91Ó°ÊÓ

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 Ó¬, we can find thelinear speed vand angular speed Ó¬ of earth at latitude and at equator. The formulas are given below.

1. The linear speed vis$v=\mathrm{Ó\neg }r$
2. The angular speed of earth $\mathrm{Ó\neg }$is$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{T}$

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

$v=\mathrm{Ó\neg }r$

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$R=6.4×{10}^6 \text{m}$

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

We know that earth makes one rotation per day and

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

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

$\begin{array}{rcl}\mathrm{Ó\neg }& =& \frac{2\mathrm{Ï€}}{8.64×{10}^4 \text{s}}\\ & =& 7.3×{10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Hence the angular speed is, $7.3×{10}^{-5}\raisebox{1ex}{${\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

$$\begin{gathered} v=\mathrm{Ó\neg }r \\ v=\mathrm{Ó\neg }R\cos {40}^0 \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} v=7.3×{10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s×{\displaystyle 6}{\displaystyle .}{\displaystyle 4}{\displaystyle ×}{\displaystyle {{\displaystyle 10}}^6}{\displaystyle  }{\displaystyle \text{³¾Ã—}}{\displaystyle \cos }{\displaystyle }{\displaystyle {{\displaystyle 40}}^0}$}\right. \\ v=3.5×{10}^2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

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

Step 3: (c) Calculation for the role="math" localid="1660897430122" $\mathrm{Ó\neg }$

$\mathrm{Ó\neg }=7.3×{10}^{-5}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Hence the value of $\mathrm{Ó\neg }$is, $7.3×{10}^{-5}\raisebox{1ex}{${\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

$$\begin{gathered} v=\mathrm{Ó\neg }r \\ v=\mathrm{Ó\neg }R \end{gathered}$$

Substitute all the value in the above equation.

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

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