time - Gravity on the International Space Station - General Relativity perspective

My question is an extension to this one: Gravity on the International Space Station.

If all the outside views of the ISS was sealed, then the crew inside would not be able to tell whether they were in orbit around the earth in orbital speed or free floating in space beyond the orbit of Neptune, right?

How would time dilation due to gravitation fields be affected? Supposing you have three atomic clocks: 1 - One on surface of the Earth, at Sea Level, 2 - One in the ISS, 3 - One in deep space beyond the orbit of Neptune.

At what speed would each clock run compared to the other two?

Answer

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.