Question

Suppose that, while lying on the beach near the equator watching the sun set over a
calm ocean, you start a stop watch just as the top of the sun disappears. You then
stand, elevating your eyes by a height H=1.70m, and stop the watch when the top of
the sun again disappears. If the elapsed time is t=574s, what is the radius of the
earth?

Answer 1

Answer on Question #50137, Physics, Mechanics | Kinematics | Dynamics

Suppose that, while lying on the beach near the equator watching the sun set over a calm ocean, you start a stop watch just as the top of the sun disappears. You then stand, elevating your eyes by a height $H = 1.70m$ , and stop the watch when the top of the sun again disappears. If the elapsed time is $t = 574s$ , what is the radius of the earth?

Solution:

As you see, the earth rotated $\theta$ degrees in 574 secs. We can calculate $\theta$ . If the earth turns $2\pi$ degrees in 86400 seconds (24 hours), how much does it turn in 574 s? This gives

$$\theta = \frac {5 7 4}{8 6 4 0 0} * 2 \pi = 0. 0 4 1 7 4 \mathrm {r a d}$$

There's a right triangle.

$$\cos \theta = \frac {r}{r + 1 . 7}$$ $$\frac {r}{r + 1 . 7} = \cos (0. 0 4 1 7 4) \approx 0. 9 9 9 1 2 9 0 1 3$$ $$r = \frac {1 . 7 * \cos \theta}{1 - \cos \theta} = \frac {1 . 7 * 0 . 9 9 9 1 2 9 0 1 3}{1 - 0 . 9 9 9 1 2 9 0 1 3} = 1 9 5 0 m$$

Answer: $r = 1950 \, m$