# Astronauts and Time Dilation

Do astronauts experience any effects of time dilation? Based on my (very limited) understanding, both a change in speed and in gravity should have an effect - and astronauts experience both. The change in speed is still not even close to the speed of light - and so I don't imagine it has much of an effect. Is this true? Does the change in gravity result in time dilation in this case?

Does the change in gravity result in time dilation in this case?

It sure does!

The change in speed is still not even close to the speed of light - and so I don't imagine it has much of an effect.

It turns out that both relativistic effects have similar magnitudes; in the first equation below both terms have a $$1/c^2$$.

In order to stay in space for long periods, astronauts need to move in orbits. In the ISS orbiting the Earth astronauts move at about 7700 meters/sec or about 17,500 miles per hour.

So relative to Earth's surface the ISS astronauts are both less deep in Earth's gravitational well which makes their time "faster" and moving fast which makes their time "slower" relative to someone on Earth's surface. The cancellation is only partial; the gravitational effect "wins".

The equations below are from Which astronaut has experienced the largest relativistic shift in time (relative to Earth's surface)? The first is a first order approximation to a more general (and longer) GR expression. $$f$$ is the frequency of a clock or an oscillator as a proxy for a person's "time". The first term is for the gravitational effect, larger $$r$$ means time moves less "slowly". The second term is for the velocity effect, faster means time moves more "slowly".

$$\frac{\Delta f}{f} \approx -\frac{GM}{r c^2} - \frac{v^2}{2c^2},$$

For the astronaut in orbit you can use the vis-viva equation for a circular orbit:

$$v^2 = GM\left( \frac{2}{r} - \frac{1}{a} \right) = \frac{GM}{a},$$

which after defining the orbital altitude $$h = a - R_E$$ gives:

$$\frac{\Delta f_{orb} - \Delta f_{surf}}{f} \approx -\frac{GM}{c^2} \left( \frac{1.5}{h+R_E} - \frac{1}{R_E} \right).$$

I've worked out an example there for Cosmonaut Sergei Krikalev; the World's Most Prolific Time Traveller) , with the following results:

Δt (sec) = (3.00E-05 - 1.33E-08 h(km)) × ΔT (days)

So if you spend 400 days at 350 km and 400 days at 400 km, that's 0.020 seconds.