Calculate Radius of Earth Using Time to Set Sun: Homework Solution

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The discussion revolves around calculating the radius of the Earth using the time it takes for the Sun to set from different heights. The initial calculation yielded an angle of 0.037956 degrees, resulting in a time of 9.1 seconds for the Sun to disappear again. However, another participant suggested that the correct time should be 5.8 seconds, indicating a potential error in the initial approach. They emphasized the need to correctly apply the angle ratio and consider the distance to the horizon for accurate calculations. The conversation highlights the importance of using proper trigonometric relationships and clarifying the calculations involved.
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Homework Statement


The Sun sets, fully disappearing over the horizon as you lie on the beach, your eyes 30 cm above the sand. You immediately jump up, your eyes now 170 cm above the sand, and you can again see the top of the Sun. If you count the number of seconds ( = t) until the Sun fully disappears again, you can estimate the radius of the Earth. Use the known radius of the Earth to calculate the time t.


Homework Equations





The Attempt at a Solution



First, I set up a ratio. change in angle theta/360 = t / 86400 seconds in a day. I solved for the change in the angle theta by using inverse cos (radius of earth/ radius of Earth + H, where H is the difference in between standing up and laying down 1.7m-3m). Solving this, I got an angle of .037956 deg. Then the ratio .037956/360 = t/86400, I found t to be 9.1 seconds. This is an incorrect answer, and I was hoping someone point out something I'm missing. I know that when I'm laying down, my view is tangent to the surface of the earth. Advice/diagrams would be much appreciated.
 
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no one?
 
rwx1606 said:
no one?

What is your distance to the horizon?

Can you use the approximation that 1.4 m / Distance to Horizon will be the Sine of the angle that your eye makes with the horizon in radians?

Then won't that ratio to circumference correspond with the time measured ratio to a revolution?
 
Depending on the latitude you may have a further correction to make, so I would want to say you are on a beach in the equatorial tropics and not on some sunny shore on the North Sea.
 
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The change in angle will be the difference in horizon distances divided by the radius of the Earth. Thats easier because you can use small angle approx. for tan. The way you've done it is possible but you have to take the inverse cosine of each ratio separately and then subtract.
 
The equation of angle theta/360 = t / 86400 should be angle theta/360(degree) = t / 86400, so the answer should be 5.8s. It is lucky that your question has the same number as mine in the masteringphysics, so I'm pretty sure that 5.8s is the answer.
 
Yawen said:
The equation of angle theta/360 = t / 86400 should be angle theta/360(degree) = t / 86400, so the answer should be 5.8s. It is lucky that your question has the same number as mine in the masteringphysics, so I'm pretty sure that 5.8s is the answer.

Wait what?

I'm having the exact same issues at the thread starter. I got angle .03795 degrees, which gives me an answer of 9.1 seconds. But it's 5.8 s.

Can you explain how you got 5.8?
 
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