Question #64452

While lying on the the beach near the equator 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 of 1.70 m and see the sun again. You stop the watch when the top of the sun disappears again. if the elapsed time is 574 s, what is the radius of the earth?
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Expert's answer

2017-01-09T13:40:14-0500

Answer on Question #64452-Physics-Classical Mechanics

While lying on the beach near the equator 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 of 1.70m1.70\mathrm{m} and see the sun again. You stop the watch when the top of the sun disappears again. If the elapsed time is 5.74 s, what is the radius of the earth?

Solution

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 dd be the distance from point B to your eyes. From Pythagorean Theorem, we have


d2+r2=(r+h)2=r2+2rh+h2d ^ {2} + r ^ {2} = (r + h) ^ {2} = r ^ {2} + 2 r h + h ^ {2}


or d2=2rh+h2d^2 = 2rh + h^2 where rr is the radius of the Earth. Since rhr \gg h , the second term can be dropped, leading to d22rhd^2 \approx 2rh . Now the angle between the two radii to the two tangent points AA and BB is θ\theta , which is also the angle through which the Sun moves about Earth during the time interval t=5.74st = 5.74 \, \text{s} . The value of θ\theta can be obtained by using


θ360=t24h\frac {\theta}{3 6 0 {}^ {\circ}} = \frac {t}{2 4 h}


This yields


θ=(360)(5.74s)(24h)(60minh)(60smin)=0.02392\theta = \frac {(3 6 0 {}^ {\circ}) (5 . 7 4 \mathrm {s})}{(2 4 h) \left(6 0 \frac {\min}{h}\right) \left(6 0 \frac {s}{\min}\right)} = 0. 0 2 3 9 2 {}^ {\circ}


Using d=rtanθd = r \tan \theta , we have d2=r2tan2θ=2rhd^2 = r^2 \tan^2 \theta = 2rh , or


r=2htan2θ=21.70tan20.02392=19.5106m.r = \frac {2 h}{\tan^ {2} \theta} = \frac {2 \cdot 1 . 7 0}{\tan^ {2} 0 . 0 2 3 9 2 {}^ {\circ}} = 1 9. 5 \cdot 1 0 ^ {6} m.


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