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.
(In reply to Solution to Part I
by K Sengupta)
Let f = increase in positional difference in rotation.
s = time of rotation
a = acceleration
Then, we note that f varies directly as s, whenever a is constant. ......(i)
Let the required period of time be r years.
Since there are precisely 1440 seconds in one day, in conformity with (i), we must have:
1440 = r/100
-> r = sqrt(144000) = 3794.73322.....
Consequently, the required period is approximately 3794.73322 years.