Gravity on the International Space Station - General Relativity perspective

Not only the position in the gravitational field is important, but also the velocity. Consider the Schwarzschild metric $$ \text{d}\tau^2 = \left(1 - \frac{2GM}{rc^2}\right)\text{d}t^2 - \frac{1}{c^2}\left(1 - \frac{2GM}{rc^2}\right)^{-1}\left(\text{d}x^2 + \text{d}y^2 +\text{d}z^2\right), $$ where $\text{d}\tau$ is the time measured by a moving clock at radius $r$, and $\text{d}t$ is the coordinate time measured by a hypothetical stationary clock infinitely far from the gravitational field. We get $$ \frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}}, $$ with $$v = \sqrt{\frac{\text{d}x^2}{\text{d}t^2} + \frac{\text{d}y^2}{\text{d}t^2} + \frac{\text{d}z^2}{\text{d}t^2}}$$ the orbital speed of the clock in the gravitational field (assuming a circular orbit, so that $r$ remains constant).

For Earth, $GM=398600\;\text{km}^3/\text{s}^2$ (see wiki).

Let us first calculate the time dilation experienced by someone standing on the equator. We have $r_\text{eq}=6371\,\text{km}$ and an orbital speed (due to the Earth's rotation) of $v_\text{eq}=0.465\,\text{km/s}$. Plugging in the numbers, we find $$ \frac{\text{d}\tau_\text{eq}}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{r_\text{eq}\,c^2}\right) - \left(1 - \frac{2GM}{r_\text{eq}\,c^2}\right)^{-1}\frac{v_\text{eq}^2}{c^2}} = 0.99999999930267, $$ so 1 second outside Earth's gravity corresponds with 0.99999999930267 seconds on the equator.

The ISS orbits the Earth at an altitude of $410\,\text{km}$, so that $r_\text{ISS}=6781\,\text{km}$, and it orbits the Earth with a speed of $v_\text{ISS}=7.7\,\text{km/s}$, and we get $$ \frac{\text{d}\tau_\text{ISS}}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{r_\text{ISS}\,c^2}\right) - \left(1 - \frac{2GM}{r_\text{ISS}\,c^2}\right)^{-1}\frac{v_\text{ISS}^2}{c^2}} = 0.999999999016118. $$ The relative time dilation between someone on the equator and someone in the ISS is thus $$ \frac{\text{d}\tau_\text{eq}}{\text{d}\tau_\text{ISS}} = \frac{0.99999999930267}{0.999999999016118} = 1.00000000028655, $$ so 1 second in the ISS corresponds with 1.00000000028655 seconds on Earth. In other words, ISS astronauts age slightly less than people on Earth.