# 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, 2018 at 11:18
• Now the record holder is not Krikalev but Gennady Padalka with 878 days en.wikipedia.org/wiki/… Aug 28, 2018 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. Aug 28, 2018 at 15:01
• @Heopps To be clear to others, record holder for most time in space, not necessarily largest relativistic shift in time. Aug 28, 2018 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, 2018 at 2:18

Mir, and later, the ISS, the two space stations relevant to long stays, have both oscillated between 320km and 420km. Is that oscillation enough that Gennady Padalka's top spot on the most time in space does not secure him the top spot on the time traveller list?

Let's bring out the numbers, armed with two facts:

1. Velocity time dilation is $$\frac{\Delta t_1}{\Delta t_2} = \sqrt{1 - \frac{v^2}{c^ 2}}$$
2. Gravitational time dilation is equal to velocity time dilation applied to the escape velocity at the current location.

High altitude and great velocity therefore have the opposite effects of each other.

$$\frac{\Delta t_1}{\Delta t_2}_{velocity} = \frac{\sqrt{1 - \frac{v_{orbit}^2}{c^ 2}}}{\sqrt{1 - \frac{v_{surface}^2}{c^ 2}}}$$

$$\frac{\Delta t_1}{\Delta t_2}_{gravity} = \frac{\sqrt{1 - \frac{v_{e_{orbit}}^2}{c^ 2}}}{\sqrt{1 - \frac{v_{e_{surface}}^2}{c^ 2}}}$$

We then have four time dilation factors to consider:

1. High (420km) orbit velocity time dilation: $$1 + 3.25 \cdot 10^{-10}$$
2. High orbit gravitational time dilation: $$1 - 4.31 \cdot 10^{-11}$$
3. Low (320km) orbit velocity time dilation: $$1 + 3.30 \cdot 10^{-10}$$
4. Low orbit gravitational time dilation: $$1 - 3.33 \cdot 10^{-11}$$

Or combined:

• High orbit time dilation: $$1 + 2.82 \cdot 10^{-10}$$
• Low Orbit time dilation: $$1 + 2.97 \cdot 10^{-10}$$

Which means low altitude cosmonauts are the greatest time travellers, which makes sense since the velocity component is more important.

This also means that the difference in extra accumulated time is only about 5%.

The difference in time spent in space between Gennady Padalka and number two on the list, Yuri Malenchenko is over 6%, which means that even if Padalka spent all his orbital time in the higher orbit, and Malenchenko all his time in the lower, Padalka would still be number one on the time traveller list.

Gennady Padalka must therefore have the largest relativistic shift time, since he's on top of the time in space list and the safety margin down to number two is large enough that we don't even have to consider the year-to-year variations in space station altitude.

• Comments are not for extended discussion; this conversation has been moved to chat. Oct 23, 2020 at 13:29