# If the earth's rotational speed increased by 2% each day starting today…what would be the difference in age 20 years from now?

If the new adjusted revolution of the earth still equaled one day and 365 days still equaled one year, how old would someone be 20 years from now (20 years based on the current rotation of the earth) compared to the new rotation of the earth?

I'm looking for a formula for the summation. (It is understood that a complete revolution around the sun would not be equal to 365 days. For the sake of the equation, leap year will not be factored in..1 year=365 days. It is also assumed that human life would still be possible given the earth's new rotational speed)

Edit 1: The increase in speed will increase by 2% over the previous day's rotation, and it will happen at once at midnight. Midnight tonight will indicate the beginning of the 20 year period (at the current rotational speed)

Edit 2: I suppose a bit of suspended disbelief would be in order for this question. Someone who is adept in physics told me that human life and possibly the earth wouldn't even exist if the rotation of it increased at that amount for that length of time.

• @arc_lupus If the rotation speeds up then one rotation, and thus one full day, would continue to come and go quicker and quicker. Quicker days = quicker years, and so on, so with the new rotational speed increasing by 2% each day of the previous day's rotation, then you can see that the # of years would increase greatly compared to the current rotation of the earth. Mar 27 '15 at 15:31
• Ah, ok, I missed the "increased" part. Short idea: $\sum_{k=1}^{k=20\cdot365} n^k$ with $n=0.98$? (Result would be near to 50, i.e. if the earth speed up continues till infinity, the people are half as old as before). Mar 27 '15 at 15:32
• Should the increase be gradual or a sudden change at midnight each day? Mar 27 '15 at 15:39
• @String All of the work above seems to imply a sudden change, but if it were continuous, we would need an integral. Mar 27 '15 at 15:44
• @SUMGUY: I find sudden changes at midnight physically implausible, but as MathHype states, a continuous change will require an integral rather than a sum. Of course I find the whole thing very plausible if we consider gradual speed changes. Mar 27 '15 at 15:47

Since $speed \times time = distance$, and the distance is always constant, so on Day $1$, the speed is $1.02s$ ($s$ is the original speed), therefore the time needed for one full rotation is $t / 1.02$ ($t$ is the original time needed for one full rotation).

On Day $2$, the speed is $1.02^2s$. Therefore the time needed for one full rotation is $t / (1.02^2)$.

$\dots$

In general, on Day $x$, the speed is $1.02^xs$. The time needed for one full rotation is $t / (1.02^x)$.

Assume $20$ years is $7300$ days, now we need:

$$\sum\limits_{x=1}^y \frac{t}{1.02^x} = 7300t$$

and solve for $y$. The $t$ on both sides can be cancelled out so we are left with:

$$\sum\limits_{x=1}^y \frac{1}{1.02^x} = 7300$$

Then we can apply the formula for the sum of the first $n$ terms of a geometric series and solve for $y$.

Added: Bad News! In fact there is no solution for $y$. Using sum of geometric series, we get:

$$\frac{1}{1.02} \frac{1-(\frac{1}{1.02})^{y+1}}{1-\frac{1}{1.02}} = 7300$$

Simplifying gets:

$$50 \times (1-(\frac{1}{1.02})^{y+1}) = 7300$$

As pointed out by @HagenvonEitzen, the LHS cannot exceed $50$ no matter how large $y$ is. So if my workings are correct, it suggests that the speed of the rotation goes so fast that in such a world, the infinity in terms of time is equal to $50$ days in the world we are now living in.

• But even $\sum_{x=1}^\infty\frac1{1.02}^x=50\ll 7300$. That is, in 50 (old) days rotational speed reaches infinity Mar 27 '15 at 16:43
• So not only is it physically absurd, it is mathematically absurd too :P This is a really strange and entertaining problem. Mar 27 '15 at 16:46
• This is not mathematical absurd at all. You reach an arbitrary high speed the closer you get to 50 old days. After that you just haven't defined what happens. In Newtonian physics, accelerating to an infinite speed in a finite time can actually happen. I read this in the German translation of a Book by Ian Stewart. I believe the English version is “Another Fine Math You Have Got Me Into”. Mar 27 '15 at 18:14
• @RolfSievers Yes you are right that the speed is arbitrary high when approaching $50$ old days. I cannot believe the result until thinking about it all over again. From a mathematical viewpoint, the lesson I learned (and was reminded) is "not to under-estimate the power of exponentials". Mar 27 '15 at 18:32
• This assumes the Earth's rotation gets 2% faster every rotation, not every current-rotation-speed day. It's not clear to me which the question is asking about. Mar 27 '15 at 19:46

Use $1$ real earth year as time unit, and denote the angular velocity of the earth rotation, measured in full turns per year, by $f(t)$. Then $f(0)=365$. When the angular velocity steadily increases by $2\%$ per day we have $f(t)=e^{\lambda t}f(0)$ for a certain $\lambda>0$, and this $\lambda$ is determined by the condition $$e^{\lambda/365}=1.02\doteq e^{0.02}\ .$$ Neglecting the error here we obtain $\lambda=7.3$. Therefore the number of felt days during the next $20$ years from now is given by $$\int_0^{20} f(t)\>dt=\int_0^{20} 365\>e^{7.3\>t}\>dt={365\over 7.3}\bigl(e^{146}-1\bigr)\ .$$ Counting the age as number of felt days divided by 365 produces a felt age of $${1\over 7.3}\bigl(e^{146}-1\bigr)\doteq 3.49681\cdot10^{62}$$ years.

• I think your $f(t)$ assumes that the speed increases by $2\%$ each real earth day which equals $1/365$ of $1$ real earth year, right? I made that error too and reached the discrete equivalent of this being $8.278\cdot 10^{61}$ years using a geometric series. But the change rate increases too as the days become shorter, which is what LaBird's answer takes into account, if I am not mistaken. Mar 27 '15 at 16:57
• @String: Of course everything is not so clearly defined here. Nevertheless I think my interpretation of $f$ is correct: If you interpret $f$ as number of felt revolutions in $1$ felt year it is always $={1\over365}$. Mar 27 '15 at 17:37
• Are you then suggesting that my discrete version of your answer is correct too? Mar 27 '15 at 18:03
• @String: Your model is slightly different. Therefore it leads to a sum instead of an integral. The end results don' t differ much, as in the case of yearly interest vs. continuous interest. Mar 27 '15 at 19:00

My original wrong answer: $$\frac{1}{365}\sum_{k=0}^{7299} 1.02^k\approx 8.278\cdot 10^{61}\text{ years}$$

My original answer contained an error assuming that the increase followed the conventional days and not the new faster days. Here are my fixed versions:

Continuous version

If the speed increases gradually and by $2\%$ compared to the last midnight, it can be described via the following differential equation: $$y'=1.02^y$$ where $y(x)$ denotes the number of rotations/"new days" after $x$ conventional days. Given the initial condition $y(0)=0$ this has solution $$y(x)=-\frac{\ln(1-\ln(1.02)x)}{\ln(1.02)}\approx -50.498\ln(1-0.0198026x)$$ which has a vertical asymptote at $x=\frac{1}{\ln(1.02)}\approx 50.498$ so after approximately $50$ and a half conventional days, the earth reaches infinite rotational speed in this model.

Discrete version

The corresponding discrete model, similar to that given in LaBird's answer, should be $$x=\sum_{k=1}^y \frac{1}{1.02^{k-1}}=51\left(1-1.02^{-y}\right)$$ which also has a vertical asymptote, this time at $x=51$.

Here is a graph of the two models (continuous as red curve, discrete as blue points):

• I agree that there shouldn't be down voting without at least a partial explanation. FWIW, I do not know the answer to this question which is why I have put it out to the world, so I am relying on this community to help decide the correct answer. Mar 27 '15 at 16:21

If I well understand the problem, after $1$ day the duration of a day become $d_1=\dfrac{1}{1+k}$ where $k=2\%$. So after $n$ days the duration af a day is $d_n=(1+k)^{-n}$, and the difference with respect the old days is $\Delta_n=n-\sum_{i=1}^n({1+k})^{-n}$. Using $n=365\times 20$ we have the result.

• If it rotates faster, it achieves more "days". So subtracting $2\%$ looks like an error. Anyway, adding $2\%$ is not the opposite of subtracting $2\%$. The opposite of multiplying by $1.02$ is dividing by it corresponding to a factor of $1/1.02\approx 0.980392156862745\neq 0.98$. BTW: I did not downvote your post. Mar 27 '15 at 16:18
• @String: Maybe that I did not understand the question. But... if the Earth increase the speed of rotation, the day become shorter. After the first day it become $1-k$, after 2 days $(1-k)(1-k)$ and after $n$ days (1-k)^n. Mar 27 '15 at 16:27
• @String: Now I see my stupid mistake!!! I' edit.. Mar 27 '15 at 16:37