified Found in: Page 221 ### College Physics (Urone)

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 ${\mathbf{6}}{\mathbf{.}}{\mathbf{4}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{6}}}{\mathbf{}}{\mathbf{m}}$ 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 $7.291159×{10}^{-5}$ rad/s.

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

See the step by step solution

## Step 1: Definition of linear velocity

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.

## Step 2: Calculation of the period of the rotation of the earth

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

$\begin{array}{c}=1×24×60×60\text{ s}\\ =86400\text{ s}\end{array}$

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

## Step 3: Calculation of the angular velocity of the earth

The angular velocity can be calculated as:

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

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

## Step 4: Calculation of the linear velocity of the earth

Earth’s radius$=6.4×{10}^{6}\text{ }\mathrm{m}$.

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

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

Therefore, the linear velocity at the earth’s surface is $470\text{ }\text{m/s}$. ### Want to see more solutions like these? 