Q. 104 - Precalculus Enhanced with Graphing Utilities

Question

Linear Speed on Earth Earth rotates on an axis through its poles. The distance from the axis to a location on Earth at  40° north latitude is about 3033.5 miles. Therefore, a location on Earth at 40° north latitude is spinning on a circle of radius of 3033.5 miles. Compute the linear speed on the surface of Earth at 40° north latitude.

Step-by-Step Solution

Answer

The linear speed on the surface of the earth at  $40°$ in the north latitude is 794 miles/hours.

Step 1. Given information.

Consider the given information,

The distance from the axis to a location on Earth at  40° north latitude is about 3033.5 miles. Therefore, a location on Earth at 40° north latitude is spinning on a circle of radius of 3033.5 miles.

Step 2. Explanation to find the value.

Since-, the earth rotates 360 degrees in 24 hours. Thus,

$1°=\frac{24 \mathrm{hours}}{360°}$

Now, check for 40 degrees.

$\begin{array}{rcl}t& =& \frac{24 \mathrm{hours}}{360°}\times 40°\\ & =& \frac{{\displaystyle 24 \mathrm{hours}}}{{\displaystyle 9}}\\ & =& \frac{8}{3} \mathrm{hours}\end{array}$

Convert the angle 40 degrees into radian.

$\begin{array}{rcl}\theta & =& 40°\times \frac{\mathrm{\pi }}{180°}\\ & =& \frac{2\mathrm{\pi }}{9}\end{array}$

Step 3. Find the linear speed.

Since angular speed is  $\omega =\frac{\theta }{t}$and linear speed is $v=\frac{\theta }{t}·r$. Thus,

$v=\frac{\theta }{t}·r$

Substitute the angle $\frac{2\mathrm{\pi }}{9}$, time is $\frac{8}{3} \mathrm{hours}$ and the radius is 3033.5  miles. Substitute the values in the formula,

$\begin{array}{rcl}v& =& 3033.5 \times \frac{{\displaystyle \frac{2\mathrm{\pi }}{9}}}{{\displaystyle \frac{8}{3}}}\\ & =& 3033.5 \times \frac{6\mathrm{\pi }}{72}\\ & \approx & 794 \mathrm{miles}/\mathrm{hours}\end{array}$