(a) What is the angular speedvabout the polar axis of a point on Earth’s surface at latitude $\mathbf{40}\mathbf{°}\mathbf{ }\mathbf{\text{N}}$? (Earth rotates about that axis.)

(b) What is the linear speedvof the point? What are

(c) $\mathit{\omega }$and

(d)vfor a point at the equator?

The polar axis of a point on Earth’s surface at latitude ${40}^0 \text{N}$

By using formulas for linear speed vand angular speed ω, we can find thelinear speed vand angular speed ω of earth at latitude and at equator. The formulas are given below.

1. The linear speed vis$v=\omega r$
2. The angular speed of earth $\omega$is$\omega =\frac{2\pi }{T}$

Where, T istheperiod of earth.

The linear speed of a point on the earth’s surface depends on its distance from the axis of rotation. For linear speed, we have

$v=\omega r$

where, r is the radius of its orbit.

A point on earth at latitude of ${40}^0$ moves along a circular path of radius

$r=R\cos {40}^0$

Where, R is earth’s radius and$R=6.4\times {10}^6 \text{m}$

On the other hand,$r=R$at the equator.

We know that earth makes one rotation per day and

$1 \text{d}=24 \text{hrs}=24\times 60\times 60 \text{s}=8.64\times {10}^4 \text{s}$

So, the angular speed of the earth is given by,

$\begin{array}{rcl}\omega & =& \frac{2\pi }{8.64\times {10}^4 \text{s}}\\ & =& 7.3\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Hence the angular speed is, $7.3\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 3: (b) Calculation for the linear speed v of the point

At the latitude of ${40}^0$, the linear speed is

$$\begin{gathered} v=\omega r \\ v=\omega R\cos {40}^0 \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} v=7.3\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s\times {\displaystyle 6}{\displaystyle .}{\displaystyle 4}{\displaystyle \times }{\displaystyle {{\displaystyle 10}}^6}{\displaystyle }{\displaystyle \text{m×}}{\displaystyle \cos }{\displaystyle }{\displaystyle {{\displaystyle 40}}^0}$}\right. \\ v=3.5\times {10}^2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence the value of linear speed is, $3.5\times {10}^2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Step 3: (c) Calculation for the role="math" localid="1660897430122" $\omega$

$\omega =7.3\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Hence the value of $\omega$is, $7.3\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Step 3: (d) Calculation for the v for a point at the equator

The latitude at the equator is $0^0$. Hence, the speed is given by

$$\begin{gathered} v=\omega r \\ v=\omega R \end{gathered}$$

Substitute all the value in the above equation.

role="math" localid="1660897847675" $$\begin{gathered} v=7.3\times {10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\times 6.4\times {10}^6 \text{m} \\ v=4.6\times {10}^2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence the value of v is, $4.6\times {10}^2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.