# Does this special relativity question have enough information to solve it?

I recently encountered this question:

How long would an someone need to spend on the ISS so that their biological clock would be one day younger than their twin who stayed on Earth? The ISS is orbiting the Earth at an altitude $$h$$.

Hint: Imagine twins born on a hypothetical stationary Alien ship and sent immediately away: one twin to Earth and the other to the ISS. Then, as measured by the Aliens, the twins are both in motion, but at different speeds. Consider all the reference frames as inertial (which they are really not).

I have been thinking about it, and I am pretty sure it is too vague to solve. You can't figure out the ISS and Earth frames' relative velocity unless you assume that it is just the orbital velocity. However, the hint says that relative to a stationary alien frame the earth and ISS are both moving. I thought that maybe you need to take into account the Earth's rotation or something, but then you would need to make an assumption about whether they are moving in the same direction or not.

I also thought that maybe it wants us to find when the ISS is one day ahead as measured in the alien frame, but the alien frame is only mentioned in the hint.

I think that maybe when my teacher was writing this she thought that when the ISS is one day younger as measured in the alien frame is the same as when it will be one day younger as measured by the Earth frame. This is wrong though, right?

Main Question: Is this problem too vague to solve, or is there something I am missing?

• "The hint says that relative to a stationary alien frame the earth and ISS are both moving. I thought that maybe you need to take into account the Earth's rotation or something, but then you would need to make an assumption about whether they are moving in the same direction or not." Why would you need to make an assumption about the direction of motion? Can't you figure out the magnitudes of each one's motion relative to the distant alien frame? Sep 15, 2023 at 1:44
• This is a very tricky one! I know because I have read the solution in the first part of the article here: mathpages.com/rr/s6-05/6-05.htm. Basically, the author shows that the proper time of an orbiter depends on its speed (SR) and gravitational potential (GR). Sep 15, 2023 at 12:34
• @MichaelSeifert I can not conceive that the rotational speed will have any SR effects, but the article I linked to points out that the Earth bulges owing to that rotation, so leveling the gravitational potential at the Earth's surface, so a GR effect. Sep 15, 2023 at 12:39
• Note that the special-relativistic time dilation due to riding eastwards on Earth's equator is equal to the gravitational time dilation corresponding to an altitude change of only ten or twenty kilometers — making the Earth's geoid, whose equatorial radius is longer than its polar radius, into a surface whose time dilation is constant. According to one paper, the clock speedup due to escaping Earth's gravity well is comparable to the clock slowdown due to orbital speed. For the ISS, I think the orbital slowdown still wins; see figure 5 in the link.
– rob
Sep 15, 2023 at 13:13

I thought that maybe you need to take into account the Earth's rotation or something, but then you would need to make an assumption about whether they are moving in the same direction or not.

Since the hint says both twins are moving, you indeed need to take into account the Earth's rotation. You would not need to make an assumption about whether they are moving in the same direction or not, however, because that's Googleable (the answer is yes). The question doesn't say if you can neglect the inclination of the ISS, but you almost surely can, because it adds a significant layer of complexity (you could of course consider it anyway and impress your teacher, just as you could consider gravitational time dilation as well and impress your teacher).

I think that maybe when my teacher was writing this she thought that when the ISS is one day younger as measured in the alien frame is the same as when it will be one day younger as measured by the Earth frame. This is wrong though, right?

The two frames certainly will not measure the same 1-day time difference, but the hint suggests that the desired answer is for the alien frame to measure a 1-day time difference.

• I don't think this problem can be solved without using GR, see my comment above. Sep 15, 2023 at 12:41
• @m4r35n357 I am pretty sure the GR effect is of roughly the same order of magnitude as the SR effect (en.wikipedia.org/wiki/Hafele%E2%80%93Keating_experiment), but it's extremely improbable the OP is supposed to take that into account. Sep 15, 2023 at 13:15
• The article I linked to goes onto this in detail, you don't need to be "sure" at all ;) It depends on altitude of course. If the Earth's gravity is meant to be ignored then it is a pretty daft question! Sep 15, 2023 at 13:31

Whatever you do, just remember:

$$\gamma = (1-v^2)^{-\frac 1 2}\approx 1+ v^2/2$$

So

$$\frac{\gamma_1}{\gamma_2}\approx\frac{1+v_1^2/2}{1+v_2^2/2}$$

$$=\frac{1+ v_1^2/2 -v_2^2/2 -v_1^2v_2^2/4}{1-v_2^4/4}$$ $$\approx 1+(v_1^2-v_2^2)/2=1+\frac {v_1+v_2} 2 (v_1-v_2)$$

$$=1+\bar v \Delta v$$ That last form: average velocity times difference in velocity should allow to convert between alien or Earth frame trivially…assuming the aliens are $$v \ll c = 1$$.

• I don't think this addresses the question (i.e. "Is there enough information?"). Sep 15, 2023 at 12:44
• But it does. You just do it for both choices, since at such low velocities, the equation is super simple. At most the error will be one part in twenty four anyway. It if teach wants the earths orbital velocity included, then you just time dilate your answer as needed.
– JEB
Sep 15, 2023 at 14:10