The earth's rotation is slowing down due to the friction against the tidal bulge produced by the moon's and sun's gravitation. The variation is irregular but in general after 100 years the earth has rotated about .25° less than it would have if the rate were the same as at the beginning of the 100 years. That corresponds to one minute's worth of rotation.

1. How much longer (in seconds) is one day today than 100 years ago?
2. How long a period of time need go by for one complete rotation (day) to be missed using the original rotation rate as a standard?
Assume a constant negative acceleration.

Facts/assumptions/interpretations:

1. The total lost rotation equates to 60 seconds (time) during 100 years (orbits).
2. Constant absolute deceleration.
3. 1 year = 1 orbit = 365.25 standard days (I know this isn't exactly correct, but it's close enough)
4. 1 standard day = 60*60*24 seconds ... this is a fixed unit of time

Let X be the number of seconds (time) of rotation lost during a year compared to the previous year. X is constant since the deceleration is absolutely constant.

During year 1 X seconds are lost
During year 2 a further 2X seconds are lost
During year 3 a further 3X seconds are lost
....
During year 100 a further 100X seconds are lost

So in total X+2X+3X+...+100X seconds are lost

-> X+2X+3X+...+100X = 60
-> 5050X = 60
-> X = 0.01188 seconds

Therefore the difference in rotation between year 0 and year 100 is equivalent to 1.188 (=100*0.01188) seconds (time). So one day today is 0.00325 seconds longer than 100 years ago.
[0.00325 is an average figure for year 100, clearly day 1 will be shorter than day 365 - but this makes no difference to 3SF]

Part two of the question is equivalent to finding the smallest Q such that:

X+2X+3X+...+QX >= 86400 (=60*60*24) where X = 0.01188 as above

At the minimum X+2X+3X+...+QX = 86400
-> (Q+1)*(Q/2)*0.01188 = 86400
-> (Q+1)*Q = 14544000
-> Q^2 + Q - 14544000 = 0
-> positive factor gives Q = 3814 years

Posted by fwaff on 2003-03-10 07:23:36