Time Dilation in the ISS: 2 Seconds in a Year

[Aristarchus_]

Homework Statement

One twin is on the Earth's surface, at water level. The other is in the ISS 408km above Earth's surface. If they are separated for one year, how much longer does the twin on Earth age?

Relevant Equations

$t= \frac {t_0}{\sqrt{1- \frac{2 \gamma M}{rc^{2}}}}$ (Edited)
Where r is the distance from the centrum of Earth. The answer is supposed to be 0.007 sec

Using the above formula I get that the time goes 6.5∗10−86.5∗10−8 percent faster in ISS. Thus, this is approximately 2 seconds in a year. But the answer is much lower. Where am I making a mistake?

 

[malawi_glenn]

One year according to which reference system?

Can you show how you came up with the speed of the ISS station in your formula?

Is gamma Newtons gravitational constant?

 

[Ibix]

It would help if you showed your working, specifically the values you input, and stated the answer you were expecting. With a quick back of the envelope calculation I get around $4.4×10^{-9}\%$ difference.

Also, $6.5×10^{-8}$ of a year is about 2s, but $6.5×10^{-8}\%$ of a year is about 0.02s. Which did you mean?

 

[PeroK]

Thoughts:

Did you take the radius of the Earth into account?

Did you take the rotation of the Earth into account? If so, it matters in which direction the ISS orbits.

Did you take gravitational time dilation into account?

 

[PeroK]

Is the answer supposed to be $0.01s$?

 

[jbriggs444]

Homework Statement:: One twin is on the Earth's surface, at water level. The other is in the ISS 408km above Earth's surface. If they are separated for one year, how much longer does the twin on Earth age?
Relevant Equations:: $t= \frac {t_0} {\sqrt{1- \frac{2 \gamma M}{rc^{2}}}$

Try adding one more curly brace after that $\LaTeX$:

##t= \frac {t_0} {\sqrt{1- \frac{2 \gamma M}{rc^{2}}}}$renders as:t= \frac {t_0} {\sqrt{1- \frac{2 \gamma M}{rc^{2}}}} However, I could not immediately follow this formula. You've left out a fair number of steps. It seems that you are working to calculate relativistic Gamma and needed a$v^2$in the formula. You have calculated that from the mass of the Earth$M$and the radius of orbit$r##. Probably, as @malawi_glenn suspects, your use of $\gamma$ is intended to represent Newton's gravitational constant $G$.

So you have $\frac{GM}{r}$ which represents the potential energy deficit of a satellite with unit mass. You may know that the orbital kinetic energy ($\frac{1}{2}mv^2$) will be half of that. So $\frac{GMm}{r} = mv^2$ and $v^2 = \frac{GM}{r}$

Nicely done.

But I think that you've bungled and left a factor of 2 in there. Orbital kinetic energy is only 1/2 of the potential energy deficit. Not the whole thing.

 

[malawi_glenn]

One can also use centripetal acc

 

[jbriggs444]

One can also use centripetal acc

Yes, nicer. I just grabbed the first tool in my bag. [Second, actually. If I'd been working the problem, a Google search for "velocity of iss in meters per second" would have been first -- 7778 m/s, about right for low Earth orbit]

 

[phinds]

https://www.theguardian.com/science...ronaut-interview-space-younger-twin-endurance

 

[PeroK]

Try adding one more curly brace after that $\LaTeX$:

##t= \frac {t_0} {\sqrt{1- \frac{2 \gamma M}{rc^{2}}}}## renders as:$$t= \frac {t_0} {\sqrt{1- \frac{2 \gamma M}{rc^{2}}}}$$

I would tend to use $\Delta t_s, \Delta t_o$ for the proper times on the surface and in the ISS orbit. A useful approach is to rewrite this as:
$$\Delta t_o = \Delta t_s \sqrt{1- \frac{v^2}{c^2}} = \Delta t_s \sqrt{1- \frac{GM}{rc^2}}$$Then expand that term to first order using a binomial expansion.

 

[PeroK]

https://www.theguardian.com/science...ronaut-interview-space-younger-twin-endurance

That said, I suggest the stress of space travel would age someone more than a tenth of second!

 

[jbriggs444]

That said, I suggest the stress of space travel would age someone more than a tenth of second!

Surely the stress tensor for a body in free fall is more nearly zero than that for a [live] body at rest... in the Ukraine. :-)

 

[Aristarchus_]

It would help if you showed your working, specifically the values you input, and stated the answer you were expecting. With a quick back of the envelope calculation I get around $4.4×10^{-9}\%$ difference.

Also, $6.5×10^{-8}$ of a year is about 2s, but $6.5×10^{-8}\%$ of a year is about 0.02s. Which did you mean?

4.2*10^{-9} is the correct answer for the percentage of time dilation in relation to Earth, how did you come up with it?

 

[Aristarchus_]

Try adding one more curly brace after that $\LaTeX$:

##t= \frac {t_0} {\sqrt{1- \frac{2 \gamma M}{rc^{2}}}}## renders as:

t=t01−2γMrc2−−−−−−−√t=t01−2γMrc2​

However, I could not immediately follow this formula. You've left out a fair number of steps. It seems that you are working to calculate relativistic Gamma and needed a $v^2$ in the formula. You have calculated that from the mass of the Earth $M$ and the radius of orbit $r$. Probably, as @malawi_glenn suspects, your use of $\gamma$ is intended to represent Newton's gravitational constant $G$.

So you have $\frac{GM}{r}$ which represents the potential energy deficit of a satellite with unit mass. You may know that the orbital kinetic energy ($\frac{1}{2}mv^2$) will be half of that. So $\frac{GMm}{r} = mv^2$ and $v^2 = \frac{GM}{r}$

Nicely done.

But I think that you've bungled and left a factor of 2 in there. Orbital kinetic energy is only 1/2 of the potential energy deficit. Not the whole thing.

Yes, I used that formula but I am getting the wrong answer by plugging in the values of r = 408km + 6371km. What is it that I am missing here?

 

[Aristarchus_]

Thoughts:

Did you take the radius of the Earth into account?

Did you take the rotation of the Earth into account? If so, it matters in which direction the ISS orbits.

Did you take gravitational time dilation into account?

I did not take gravitational time dilation into account. How would one do so? Is it a separate formula from the one provided above?

 

[Aristarchus_]

One year according to which reference system?

Can you show how you came up with the speed of the ISS station in your formula?

Is gamma Newtons gravitational constant?

How would I take the speed into account? Would I need the orbital velocity or the escape velocity in the Lorentz formula?

 

[PeroK]

I did not take gravitational time dilation into account. How would one do so? Is it a separate formula from the one provided above?

I wasn't sure whether you were supposed to or not. It turns out that for the ISS the velocity-based effects are significantly greater than the gravitational effects. So, you can neglect both gravitation and the Earth's rotation. It wasn't immediately obvious to me that you could do so.

 

[malawi_glenn]

Did you take gravitational time dilation into account?

One can not do that at this level, I give exactly this problem (well I do say in what frame "1 year" is measured in). I do mention that one can take into account also GR effects and what the actual timedilation is for ISS. I also mention the Hafele & Keating experiment.

Would I need the orbital velocity or the escape velocity in the Lorentz formula?

You need the relative velocity between the two frames, and assuming that Earth frame is an inertial frame.

What is it that I am missing here?

First tell us what value you get for $v$

 

[PeroK]

First tell us what value you get for $v$

There's no reason I can see to calculate $v$.

 

[Aristarchus_]

Try adding one more curly brace after that $\LaTeX$:

##t= \frac {t_0} {\sqrt{1- \frac{2 \gamma M}{rc^{2}}}}## renders as:

t=t01−2γMrc2−−−−−−−√t=t01−2γMrc2​

However, I could not immediately follow this formula. You've left out a fair number of steps. It seems that you are working to calculate relativistic Gamma and needed a $v^2$ in the formula. You have calculated that from the mass of the Earth $M$ and the radius of orbit $r$. Probably, as @malawi_glenn suspects, your use of $\gamma$ is intended to represent Newton's gravitational constant $G$.

So you have $\frac{GM}{r}$ which represents the potential energy deficit of a satellite with unit mass. You may know that the orbital kinetic energy ($\frac{1}{2}mv^2$) will be half of that. So $\frac{GMm}{r} = mv^2$ and $v^2 = \frac{GM}{r}$

Nicely done.

But I think that you've bungled and left a factor of 2 in there. Orbital kinetic energy is only 1/2 of the potential energy deficit. Not the whole thing.

But I don't know why I'm getting the wrong answer. The value of radius should be 408km +6371km right?

 

[PeroK]

But I don't know why I'm getting the wrong answer. The value of radius should be 408km +6371km right?

What answer are you getting?

 

[malawi_glenn]

There's no reason I can see to calculate $v$.

If he/she can not get $v$ correct, then its hard to get everything else correct.

I am getting the wrong answer by plugging in the values of r = 408km + 6371km.

Would be intersted to see what value of $v$ he/she gets.

 

[Aristarchus_]

One can not do that at this level, I give exactly this problem (well I do say in what frame "1 year" is measured in). I do mention that one can take into account also GR effects and what the actual timedilation is for ISS. I also mention the Hafele & Keating experiment.You need the relative velocity between the two frames, and assuming that Earth frame is an inertial frame.

First tell us what value you get for $v$

If we have the radius of the orbit, we shouldn't need V

 

[malawi_glenn]

If we have the radius of orbit, we shouldn't need V

No but you can still calculate it and compare with the number that was given in #8
Sometimes making small steps are good to "debug" caclulations. Once you understand and make sure every ingredient is properly calculated, one can do the entire thing.

Also, what calculator are you using?

 

[Aristarchus_]

What answer are you getting?

6.53*10^{-8} in percent, which makes roughly 2 seconds of a year

 

[malawi_glenn]

6.53*10^{-8} in percent, which makes roughly 2 seconds of a year

It was already pointed out that "percent" is wrong here. Did't you read post #3?
Don't use percentages unless you have to. 0.0005 is 0.05% why is "%" more helpful here? It's just another layer of confusion.

 

[malawi_glenn]

Yes, I used that formula but I am getting the wrong answer by plugging in the values of r = 408km + 6371km. What is it that I am missing here?

No the forumula you wrote had a factor of 2 in it.

 

[Aristarchus_]

It was already pointed out that "percent" is wrong here. Did't you read post #3?

When values are plugged into the formula, I get 1.000000000653, which is t = t_0 * 1.000000000653. Then I converted this into percentage and estimated the time of the year by multiplying by 3600*24*365. But it was pointed out that 6.5*10^{-8} is not the correct percentage...How so?

 

[malawi_glenn]

When values are plugged into the formula, I get 1.000000000653, which is t = t_0 * 1.000000000653. Then I converted this into percentage and estimated the time of the year by multiplying by 3600*24*365. But it was pointed out that 6.5*10^{-8} is not the correct percentage...How so?

Why do you need to convert into %? It's pretty pointless.

You have still not answered my question in what frame 1 year is measured in. Is it 1 year on Earth or 1 year on ISS?

 

[PeroK]

6.53*10^{-8} in percent, which makes roughly 2 seconds of a year

That's definitely wrong. I get about $0.01 s$ taking into account only the speed of the ISS.

 

[Aristarchus_]

Why do you need to convert into %? It's pretty pointless.

You have still not answered my question in what frame 1 year is measured in. Is it 1 year on Earth or 1 year on ISS?

I am not sure, I am supposed to compare the two. For example how much younger or older is the twin in ISS after a year in comparison to the one on Earth

 

[PeroK]

You have still not answered my question in what frame 1 year is measured in. Is it 1 year on Earth or 1 year on ISS?

That can't be important, given we are dealing with tiny difference in proper times.

 

[Aristarchus_]

That's definitely wrong. I get about $0.01 s$ taking into account only the speed of the ISS.

I miscalculated than, can you provide your calculation?

 

[malawi_glenn]

That can't be important, given we are dealing with tiny difference in proper times.

It's not important here, but its not all about getting the correct answer.

 

[PeroK]

I miscalculated than, can you provide your calculation?

It's in a spreadsheet. That said I don't like your method. Do you know how to use the binomial theorem for the square root?

 

[malawi_glenn]

I miscalculated than, can you provide your calculation?

It has already been pointed out that in your formula, there is a factor of 2 wrong in the calculation of the speed. Have you tried again after that?

 

[Aristarchus_]

It has already been pointed out that in your formula, there is a factor of 2 wrong in the calculation of the speed. Have you tried again after that?

I think I figured out the mistake in the calculation. Now I will brainstorm a bit...

 

[malawi_glenn]

Do you know how to use the binomial theorem for the square root?

My Texas TI-82 manage fine just cranking in the entire calculation at once

 

[Aristarchus_]

My Texas TI-82 manage fine just cranking in the entire calculation at once

Another issue is getting 4.18*10^{-9} % as the percentage of how much faster time elapses on ISS than on Earth...

 

[jbriggs444]

Another issue is getting 4.18*10^{-9} % as the percentage of how much faster time elapses on ISS than on Earth...

Excuse me? What leads you to believe that the proper time on the ISS elapses more rapidly than that on Earth?

Time dilation describes how moving clocks appear to run slow (tick less elapsed proper time) compared to reference clocks in an inertial frame. In the context of Newtonian gravity and special relativity, there is only one inertial frame here.

 

[PeroK]

I would tend to use $\Delta t_s, \Delta t_o$ for the proper times on the surface and in the ISS orbit. A useful approach is to rewrite this as:
$$\Delta t_o = \Delta t_s \sqrt{1- \frac{v^2}{c^2}} = \Delta t_s \sqrt{1- \frac{GM}{rc^2}}$$Then expand that term to first order using a binomial expansion.

Let me show you a technique that you need to be familiar with. It's used all over physics and applied maths:
$$\Delta t_o = \Delta t_s \sqrt{1- \frac{GM}{rc^2}} \approx \Delta t_s(1 - \frac 1 2 \frac{GM}{rc^2})$$That's called a first order binomial expansion. From that you get:
$$\Delta t_s - \Delta t_o = \Delta t_s(\frac{GM}{2rc^2})$$And, of course, if you want to use $v$, we have:
$$\Delta t_s - \Delta t_o = \Delta t_s(\frac{v^2}{2c^2})$$Messing around with percentages is not the way to go.

 

[malawi_glenn]

Let me show you a technique that you need to be familiar with

I teach this problem for my 17y old students at high school. They don't bareley know what a derivative is at this point.

Now if OP have taken a university level class on calculus, then of course it is a good idea to invoke Mac Laurin expansion. However, modern pocket calculators can deal with this problem just fine. For lower speeds, Mac Laurin expansion is needed though.

 

[jbriggs444]

However, modern pocket calculators can deal with this problem just fine. For lower speeds, Mac Laurin expansion is needed though.

At ten digits of .999999999... before the significant digits start, one is starting to tickle the precision boundaries of 64 bit floating point (16 digits or so). I'm with @malawi_glenn. We're OK in this case but need to be aware of the limitations of our tools.

You can get complete loss of precision when dealing with numbers like 0.99999999999999998 (sixteen digits of nines) which become indistinguishable from one.

The ideas that taking a square root halves the difference between the original value and 1 and the idea that taking the reciprocal inverts the difference between the original value and 1 are pretty useful and may be accessible without calculus.

$\sqrt{0.99} \approx 0.995$
$\sqrt{1.01} \approx 1.005$
$1/0.99 \approx 1.01$
$1/1.01 \approx 0.99$
$1.01^2 \approx 1.02$
$0.99^2 \approx 0.98$

 

[PeroK]

I teach this problem for my 17y old students at high school. They don't bareley know what a derivative is at this point.

Perhaps 17 is a good age to lose one's mathematical virginity!

 

[vela]

I teach this problem for my 17y old students at high school. They don't barely know what a derivative is at this point.

They don't need to know how to derive the expansion, just how to use it.

You could also point out it's just a generalization of the binomial theorem, which they should learn about in algebra, to real exponents, i.e.,
$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$ with $n=1/2$.

 

[malawi_glenn]

They don't need to know how to derive the expansion, just how to use it.

You could also point out it's just a generalization of the binomial theorem, which they should learn about in algebra, to real exponents, i.e.,
$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$ with $n=1/2$.

We have a different system, they learn binomial formula when they are 18 in swedish high school. But only for positive integer coefficients.

If "new" math is not needed, no need of invoking it.

 

[malawi_glenn]

At ten digits of .999999999... before the significant digits start, one is starting to tickle the precision boundaries of 64 bit floating point (16 digits or so). I'm with @malawi_glenn. We're OK in this case but need to be aware of the limitations of our tools.

You can get complete loss of precision when dealing with numbers like 0.99999999999999998 (sixteen digits of nines) which become indistinguishable from one.

The ideas that taking a square root halves the difference between the original value and 1 and the idea that taking the reciprocal inverts the difference between the original value and 1 are pretty useful and may be accessible without calculus.

$\sqrt{0.99} \approx 0.995$
$\sqrt{1.01} \approx 1.005$
$1/0.99 \approx 1.01$
$1/1.01 \approx 0.99$
$1.01^2 \approx 1.02$
$0.99^2 \approx 0.98$

One can also use factorization of conjugate and calcluate in this order
$\dfrac{1}{\sqrt{1+v/c}}t_0\dfrac{1}{\sqrt{1-v/c}}$
And factorization of conjugate i teach them when they are 16 y old

 

[PeroK]

Just to clarify the overall situation here. If you look online there seems to be a lot of misinformation about this, with confusion over the significance of gravitational and velocity-based time dilation.

First, we use reference time $\Delta t$ for a clock at rest far from the Earth. From GR we have the dilated time of an object in a circular orbit of radius $r$:
$$\Delta t_o = \Delta t \big (1 - \frac{3GM}{rc^2} \big )^{\frac 1 2} = \Delta t \big (1 - \frac{3GM}{2rc^2} \big )$$This neatly splits into a velocity-based component:
$$\Delta t_o = \Delta t \big (1 - \frac{GM}{rc^2} \big )^{\frac 1 2} = \Delta t \big (1 - \frac{GM}{2rc^2} \big )$$And a gravitational component:
$$\Delta t_o = \Delta t \big (1 - \frac{2GM}{rc^2} \big )^{\frac 1 2} = \Delta t \big (1 - \frac{GM}{rc^2} \big )$$We see from this that, overall, the gravitational component is twice as significant as the velocity-based component. However, the Earth clock has velocity-based and gravitational time dilation components as well. The velocity component is negligible, as the Earth's rotational speed is less than 10% of the ISS orbital speed. But, the gravitational component is:
$$\Delta t_e = \Delta t \big (1 - \frac{2GM}{Rc^2} \big )^{\frac 1 2} = \Delta t \big (1 - \frac{GM}{Rc^2} \big )$$And, this is crudely approximately the same as the ISS gravitational component, as $R \approx r$. Which means that overall we can consider only the speed of the ISS relative to Earth to get a good approximation. I get a figure of $0.01s$ per year.

Note that for the second approximation (taking gravity into account), I get only $0.009s$ per year difference between the ISS and Earth.

I must confess I couldn't find anything on line that agrees with this. And, the OP's quoted answer seems to be $0.013s$ per year. But, I don't get that answer.

 

[PeroK]

PS actually Wikipedia agrees with me:

https://en.wikipedia.org/wiki/Time_dilation#Time_dilation_caused_by_a_relative_velocity

 

[malawi_glenn]

@Aristarchus_ have you covered time dilation effects due to gravity in your class? Have you covered MacLaurin series in any class prior to this class?

And, the OP's quoted answer seems to be 0.013s per year.

The answer is supposed to be 0.007 sec

4.2*10^{-9} is the correct answer for the percentage of time dilation in relation to Earth

Pretty hard for me to understand what the answer OP has been given. These two numbers are not consistent.