Imagine a metal hoop fitted tightly around the Earth at the equator. We now cut the hoop and insert a piece exactly 10 ft long. Now, we hold the hoop in a position so that it is exactly concertric with the equator. How far above the survace will the hoop now be?
Students usually begin by drawing the situation (figure 1). In this figure, C is the original circunference of the hoop and r is the radius; C’ is the circumference of the hoop with the 10 ft added, R is the new radius, and x is the distance (in feed) of the new hoop above the Earth’s survace:
(original circle) C = 2πr
(new circle) C’ = 2π
R= C + 10
C’ + 10 = 2π (r + x)
10 = 2π x
5/π = x
1, 6’ = x
Let’s solve this problem by considering the extreme case. Let the original circle shrink until r equals 0 (Figure 6.6). now, the original “ hoop” has become point O, with C = 0.
The distance we seek is the length of the radius of a circle with a circumference of 10 ft. Thus ,
10 = 2πr
5/π = r
1,6 = r
The distance above the surface is 1,6 ft.