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Laying on the beach while sunseting question

  1. Aug 29, 2012 #1
    1. The problem statement, all variables and given/known data
    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 = 1.58 m, and stop the watch when the top of the sun again disappears. If the elapsed time is t = 10.9 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.


    2. Relevant equations
    86400 s=1 day

    3. The attempt at a solution
    I attempted to solve this by setting the angle, but I think I got the wrong angle. (.0454°)
    cos(0.0454°)=((r)/(r+h)) h=1.58
    When I solved for the radius I got 1.5e^3, but it is not the right answer. Please help me figure where I went wrong! Thank you!
     
  2. jcsd
  3. Aug 29, 2012 #2

    ehild

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    Hi djkinney,

    how did you get the angle?

    ehild
     
  4. Aug 29, 2012 #3
    Hey,
    Well I'm not quite sure as to how to get the angle. I tried two different ways so far: First, I tried: 10.9s/86400s=1.26e-4°. The second was me trying something crazy and doesn't make any sense: 10.9s*(1min/60s)*(1hr/60min)*(360/24hr)=.0454°
     
  5. Aug 29, 2012 #4

    ehild

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    The first method will be all right, but it is not degrees. 1.26e-4 is the ratio to the total revolution, which corresponds to 360 degrees.

    ehild
     
  6. Aug 29, 2012 #5
    Okay, so that means that 1.26E^-4*360°=.04536° for the angle? Then I would still be getting my final answer as 1.5E^3
     
  7. Aug 29, 2012 #6

    ehild

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    Well, what is that angle and how is it related to the radius of the planet and your height?

    ehild
     
  8. Aug 29, 2012 #7

    CWatters

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    Cos(Angle) = R/(R+1.58)

    Angle = 360 * 10.9/(24 * 60 * 60)
     

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    Last edited: Aug 29, 2012
  9. Aug 29, 2012 #8
    Okay, so when I solve for the angle I get .0454°
    Then take the cos of the angle in which I get ≈ .9897
    Then I set that equal to R/(R+h) h=1.58 → .9897=R/(R+1.58)
    Solve for R by multiplying .9897(R+1.58)=R
    Then get like terms of one side 1.59=R-R(.9897)
    A R can be taken out 1.59=R(1-.9897)
    Then I divide 1.59/(1-.9897)=R
    R=154m
    Is this the correct setup?
     
  10. Aug 29, 2012 #9

    CWatters

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    I get

    R = 1.58 Cos(Angle) / (1-cos(angle))

    I make Cos(angle) = 0.9999996858

    and

    R= 5,029,238 meters

    The answer is error prone because the denominator involves subtracting two similar numbers. Might be best to give an answer such as approx 5,000,000 meters?
     
    Last edited: Aug 29, 2012
  11. Aug 29, 2012 #10
    Wow, I would've never done that. Even my calculus professor didn't tell me that. Thank you so much guys for your help!
     
  12. Aug 29, 2012 #11

    CWatters

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    Tip: It's usually a mistake to substitute actual values given in the problem until you have rearranged the equations.

    The steps I did were..

    Cos(angle) = R/(R+1.58)

    Cos(angle) * (R+1.58) = R

    R*Cos(angle) + 1.58*Cos(angle) = R

    1.58*Cos(angle) = R - R*Cos(angle)

    1.58*Cos(angle) = R (1 - Cos(angle))

    R = 1.58*Cos(angle) / (1 - Cos(angle))

    but there might be a shorter way!

    I then put the result of Cos(360*10.9/24*60*60) in the memory of the calculator to preserve as many digits as possible while calculating the actual result.
     
  13. Aug 29, 2012 #12
    The problems with roundoff error can be removed by using a Taylor series expansion:

    1-cosθ ~ θ2/2
     
    Last edited: Aug 29, 2012
  14. Aug 29, 2012 #13
    I have no preference about length of finding a solution as long as it's the correct one. Thank you so much once again!
     
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