homework and exercises - If an astronaut had stationed in International Space Station for the duration of mission, 17 years, would he be older?

Today the NASA International Space Station started the 100000 orbit after 17 years in the space.
I just wonder if there were a team of astronauts which were in the Lab for all the duration of last 17 years, how much older they would be due to the combination of Einstein's especial theory of relativity and general theory of relativity?
Here is the link to NASA's youtube video. http://youtu.be/g5chOA-WEuw

Answer

The first answer has all the results, but I will try to show some calculations, cause I have been writing them since there was no answer.

It is known from General Theory of Relativity (GTR) that the closer you are to a massive object - the slower the time goes. On the other hand Special Theory of Relativity (STR) gives us the next statement: the faster you move - the slower the time goes.

I can give here only a half of an answer due to my low knowledge of GTR.

Let's compare relative speeds of the ISS and a person on the surface of the Earth (on the equator):

We know that the orbit of ISS is approximatele 400 km. Knowing this we can calculate its speed: $$\frac{mV_1^2}{r+400000}=G\frac{mM}{(r+400000)^2}$$ where $r$ is a radius of the Earth, $M$-its Mass, $m$-mass of the ISS, $G$-Gravitational constant. From this we get $V_1=\sqrt{\frac{GM}{r+400000}}$

Also we can approximate the speed of a person on the surface of the Earth:$$V_2=r\frac{2\pi}{T}$$

So from theese two equations we can calculate the time dilation per second relatively to the center of the Earth for each case: $$\tau_1=\frac{1}{\sqrt{1-\frac{V_1^2}{c^2}}}, \tau_2=\frac{1}{\sqrt{1-\frac{V_2^2}{c^2}}}$$

Then we can calculate the difference and multuply it on 17 years:$$|\Delta \tau|*17*365*24*60*60=0.174634$$

For better understanding we can look at the problem from different angle i.e. Lorentz contraction (effect of STR) or contraction of the distances in the direction of speed:

Let the orbit length be $l=2 \pi (r+400000)$ With every turn the ISS will cover the distance equal to $l_1=l \sqrt{1-\frac{V_1^2}{c^2}}$. Then we just calculate the differenc between 2 times $$T_1=\frac{100000l}{V_1}, T_2=\frac{100000l_1}{V_1}$$ We get that $\Delta T \approx 0.17$

Note:we have only paid respect to time dilation due to STR. The GTR requires more complex calculations but it will shift our answer a bit to the direction of zero, because it will slow the time more for those who are closer to the center of the Earth.