Originally Posted by Fbone
Since my last question was a dud. Here's a math one as this should be universal.
A string is tied tightly around the Earth's equator.
An additional 10 feet is spliced in.
If the slack is evenly distributed, how high off the Earth's surface will the string be?
Call the Earth's radius r feet. The string length around the world is τr feet. (Where τ is 2π.)
Add 10 feet. New string length is τr+10. Radius of new circle with circumference τr+10 is (τr+10)/τ = r+10/τ.
So it will be 10/τ feet above the surface. That's about 1.6 feet.