# Trig. Question

• May 26th 2007, 01:31 PM
mathkid2007
Trig. Question
Suppose a metal band stretched around the Earth at the equator. The length of this circular band would be 24901.143 miles (131478035 feet). If this band were to be lengthened by 6 feet and the band were suspended evenly around the Earth above the equator, would you expect there to be a noticeable space between the new band and the surface of the Earth? Find the distance between the lengthened metal band and the surface of the Earth in feet.

I have no clue how to start this so if somebody can just tell me how to get started I should be able to finish...Thanks!
• May 26th 2007, 01:34 PM
Jhevon
Quote:

Originally Posted by mathkid2007
Suppose a metal band stretched around the Earth at the equator. The length of this circular band would be 24901.143 miles (131478035 feet). If this band were to be lengthened by 6 feet and the band were suspended evenly around the Earth above the equator, would you expect there to be a noticeable space between the new band and the surface of the Earth? Find the distance between the lengthened metal band and the surface of the Earth in feet.

I have no clue how to start this so if somebody can just tell me how to get started I should be able to finish...Thanks!

Use the formuala for the circumference of a circle for the old and new band and solve for the radius of each. the difference in the radius will give the desired distance
• May 26th 2007, 03:55 PM
Soroban
Hello, mathkid2007!

This is a classic (old) problem with a surprising answer.

Quote:

Suppose a metal band stretched around the Earth at the Equator.
The length of this circular band would be 24901.143 miles (131478035 feet).
If this band were to be lengthened by 6 feet and the band were suspended evenly
around the Earth above the equator, would you expect there to be a noticeable space
between the new band and the surface of the Earth?
Find the distance between the lengthened metal band and the surface of the Earth in feet.

Never mind the numbers they gave us . . .

Let $R$ be the radius of the earth.
Then the circumference around the Equator is: . $C \:=\:2\pi R$

Now $C$ is increased by 6 feet.
There will be a corrsponding increase in the radius; call it $x$.

We have: . $2\pi(R + x) \:=\:C + 6$

. . Then: . $\underbrace{2\pi R}_{\text{this is }C} + 2\pi x \:=\:C + 6$

. .Hence: . $C + 2\pi x \:=\:C + 6\quad\Rightarrow\quad 2\pi x \:=\:6$

Therefore: . $x \:=\:\frac{6}{2\pi} \:\approx\:0.955\text{ feet }\:\approx 11.5\text{ inches}$

The band will be about a foot above the surface of the earth.

Note: Since the radius $R$ drops out of the problem,
. . . . .the original radius does not matter.

If you wrapped a string around a basketball, then extended the string by 6 feet,
. . the new string would be about a foot from the basketball.