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Thread: Earth slowing physics problem

  1. #1 Earth slowing physics problem 
    Forum Bachelors Degree Demen Tolden's Avatar
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    Hello guys. I'm going to warn you and say that this is a homework problem. I don't really need the answer. I just want to understand what I am doing wrong.

    Problem 18:
    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 66 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.)?

    My work:

    I believe that a forumla that would be useful for this problem is the "partial sum formula."
    http://en.wikipedia.org/wiki/1_%2B_2..._%C2%B7_%C2%B7

    What I've done is multiply what I believe to be rate of daily gain (r) by the partial sum formula where n = the total number of days in 66 centuries.

    My math:







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  3. #2  
    Forum Bachelors Degree Demen Tolden's Avatar
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    oops, hours conversion:



    still wrong though


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  4. #3  
    Universal Mind John Galt's Avatar
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    I must be completely misunderstanding the question. You state that the Earth slows by 1ms per century, therefore - surely - in 66 centuries it will slow by 66ms. Either I am thick, or you have stated the problem badly.
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  5. #4  
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    For one thing there are a thousand milliseconds in a second not a million.

    Ophiolite, they are asking for the accumulated error over the 66 centuries. It's not 1 ms per century. 1 ms is the change in the lenght of 1 day at the end of the century.
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  6. #5  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Ohhh! I was looking at microsecond, not millisecond!

    Thanks for the help Harold. That was my mistake.
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  7. #6  
    Universal Mind John Galt's Avatar
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    Quote Originally Posted by Harold14370
    Ophiolite, they are asking for the accumulated error over the 66 centuries. It's not 1 ms per century. 1 ms is the change in the lenght of 1 day at the end of the century.
    Ah. That's allright then. It was the 'thick' alternative.
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  8. #7 Re: Earth slowing physics problem 
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    Quote Originally Posted by Demen Tolden
    Hello guys. I'm going to warn you and say that this is a homework problem. I don't really need the answer. I just want to understand what I am doing wrong.

    Problem 18:
    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 66 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.)?

    My work:

    I believe that a forumla that would be useful for this problem is the "partial sum formula."
    http://en.wikipedia.org/wiki/1_%2B_2..._%C2%B7_%C2%B7

    What I've done is multiply what I believe to be rate of daily gain (r) by the partial sum formula where n = the total number of days in 66 centuries.

    My math:








    1 day, times 365.25 days, equals 365.25 days, times 100 years equals 36,525 days in a century or 1/36,525th of a microsecond a day increase.

    66 centuries times 36,525 days means there are 2,410,650 days in 66 centuries. Each day increments by 1/36,525 of a micro second from zero increase on day zero, to 66 micro seconds on last day. That means that there are 66 times 36,525, increments of 1/36,525ths micro seconds, or 2,410,650 1/36525ths microsecond increases at the end of 66 centuries.

    So if you take 2,410,650 and square it to get the ramp, you get 5,811,233,422,500 if you divide that in half you get 2,905,616,711,250 total, 1/36,525 micro seconds over 66 centuries.

    2,905,616,711,250 divided by 36,525 Equals 79,551,450 micro seconds. Equals 79.55145 seconds.

    I doubt any clock is going to measure that. And I certainly will not be here to check. Ha-ha.


    This is dependent on which calender you use. You may also have to work out leap year. You might experience an extra leap year in 100 years.


    This is how I think about it. Pretty scary.


    Sincerely,


    William McCormick
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  9. #8 Re: Earth slowing physics problem 
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    Quote Originally Posted by William McCormick
    1 day, times 365.25 days, equals 365.25 days, times 100 years equals 36,525 days in a century or 1/36,525th of a microsecond a day increase.

    66 centuries times 36,525 days means there are 2,410,650 days in 66 centuries. Each day increments by 1/36,525 of a micro second from zero increase on day zero, to 66 micro seconds on last day. That means that there are 66 times 36,525, increments of 1/36,525ths micro seconds, or 2,410,650 1/36525ths microsecond increases at the end of 66 centuries.

    So if you take 2,410,650 and square it to get the ramp, you get 5,811,233,422,500 if you divide that in half you get 2,905,616,711,250 total, 1/36,525 micro seconds over 66 centuries.

    2,905,616,711,250 divided by 36,525 Equals 79,551,450 micro seconds. Equals 79.55145 seconds.

    I doubt any clock is going to measure that. And I certainly will not be here to check. Ha-ha.


    This is dependent on which calender you use. You may also have to work out leap year. You might experience an extra leap year in 100 years.


    This is how I think about it. Pretty scary.


    Sincerely,


    William McCormick
    William, you did the same thing as Demen and used microseconds instead of milliseconds. You actually got the same answer as Demen.

    79,551,450 milliseconds/1000 = 79551 seconds
    79551/3600 seconds per hour = 22.1 hours which is what Demen would have gotten after correcting microseconds to milliseconds.
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  10. #9 Re: Earth slowing physics problem 
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    Quote Originally Posted by Harold14370
    Quote Originally Posted by William McCormick
    1 day, times 365.25 days, equals 365.25 days, times 100 years equals 36,525 days in a century or 1/36,525th of a microsecond a day increase.

    66 centuries times 36,525 days means there are 2,410,650 days in 66 centuries. Each day increments by 1/36,525 of a micro second from zero increase on day zero, to 66 micro seconds on last day. That means that there are 66 times 36,525, increments of 1/36,525ths micro seconds, or 2,410,650 1/36525ths microsecond increases at the end of 66 centuries.

    So if you take 2,410,650 and square it to get the ramp, you get 5,811,233,422,500 if you divide that in half you get 2,905,616,711,250 total, 1/36,525 micro seconds over 66 centuries.

    2,905,616,711,250 divided by 36,525 Equals 79,551,450 micro seconds. Equals 79.55145 seconds.

    I doubt any clock is going to measure that. And I certainly will not be here to check. Ha-ha.


    This is dependent on which calender you use. You may also have to work out leap year. You might experience an extra leap year in 100 years.


    This is how I think about it. Pretty scary.


    Sincerely,


    William McCormick
    William, you did the same thing as Demen and used microseconds instead of milliseconds. You actually got the same answer as Demen.

    79,551,450 milliseconds/1000 = 79551 seconds
    79551/3600 seconds per hour = 22.1 hours which is what Demen would have gotten after correcting microseconds to milliseconds.

    Yea, but I don't read your guys math or understand it, so I worked it out according to my way of doing things. I figured I would share my strange thought processes with you guys. Maybe you would have found a flaw with my methods. And I could improve my math.

    What is it actually demen meant? That it was 1/1000th of second increase per day over a century? I thought he actually said 1/1,000,000th of a second.

    Now I think I understand what happened. I thought that he knew the question and stated 1/1,000,000 and just put the "m" in by mistake. You are saying that the "m" is what was stated in the original question. Now I understand.

    At least my math was right.

    Sincerely,


    William McCormick
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  11. #10 Re: Earth slowing physics problem 
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    Quote Originally Posted by Harold14370
    Quote Originally Posted by William McCormick
    1 day, times 365.25 days, equals 365.25 days, times 100 years equals 36,525 days in a century or 1/36,525th of a microsecond a day increase.

    66 centuries times 36,525 days means there are 2,410,650 days in 66 centuries. Each day increments by 1/36,525 of a micro second from zero increase on day zero, to 66 micro seconds on last day. That means that there are 66 times 36,525, increments of 1/36,525ths micro seconds, or 2,410,650 1/36525ths microsecond increases at the end of 66 centuries.

    So if you take 2,410,650 and square it to get the ramp, you get 5,811,233,422,500 if you divide that in half you get 2,905,616,711,250 total, 1/36,525 micro seconds over 66 centuries.

    2,905,616,711,250 divided by 36,525 Equals 79,551,450 micro seconds. Equals 79.55145 seconds.

    I doubt any clock is going to measure that. And I certainly will not be here to check. Ha-ha.


    This is dependent on which calender you use. You may also have to work out leap year. You might experience an extra leap year in 100 years.


    This is how I think about it. Pretty scary.


    Sincerely,


    William McCormick
    William, you did the same thing as Demen and used microseconds instead of milliseconds. You actually got the same answer as Demen.

    79,551,450 milliseconds/1000 = 79551 seconds
    79551/3600 seconds per hour = 22.1 hours which is what Demen would have gotten after correcting microseconds to milliseconds.

    Why did he come up with a slightly different answer then both of us? I am talking about the numeric value?

    Input error? Or a formula difference?


    Sincerely,


    William McCormick
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  12. #11  
    Forum Bachelors Degree Demen Tolden's Avatar
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    I believe that the answer I have now, 22.097634 hours is the correct answer my book has for this problem. I got this answer after fixing my millisecond-microsecond error.
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  13. #12 Re: Earth slowing physics problem 
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    [quote="William McCormick"]
    Quote Originally Posted by Harold14370
    Why did he come up with a slightly different answer then both of us? I am talking about the numeric value?

    Input error? Or a formula difference?
    He didn't. I just rounded it off.
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  14. #13 Re: Earth slowing physics problem 
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    Quote Originally Posted by Harold14370
    He didn't. I just rounded it off.


    He shows 79551483 as the answer.

    I come up with 79,551,450 without rounding or even hitting any division problems that caused decimal places. I am just curious if there is some basic flaw in that formula?



    Sincerely,


    William McCormick
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  15. #14  
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    Here is what is wrong with my methods.

    When I get the squared number divided by two. I have to add on 1/2 a micro or milli second for each day passed. Then my answer matches yours.


    The last time I did this I was nine years old.








    Sincerely,


    William McCormick
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