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

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 = 1.31 m, and stop the watch when the top of the sun again disappears. If the elapsed time is t = 11.7 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

The Key idea here is that just as the Sun disappears, your line of sight to the top of the sun is tangent to the planet's surface. Two such lines of sight are possible. First, when your eyes are are a height H above the ground (while you are standing) and second, when you are lying down.Let the radii be R and the distance between the points of contact of the tangents on the planet be D

Now, from the Pythagorean theorem, we have :

D^2 + R^2 = (R+H)^2 = R^2 +2RH +H^2
or D^2 = 2RH +H^2

Since the height H is so much smaller that the planet's radius R, the term H^2 can be neglected... Now we have :

d^2 = 2RH ..................................(i)

The angle between the radii to the tangent ponts is A, which is also the angle through which the Sun moves about the planet during the elapsed time, t= 10.5 s. During a full day, which is approximately 24h, the Sun moves through an angle of 360 degrees about the planet. This allows us to write :

A/360 = t/24h

which t= 11.7s, gives us

A = [(360)(11.7s)}/[(24h)(60min/h)(60 s/min)} = 0.04875 degrees

now, since D = R tan A ,

Substituiting this for D in (i) gives us,

R^2tan^2A = 2RH

R = 2H/tan ^2 A

Substituiting A = 0.04875 and H = 1.31m, we find,

R = [(2)(1.31)] / tan^2 (0.04875)= 3.6 X 10^6 m (Answer)

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