Chapter 6.1, Problem 106AYU

To determine

To find:the linear speed of the earth and expressinmiles perhour.

Answer to Problem 106AYU

The linear speed of the earth is $v\approx 6.66\times {10}^4\frac{\text{miles}}{\text{hour}}$

Explanation of Solution

Given information:

The givenmean distance of the earth from the sun (radius) is,

  $\text{r=9}\text{.29}\times {\text{10}}^7\text{miles}$

The orbit of the earth around the suntotakes $1$ revolution in $365$ days.

Concept used:

The object travels around the circles;

That an object moves around a circles of radius $r$ at a constant speed .If $s$ is the distance travelled in time $t$ around this circles, then the linear speed $v$ of the object its defined as.

  $v=\frac{s}{t}$

The relation between linear speed and angular speed;

  $v=r\omega$

Here $\omega$ is measured in radius per unit time.

The angular speed $\omega$ of this object is the angle $\theta$ (measures in radians) swept out. Divided by the elapsed time $t$ that is

  $\omega =\frac{\theta }{t}$

The earth rotates around the sun to complete the one revolution in $365\text{days}$ .

  $1\text{revolution}=2\pi \text{radians}$

The central angle of the earth from the sun is

  $\theta =2\pi \text{radians}$

Convert the days in hours.

  $\begin{array}{l}t=365\text{days}\\ 365\text{days}=365\times 24\text{hours}\\ \left(1\text{day=24}\text{hours}\right)\\ t=8760\text{hour}\end{array}$

The linear speed is.

  $v=\frac{s}{t}$

The distance travelled with radius is calculated as,

  $s=r\theta$

The radius is $\text{r=9}\text{.29}\times {\text{10}}^7\text{miles}$ the angle is $\theta =2\pi \text{radians}$

The distance of the earth from the sun is,

  $\begin{array}{l}s=9.29\times {10}^7\times 2\pi \\ s=18.58\pi \times {10}^7\text{miles}\end{array}$

The linear speed is calculated as,

  $\begin{array}{c}v=\frac{s}{t}\\ \text{=}\frac{18.58\pi \times {10}^7}{8760}\frac{\text{miles}}{\text{hours}}\\ v=6.66\times {10}^4\frac{\text{miles}}{\text{hours}}\end{array}$

Therefore, the linear speed of the earth is $v\approx 6.66\times {10}^4\frac{\text{miles}}{\text{hours}}$