Question

(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?

Short answer

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 $470\text{m/s}$.

Step-by-step solution

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.

Calculation of the period of the rotation of the earth

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

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

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

Calculation of the angular velocity of the earth

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 $7.275\times {10}^{-5}\text{\hspace{0.33em}rad/s}$.

Calculation of the linear velocity of the earth

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

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

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

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