Calculating Cumulative Effect of Slowing Earth's Rotation Over 20 Centuries

• NYmike
In summary, the question asks for the cumulative effect on the measure of time over 20 centuries if the length of the day increases by 0.001 seconds per century. The answer is 2 hours, and there is a quicker way to solve this problem by using integration instead of a 20-step summation.

NYmike

Hello. I am stuck on this physics problem that i have for homework. Its the wording of the question that's odd to me. I am not sure exactly where to begin.

Assuming the length of the day uniformly increases by 0.001 sec. per century, calculate the cumulative effect on the measure of time over 20 centuries. Such slowing of the Earth's rotation is indicated by observations of the occurrences of solar eclipses during this period.

So after one century, the length of each day will be 24 hours and .001 s? And after the second century the length of each day will be 24 hours and .002 s? Is there a quicker way to do this than doing a 20-step summation!?? I started off by multiplying .001 x 365, because that will give you the total amount of added time after a year ... however after that it would be long summation, and I am sure that there must be a quicker way.

(FYI, the answer is 2 hours if you want to check your work :P)

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NYmike said:
Hello. I am stuck on this physics problem that i have for homework. Its the wording of the question that's odd to me. I am not sure exactly where to begin.

So after one century, the length of each day will be 24 hours and .001 s? And after the second century the length of each day will be 24 hours and .002 s? Is there a quicker way to do this than doing a 20-step summation!?? I started off by multiplying .001 x 365, because that will give you the total amount of added time after a year ... however after that it would be long summation, and I am sure that there must be a quicker way.

(FYI, the answer is 2 hours if you want to check your work :P)

One thing that worries me is the word continously, but disregarding that you are on the right track. But you will be off by a factor of 100.

So you have 100 years in every century with 365 days in every year (we are disregarding the fractional day here).

$$36500 \times .001 + 36500 \times .002 + 36500 \times .003 \ldots 36500 \times .019$$

notice it only goes up to .019 since the first year has no additional time.

This is then equal to:

$$36500 \times (.001 + .002 + .003 \ldots .019)$$

Then divide this number by 3600 (the number of seconds in an hour) and you should get the correct answer.

The proper way to do this is by integration since the change is continuous.
Let the length of a day in seconds D be a continuous function of time t. D(t) is linear. The number of days in a century is about 365.25*100 = dc. So if the length of the day at t=0 is 24*3600, after dc days D = 24*3600+.001. The slope of the graph of D(t) vs. t is:

$$\frac{D(dc)- D(0)}{dc}$$ so:

$$D(t) = D(0) + \frac{D(dc)- D(0)}{dc}t$$

Integrating D(t) from 0 to 20 centuries gives us the length (L(t)) in days:

$$L(t) = \int_0^t (D(0) + \frac{D(dc)- D(0)}{dc}t) dt$$

$$L(t) = D(0)t + \frac{D(dc)- D(0)}{2dc}t^2$$

$$L(20dc) = D(0)(20dc) + \frac{D(dc)- D(0)}{2dc}(20dc)^2$$

If you work that out and subtract D(0)*20dc (the measured time), you will get the difference between actual accumulated time and measured time.

eg.

$$L(20dc) = D(0)(20dc) + \frac{.001}{2dc}(20dc)^2$$

$$L(20dc) = D(0)(20dc) + .001(200dc)$$

$$L(20dc) - D(0)(20dc) = 7305$$

So the difference is 7305 seconds or just over 2 hours.

AM

Edit: My earlier post used seconds rather than days.

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Norman said:
One thing that worries me is the word continously, but disregarding that you are on the right track. But you will be off by a factor of 100.

So you have 100 years in every century with 365 days in every year (we are disregarding the fractional day here).

$$36500 \times .001 + 36500 \times .002 + 36500 \times .003 \ldots 36500 \times .019$$

notice it only goes up to .019 since the first year has no additional time.

This is then equal to:

$$36500 \times (.001 + .002 + .003 \ldots .019)$$

Then divide this number by 3600 (the number of seconds in an hour) and you should get the correct answer.

Yeah, I realized that this is quite easy once you look at it this way:

$$36500 \times (.001 + .002 + .003 \ldots .020) = 36500 \times 10^{-3} \times \sum_{x=0}^{20} x \rightarrow 36500 \times 10^{-3} \int_0^{20} x dx$$

all we are doing here is going from the discrete sum to the continuous integration (I know, I know this is not mathematically rigorous, I left out the limiting procedure, but hey, I am a physicist)

You get the same answer as Andrew.

1. What is the cumulative effect of slowing Earth's rotation over 20 centuries?

The cumulative effect of slowing Earth's rotation over 20 centuries is a decrease in the length of a day. This means that the Earth's rotation has slowed down, causing each day to be slightly longer than the previous one.

2. How is the cumulative effect of slowing Earth's rotation measured?

The cumulative effect is typically measured by calculating the difference between the current length of a day and the length of a day 20 centuries ago. This difference is then divided by the number of days in 20 centuries to determine the average decrease in length per day.

3. What factors contribute to the slowing of Earth's rotation?

The main factor contributing to the slowing of Earth's rotation is tidal friction, which is caused by the gravitational pull of the Moon and the Sun. Other factors include changes in Earth's mass distribution and changes in the planet's internal structure.

4. How does the slowing of Earth's rotation affect our daily lives?

The slowing of Earth's rotation has a very small impact on our daily lives. The change in the length of a day is only a few milliseconds per century, which is imperceptible to humans. However, it can affect certain astronomical and navigational calculations that require precise timing.

5. Is there any way to reverse the slowing of Earth's rotation?

There is currently no way to reverse the slowing of Earth's rotation. However, in the distant future, it is possible that the Earth's rotation could speed up again due to the redistribution of mass and energy within the planet. This phenomenon is known as tidal acceleration.