# Which astronaut has experienced the largest relativistic shift in time (relative to Earth's surface)?

Following this answer and then this question (where I've linked to Cosmonaut Sergei Krikalev; the World's Most Prolific Time Traveller) I've noticed that currently Krikalev does not hold the precisely longest record in space.

Relativistic effects include both velocity, and gravitation. Moving fast relative to the surface of the Earth can have an effect, but so can being farther from it, or higher in the Earth's gravitational well.

So since space stations tend to move up or down in their orbits over the years, whomever holds the record for the longest time in space does not necessarily hold the record for the largest relativistic shift in time, relative to the Earth's surface.

Question: Has this ever been worked out explicitly? Is there a known record holder for the farthest "time traveller"? Would someone like to try now?

According to that link, Krikalev's travel is about 20 milliseconds into the future, and the average half-year stint on the ISS is about 7.

From here (or here if you are ambitious) the lowest order terms to the relativistic frequency shift of a clock in orbit around a gravitational body are:

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

where the first term is the gravitational shift and the second is time dilation, and ignores gravity terms from $$J_2$$ and higher (roughly a part per thousand LEO).

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:

$$\frac{\Delta f_{orb}}{f} \approx -\frac{GM}{r c^2} - \frac{GM}{2c^2} = -\frac{GM}{c^2} \frac{1.5}{h+R_E}.$$

For the astronaut on the surface let's ignore the much lower velocity since the dependence is quadratic:

$$\frac{\Delta f_{surf}}{f} \approx -\frac{GM}{c^2} \frac{1}{R_E}.$$

So the double difference between orbit shift and surface shift is then:

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

If you expand that for LEO, and remember that the "jump into the future" means the astronaut's clock ran slower because the time dilation (slowing) at orbital velocity is a much larger effect than the speed up due to being higher in the gravitational well, then you get:

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.

• @user3528438 in sports for example there are all kinds of desirable, as well as undesirable records, and their holders don't get to choose if they want to hold them or not. Being an astronaut means a lot of your spaceflight data is public. – uhoh Aug 28 '18 at 11:18
• Now the record holder is not Krikalev but Gennady Padalka with 878 days en.wikipedia.org/wiki/… – Heopps Aug 28 '18 at 12:36
• Don't forget that being higher in Earth's gravitational well means that your clock runs faster than Earth, while high velocity means your clock runs slower than Earth. This means the effects balance each other out to an extent. – called2voyage Aug 28 '18 at 15:01
• @Heopps To be clear to others, record holder for most time in space, not necessarily largest relativistic shift in time. – called2voyage Aug 28 '18 at 15:03
• @RussellBorogove Since the top five time in space records fall within a +/-10% window, let's decide quantitatively if "basically the same speed and altitude" is really the right way to look at this or not. I don't know how to do the calculation correctly (otherwise I would have answered Parker Solar Probe passing extremely close to the Sun; what relativistic effects will it experience and how large will they be?) let's see if we can find a source or user who does! – uhoh Aug 29 '18 at 2:18