Chapter 6.1, Problem 113AYU

To determine

To find: The travel speed on the surface of Earth at the equator to keep up with the Sun.

Answer to Problem 113AYU

The travel speed on the surface of Earth at the equator to keep up with the Sun is $1036.72\text{miles}/\text{h}$

Explanation of Solution

Given information:

The radius of Earth is $3960\text{miles}$ .

Concept used:

An object moves around a circular path 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 is defined as,

  $v=\frac{s}{t}$

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 relation between linear speed and angular speed;

  $v=r\omega$

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

The travel speed on the surface of Earth at the equator to keep up with the Sunis calculated as,

  $\begin{array}{c}v=\frac{2\pi r}{t}\\ =\frac{2\pi \left(3960\right)}{24}\\ =1036.72\text{miles}/\text{h}\end{array}$

Thus, thetravel speed on the surface of Earth at the equator to keep up with the Sunis $1036.72\text{miles}/\text{h}$ .