Solve a problem about a metal hoop fitted tightly around the Earth at the equator

Problem.

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?

Solution :

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

 Figure 1

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.

Figure 4

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.

Tentang matitaputtychristi

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