# 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?

• "Free floating?" Objects in space don't "float", they fall. The ISS falls in an orbit around the Earth, the Earth (and Neptune) falls in an orbit around the Sun, the Sun falls in an orbit around the galactic core, and the galaxy as a whole is falling toward something, somewhere. – Solomon Slow Jul 7 '15 at 14:03

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.

• @BrandonEnright Can you explain your edit rollback? Special relativistic effect dominates over general relativistic for the ISS. Less time passes for the faster-moving astronauts than their buddies on the ground. The last sentence of this answer implies the opposite. – pericynthion Feb 27 '15 at 20:51
• @pericynthion I believe the last sentence is actually correct as written. I too thought time was slightly slower on the ISS which is why I erroneously approved the edit in the first place. I think the final sentence does properly describe Teq over TISS though. – Brandon Enright Feb 27 '15 at 20:59
• I agree the last equation is correct, but the last sentence contradicts it. "1 second on Earth corresponds to 1.00000000028655 seconds on the ISS" is saying that a clock carried to the space station and back will show a later time than a clock that stayed on the ground. That's backwards. – pericynthion Feb 27 '15 at 21:09
• @EmilioPisanty and everyone else: you are correct, the last sentence was wrong, I've corrected it now. Thanks. – Pulsar Jul 17 '15 at 1:44
• Shouldn't the ISS be running faster than earth since its at a higher altitude. The fact that it runs slower is due to SR, not GR. – ps95 Dec 13 '15 at 12:57

Clocks tick slower at lower altitudes. So 1. On the surface of the Earth will be the slowest. Now since the ISS has no way of knowing whether it is in orbit or in deep space, you might think that clock 2 and 3 should tick at the same rate. But instead clocks 2 and 3 will just feel like as if they were ticking at the same rate. Astronauts at 2 and 3 will not feel anything unusual with their clocks, yet if you bring clock 2 and 3 closer after some time, you will notice that more time hase elapsed on 3. So how is this possible? Well, the idea behind relativity is that you would never notice if your time goes slower. And that will be the case for your 3 atomic clocks.

I hope this clarifies.

time is slower deeper in a gravity well, an object higher up has faster time for less gravity, however if it is in orbit it is moving fast so time slows. Both dilation factors have to be taken into account, so time is slightly faster on ISS than on surface, even though it moves fast. any way according to Einstien

There is a slight error in the accepted answer by Pulsar that I want to point out. The Schwarzschild metric is not written correctly. The term:

$$\left(1-\frac{2GM}{rc^2}\right)^{-1}$$

only applies in the "radial direction" moving in towards or away from the central mass. When you are moving in a pure non-radial direction the term is replaced by a simple "1".

As we can assume the ISS to move in a pure non-radial fashion the expression for time dilation on ISS should actually be:

$$\frac{\text{d}\tau}{\text{d}t} = \sqrt{ 1 - \frac{2GM}{rc^2} - \frac{v^2}{c^2}},$$

You can use this expression for ISS and for the sea level time measurer, setting in values for the radial distance to the centre of the Earth and the velocity relative to the center of the Earth. For the measurer beyond Neptune you can use the same expression, but first you would have to use values for the radial distance to the Sun and the velocity relative to the Sun for both the "Neptune-measurer" and the "near-Eart- measurers" and then add the contribution from the Earth for the near-Earth measurers.