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Suppose that, while lying on a beach near the equator of a far-off planet watchi

ID: 1639779 • Letter: S

Question

Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a stopwatch just as the top of the sun disappears. You then stand, elevating your eyes by a height H = 1.91 m, and stop the watch when the top of the sun again disappears. If the elapsed time is t = 10.1 s, what is the radius r of the planet to two significant figures? Notice that duration of a solar day at the far-off planet is the same that is on Earth. (Hint: See Sample Problem 1-4.)

Explanation / Answer

When the Sun first disappears while lying down, your line of sight to the top of the Sun is tangent to the Earth’s surface at point A shown in the figure. As you stand, elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth’s surface at point B.

let d be the distance from point B to to your eyes from pythogorean theorem

d^2 + r^2 = ( r+h)^2 = r^2 + 2rh + h^2

Now the angle between the two radii to the two tangent points A and B is , which is also the angle through which the Sun moves about Earth during the time interval t = 11.1 s. The value of can be obtained by using

theta/ 360 = t/ 24h

theta = ( 360) (10.1)/ 86400 s

=0.0420 degree

d = r tan theta

d^2 = r^2 tan^2 theta = 2rh

r = 2h/ tan^2 theta

= 2 ( 1.91)/ tan^2 0.0420

=7.08 * 10^6 m

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