Question

II A 3000 -m-high mountain is located on the equator. How much faster does a climber on top of the mountain move than a surfer at a nearby beach? The earth's radius is $6400\text{km}$.

Short answer

The speed of a climber on top of mountain is $0.208\text{m}/\text{s}$faster than the speed of a surfer at nearby beach.

Step-by-step solution

Step 1. Given information

The$3000\text{m}$ height mountain is located on the equator of the earth, the radius of earth is$6400\text{km}$.

Step 2. Explanation

The difference in speed is,

$\Delta v=v_1-v_2\mathrm{...}\left(\text{I}\right)$

Here, $\Delta v$is the difference in the speed, $v_1$is the speed of the climber in the top of the mountain, $v_2$is the speed of the surfer at nearby beach.

The speed of the climber in the top of the mountain is,

$v_1=\frac{2\pi (r+h)}{24\text{h}}$

Here, $r$is the radius of the earth, and $h$is the height of the mountain.

The speed of the surfer at a nearby beach is,

$v_2=\frac{2\pi r}{24\text{h}}$

Substitute $\frac{2\pi (r+h)}{24\text{h}}$for $v_1$and $\frac{2\pi r}{24\text{h}}$for $v_2$in the equation (I) to find $\Delta v$

$\Delta v=\frac{2\pi (r+h)}{24\text{h}}-\frac{2\pi r}{24\text{h}}$

$\frac{2\pi \left(\left(6400\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)+3000\text{m}\right)}{\left(24\text{h}\times \frac{60\min }{1\text{h}}\times \frac{60\text{s}}{1\text{min}}\right)}-\frac{2\pi \left(6400\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)}{\left(24\text{h}\times \frac{60\text{min}}{1\text{h}}\times \frac{60\text{s}}{1\text{min}}\right)}$

$\begin{array}{l}=465.63\text{m}/\text{s}-465.42\text{m}/\text{s}\\ =0.208\text{m}/\text{s}\end{array}$

Therefore, the speed of a climber on top of mountain is$0.208\text{m}/\text{s}$ faster than the speed of a surfer at nearby beach.