Question

Speed at the Equator The earth rotates about its axis once every 23h 56 min 4s, and the radius of the earth is. Find the linear speed of a point on the equator in$\mathbf{\text{mi}}\mathbf{/}\mathbf{\text{h}}$.

Short answer

$\mathbf{1039}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{\text{miles/hr}}$.

Step-by-step solution

Step 1. Given information.

Value of $1\text{\hspace{0.17em}hr}=60\text{min}$and$1\text{\hspace{0.17em}\hspace{0.17em}min}=60\text{\hspace{0.17em}sec}$.
The given values are:
The radius of the earth is$r=3960\text{\hspace{0.17em}\hspace{0.17em}miles}$and time in 1 rotation is$23\text{\hspace{0.17em}hr\hspace{0.17em}\hspace{0.17em}}56\text{min\hspace{0.17em}\hspace{0.17em}}4\text{sec}.$.
To find:
Linear Speed of a point on the equator.

Step 2. Formula Used

Convert total time in hr and find the angular speed $w=\frac{\theta }{t}$and linear speed $v=rw$.

Step 3. Finding the value of linear speed.

$\text{Totaltimeinhr\hspace{0.17em}}t=23+\frac{56}{60}+\frac{4}{60\cdot 60}$

$t=23.9344\text{h}$

As we know that
Angular speed$w=\frac{\theta }{t}$

$w=\frac{2\pi }{23.9344}\text{rad/hr}$

Since,

The linear Speed$v=rw$

$r=3960$

Put the values in this formula of linear speed

Thus,

The linear speed of a point is$v=3960\cdot \left(\frac{2\pi }{23.9344}\right)$

The linear speed of a point is$v=1039.55\text{\hspace{0.17em}\hspace{0.17em}miles}/\text{hr}$