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Very hot question(Answering will get you 5 dollars, just kidding.)

  1. Sep 17, 2007 #1
    Very hot question(Answering will get you 5 dollars, just kidding.

    Because Earth's rotation is gradually slowing, the length of each day increases: The day at the end of 1.0 century is 1.0 ms longer than the day at the start of the century. In 24 centuries, what is the total (in hours) of the daily increases in time (that is, the sum of the gain on the first day, the gain on the second day, etc.)?
  2. jcsd
  3. Sep 17, 2007 #2

    Chi Meson

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    Are you asking us to do your work for you?

    I've got an answer. Do you have a particular method that you are expected to use? I used simple logic, using blocks and triangle to represent accumulated time.
    Last edited: Sep 17, 2007
  4. Sep 17, 2007 #3
    Well you need to use arithematic sequence and I got 2.81 hours total change at the end of the given tine. Maybe I understood the question wrong. Sn= n/2(2a+(n-1)d)

    d=1(i think i am not sure)

    Please help.
  5. Sep 17, 2007 #4


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    Your formula is for the sum a+(a+d)+(a+2d)+...+(a+(n-1)d). Since you are only summing the differences, you can put a=0. d should be the difference in the length of the day between two successive days. What is this? Hint: it's not 1ms.
  6. Sep 18, 2007 #5
    For one CENTURY, the increase is one ms per day. You have to find the difference per DAY.
  7. Sep 18, 2007 #6
    its changing the difference per day at the end of the centuary changing.
    .................|(1 millisecond)

    rate at which the time is changing is the difference per day. I dont know what that is.
    Its a tough question.

    We changed the question and used the value of 48 centuries and have received a value of 11.92 hours of cumulative change we used 48 as the difference. The answer was wrong.

    The thing is I do not know what the difference per day is.
  8. Sep 18, 2007 #7
    All I know is the difference is per day. It is a rate of change. Damn I am sorry if I cannot provide you with more information.
  9. Sep 18, 2007 #8


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    Ok, say the day at the beginning is 80000000 ms. At the end of a century it is 80000001 ms. What is it on the second day? Interpolate. It's just a hair over 80000000 ms. How big a hair?
  10. Sep 18, 2007 #9
    the d(difference) is 1/(365* 100*60*60). If your value for n= x*365*100(where x is the value of centuries).

    Thanks Dick. The answer is 2.92 cumulative hours. (This was an interesting question)
  11. Sep 18, 2007 #10


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    Putting some units on things would help. I couldn't figure out your d for a bit. Now I realize that you wrote it in units of millihours.
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