# Homework Help: Time dilation difference of astronaut problem

1. Apr 7, 2007

### anarine24

I have a question regarding time dilation: An astronaut travels at a speed of 7800 m/s relative to the earth. According to a clock on earth, the trip lasts 15 days. Determine the difference (in seconds) between the time recorded by the earth clock and the astronaut’s clock.

Now I took the formula t = t0/sqrt(1-v^2/c^2), where v=7800m/s and t=15 days, or 1.3E6s. I plugged those in (with c=3E8m/s) to find t0. Since the question asked for the difference between the two times, I did t-t0 and I got an answer of 0s, because my t0 turned out to be the same as t. I was also told with the problem that the answer is 4.4E-4 seconds, but I'm not seeing how. I know that the time measured by the earth (t) is supposed to be higher than the time measured by the astronaut (t0), so there must be some difference of times.

Anyway, I'm stuck and it's probably easy but I just can't see it right now. Any help would be greatly appreciated!!!

Last edited: Apr 7, 2007
2. Apr 7, 2007

### HungryChemist

if you're punching in those number with your calculator, it is most likely that you're calculator is not showing you the digits that are necessary to distinguish b/t t0 and t since the difference is very small as you can infer from the fact that your space ship has a velocity that are much smaller compare to that of speed of light.

Suggestion: Use Mathematica with added command such as FullForm. It will return a value with all the digits that is saved.

3. Apr 7, 2007

### daniel_i_l

How is t0 the same as t if t = t0/sqrt(1-v^2/c^2)? But i think that your problem is that since 7800m/s is so much less than the speed of light the gamma factor is very close to one so the difference in times is very close to 0. You know that this has to be true or else you'd notice the effects of SR in your everyday life.

Last edited: Apr 7, 2007
4. Apr 7, 2007

### nrqed

You did it right. The problem is just a matter of significant figure. Maybe you used a calculator with poor precision and it rounded off too much. Either you must use a calculator which handles more significant figures or you must use a Taylor series which gives that for v much smaller than c (which is the case here), one over the square root of 1-v^2/c^2 may be approximated by

$$\frac{1}{\sqrt{1-v^2/c^2}} \approx 1+ \frac{v^2}{2 c^2}}$$

so the difference of time between the two frames is approximately $t_0 \times \frac{v^2}{2 c^2} = 4.39 \times 10^{-4} s$

5. Apr 7, 2007

### anarine24

Oh thank you so much b/c I was wondering what I did wrong!!! I used a TI-83 graphing calculator to solve this problem, so it must have rounded the number off too much.

6. Apr 7, 2007

### nrqed

I am surprised it did not give you the correct answer, though. Try plugging in $\frac{1}{\sqrt{1-v^2/c^2}} -1$ in your calculator and see what you get...I have a cheap calculator and it gives me 3.38 x10^(-10) which is correct.

One thing you should not do in calculations at nonrelativistic speeds is to calculate an intermediate step and then by hand reenter a rounded off value for the next step (for example, your calculator gives 0.999999 for a step and you use 1 for the next step). You must always store the answer in your calculator and reuse that for the next step. The calculator will keep the correct number (usually..unless you go to very small speeds in which case you either need a specialized software like Maple or Mathematica or you need the Taylor trick I mentioned).

Patrick

7. Apr 7, 2007

### nrqed

I just realized something....

Now I realize that you must have calculated directly "t" with your calculator, right? Yes, it shows up the same answer because the calculator may only show a finite number of digits. The answer is 1 300 000.00039 but the calculator (mine at least) only shows 1300 000.

This is why you must calculate the DIFFERENCE $\Delta t$. That number will show up correctly to be 4.39 times 10^-4. Try it!!

Patrick

Last edited: Apr 7, 2007
8. Apr 7, 2007

### anarine24

I just tried it and it works now. I calculated it directly with the calculator and I got about the same answer you got. Before, when I took the time difference, I ended up with 1.3E6 - 1.3E6, instead of 1.3E6 - 1 300 000.00039. It was the .00039 that messed me up. Again, thanks alot for your help!!

9. Sep 30, 2011

### Bob1936

Hi:

I'm new to the forum, and I hope I'm doing this right.

I've seen the time dilation formula on many websites. The article I'm refering to is on the site:
http://www.fourmilab.ch/cship/timedial.html

This is an excellent site. I understand the article. I can even get the same results using the formula.

There is one thing that I can't understand. He mentions cosmonauts spending a year on the space station "Mir." He mentions the velocity as about 7700 Meters per second, and, using the formula gives a time dilation factor of 1.00000000033. I get the same thing.

Now here is what I can't understand. He states that as a result of this, for every second on earth, for the cosmonauts, it is 3 NANOseconds less, and, while this is not much, in
one year, that will add up to 3.8 seconds.

Wouldn't it take a billion nanoseconds to equal one second? Since there are 31,365,000 seconds in one year, how can this add up to even one second? What am I missing here?
Three times the number of seconds in one year wouldn't even add up to one billion.

What am I not taking into consideration? Any help would be appreciated.