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What is the radius r of Earth?

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data


    Suppose that, while lying on a 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 H=1.70 m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t=11.1s, what is the radius r of Earth?


    2. Relevant equations

    Does the Earth's rotation speed come into account?


    3. The attempt at a solution

    This problem is baffling me, as I can't quite grip what to do or where to start. The Earth is also not flat either which introduces another problem. I am pretty sure there is going to be an equation with r in it. Then another one with r+1.7 in it. The other variable, I am not sure of but I know there has to be atleast another one. If someone could please draw a picture and upload, I immensely appreciate it as once I get a visual for what I am doing, I understand a whole lot more, but that is not the case with this problem. Thanks guys.
     
  2. jcsd
  3. Aug 25, 2010 #2

    ehild

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    Re: Sunset

    Here is a picture for you. The red lines are the rays of sun, they are parallel as the sun is very far away. The blue dot is your head. If you stayed lying, you would not see the sun after sunset as the Earth screens the rays, but you can still see them when you stand up. The Earth rotates so when you stand you go round along a circle of radius R+1.7 m.

    ehild
     

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  4. Aug 25, 2010 #3
    Re: Sunset

    Thank you, that pic helped me right away and I got it. I never thought about the sun's rays hitting the Earth, I tried to picture a beach and looking west and that way could not account for the Earth rotating. Thanks so much. Is the answer 5217957.989m? I got that but that is not the radius of the Earth...
     
  5. Aug 25, 2010 #4

    ehild

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    Re: Sunset

    The radius of Earth is 6370 km, you got 5218 km, it is not bad. I hope the problem did not want to take more things into account.

    ehild
     
  6. Aug 25, 2010 #5
    Re: Sunset

    Actually, I checked the answer to the problem and what I got is correct, so I guess it was not related to the actual radius. I solved it like this: 11.1s/(3600*24)s=x/360, and so the angle was 11.1/240 degrees. There was a right triangle with hypotunese r+1.7, and a leg r. The angle between these two was 11.1/240 degrees and by using trig, I got the other leg to rtan(11.1/240). Then r^2+3.4r+2.89=r^2+.00087r^2, r= 5217957.989m.
     
  7. Aug 25, 2010 #6

    ehild

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    Re: Sunset

    OK, well done! The light bends in the atmosphere because of refraction and the sun is seen for some time after it is actually below the horizon, so this method is not too accurate.


    ehild
     
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