# Simple but poorly worded unit conversion question

## Homework Statement

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 54 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. Relevant equations
Sum of an arithmetic sequence (see below)

## The Attempt at a Solution

I've seen this question "answered" a number of times via a google search (for example, https://www.physicsforums.com/showthread.php?t=208510"), but nobody seems to ask the more fundamental question: What are they even asking? Most attempts at answering the question use an arithmetic sequence. For example, if the day is 1ms longer at the end of each century, then at the first day it was 1/36525 ms (365.25 days/year, 100 years/century) longer than at the beginning of the first century, and at the second day it was 2/36525 ms longer than at the beginning of the first century, and more generally at the n'th day it was n/36525 ms longer than at the beginning of the first century. So the sum of all these gains is (1 + 2 + 3 + ... + k) / 36525, where k is the total number of days in 54 centuries.

I have three problems with all of these solutions.

First of all, it makes no sense. The answer you get is meaningless in the context of physics and has no physical interpretation that I can come up with. This is normal in a math book to have problems with no physical interpretation, but I'm suspicious of seeing a physics problem like this.

Second, I'm not sure what this formula is that people are using for the sum of an arithmetic sequence. In the link above, for example, someone claims that the sum is (n/2) · [2·a1 + (n-1)·d]. What are a1 and d? And what happened to n*(n+1) / 2?

Third, the wording of the question says "what is the total (in hours) of the daily increases in time". The daily increases. The increases between each day. The increases between each day are constant, 1/36525 ms. So the sum of the daily increases for a century should be 1ms, and the sum of the daily increases for 54 centuries should be 54 ms, or 1.5 × 10^-5 hours.

Anyone have any insight on what the heck this question is talking about?

Thanks

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Second, I'm not sure what this formula is that people are using for the sum of an arithmetic sequence. In the link above, for example, someone claims that the sum is (n/2) · [2·a1 + (n-1)·d]. What are a1 and d? And what happened to n*(n+1) / 2?
A term in an arithmetic sequence is given by $$u_{n} = a + (n-1) d$$.
a is the constant base term, while d is the common difference between consecutive terms.
The $$\frac{n(n+1)}{2}$$ that you cite is only true for the sum 1+2+3+...n, which would constitute an arithmetic sequence $$u_{n} = 1 + (n-1) (1) = n$$. Substituting a and d = 1 into the general formula thus yields $$\frac{n(n+1)}{2}$$.

A term in an arithmetic sequence is given by $$u_{n} = a + (n-1) d$$.
a is the constant base term, while d is the common difference between consecutive terms.
The $$\frac{n(n+1)}{2}$$ that you cite is only true for the sum 1+2+3+...n, which would constitute an arithmetic sequence $$u_{n} = 1 + (n-1) (1) = n$$. Substituting a and d = 1 into the general formula thus yields $$\frac{n(n+1)}{2}$$.
Ahh, I'm always bad at remembering formulas. I tend to just stick with a single formula whenever possible and then just massage everything into the right format to make it work.

Thanks, now just looking for any suggestions on the problem itself

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 54 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.)?
As I read the question I just see it as your 3rd point. The sum of the gain of each of the days is simply 54ms. As the question gives you the increase per century independent of the length of the day I don't see why you are involving sumations at all. The increase per century is 1ms independent of the lenght of the day during the century as I see it.

ideasrule
Homework Helper
Anyone have any insight on what the heck this question is talking about?
I'm pretty sure I know what it means.

Suppose you have a cesium atomic clock that can keep time much better than Earth's rotation. (Actually, you don't have to suppose; they exist.) After 54 centuries, the clocks will say that the time is such-and-such. Because Earth's rotation has been slowing down, astronomical observations will indicate a different time. The clocks might say that it's noon in England, for example, even though the Sun has yet to rise. The question asks: what is the difference between the clock's time and the time determined by Earth's rotation?

This question is extremely meaningful because it turns out that the slowing of Earth's rotation is more than detectable with atomic clocks. Every few years, a "leap second" is added to the time to compensate for decreases in the length of the day. Many effects contribute to the decrease: the most significant are tidal interactions with the Moon and glacier rebound, but other effects are of a similar order of magnitude. That's why leap seconds don't get inserted at regular intervals: because Earth's rotation is not slowing at a constant rate. However, for this question, you have to assume that it is; you'll be able to get a reasonably accurate answer because the extremely regular tides generated by the Moon dominate other factors.

If you're interested, you can read http://tycho.usno.navy.mil/leapsec.html [Broken] for a more detailed explanation of the leap seconds that inspired your question.

About the question itself: using the arithmetic series formula is only an approximation, but it's more than accurate enough considering the uncertainty in the rate that Earth's rotation slows. What you're assuming is that Earth rotates at exactly the same rate for one revolution, then instantly changes its rotation speed and rotates exactly 1 revolution at the new speed, then repeats the process. In reality, rotation speed is slowing constantly and not just at year-end, and to rigorously account for that you'd have to use integration. For this problem, integration happens to be easier. Suppose you have a small period of time dt. After time t, dt has lengthened by ktdt, where k is 1 ms/century. Its length would then be (1+kt)dt. Integrating that gives t+kt2/2, so kt2/2 would be the time difference between real time and solar time.

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As I read the question I just see it as your 3rd point. The sum of the gain of each of the days is simply 54ms. As the question gives you the increase per century independent of the length of the day I don't see why you are involving sumations at all. The increase per century is 1ms independent of the lenght of the day during the century as I see it.
I actually agree with you, but myself and a friend both solved the problem independently, and I did it the 54ms way, and he did it the summation way. Neither of us could understand why the other was doing it the way they did, so we searched on google and found many "solutions" to the problem. Almost all used the summation way, which is really strange to me. I guess this was why I got my undergrad degree in math, so I dont' have to worry about lack of precision in the wording of problems :)