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Time of sunset using Radius of earth

  1. Aug 23, 2009 #1
    URGENT Time of sunset using Radius of earth

    1. The problem statement, all variables and given/known data

    The Sun sets, fully disappearing over the horizon as you lie on the beach, your eyes 20 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.

    2 sig figs
    2. Relevant equations



    3. The attempt at a solution
    radius = 6400m
    radius + height = 6400.015
    cos(R/(R + h)) = .54 radians
    (time/ 1 day) = radians/ 2pi
    time = ? I dont think im doing this right because im getting like 73000 seconds with this method.
     
    Last edited: Aug 23, 2009
  2. jcsd
  3. Aug 23, 2009 #2

    tiny-tim

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    Hi George3! :smile:
    Small world! :wink:
     
  4. Aug 23, 2009 #3
    I meant km
     
  5. Aug 23, 2009 #4

    D H

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    What is 6400 km + 150 cm?

    Edit
    Why are you taking the cosine of R/(r+h)?
     
    Last edited: Aug 23, 2009
  6. Aug 23, 2009 #5
    I was taking the cosine of(R/R+h) to find the angle that the earth moved. Do you have any idea how to tackle this problem?
     
  7. Aug 23, 2009 #6

    D H

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    Drawing a picture is often a good start.

    Since you are taking the cosine of R/(R+h), it must represent some angle. What is it? (Hint: There is no such angle.)
     
  8. Aug 23, 2009 #7
    I drew a picture, the line of sight when you are lying down is tangent to the earth's surface. This would make a right angle with the radius of the earth. Where am I going wrong?
     
  9. Aug 23, 2009 #8

    D H

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    R/(R+h) is the ratio of two sides of your triangle. It is not an angle. So why are you taking the cosine of this ratio? You are on the right track but you are using the wrong function.
     
  10. Aug 23, 2009 #9
    For this problem I did this:
    arccos(6400000m/6400001.5m) = .00021radians
    .00068radians = .039degrees
    .039degrees / 360degrees * 86400secs = 9.41secs
    Is this right?
     
    Last edited: Aug 23, 2009
  11. Aug 23, 2009 #10

    D H

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    Much better.
     
  12. Aug 24, 2009 #11
    This is an online assignment and I only have one more chance to get it right. Should I turn this in or keep working?
     
  13. Aug 24, 2009 #12

    ideasrule

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    Wait. I don't know how strict your assignment's requirements are, because I haven't seen it, but there's one thing you haven't considered.

    You've been assuming that the Sun moves across the sky at 360 degrees per 24 hours, which is (at least approximately) correct, but does it lose altitude at that rate? Think about the Sun's motion: it rises near the east but not exactly due east, arcs its way towards the north or south and reaches due north/south at solar noon, and sets in the west. It doesn't rise in the east and plop right down in the west, so there's a horizontal component of its motion.

    You've calculated the minimum possible value of "t". At the poles, where the Sun doesn't set for six months, "t" is on the order of hours or days. If the Earth didn't revolve around the Sun, "t" would be infinite.
     
  14. Jan 26, 2010 #13
    I highly doubt we need to take into consideration of the differences in altitude (just as we prob. don't have to account for the fact that atmosphere bends the light, etc). My attempt at this problem is similar, but with law of sine. And honestly, I can't get this problem either (also at one attempt left).

    Lol.

    I'll see.
     
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