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Relativistic Effects On International Space Station (ISS)

  1. Nov 28, 2014 #1
    On the ISS, special relativity dictates the station's clocks run slower than clocks on Earth because of the high velocity, but general relativity dictates that the station's clocks run faster than clocks on Earth because of the lesser gravity. Which effect is predominant, and do the station's inhabitants age faster or slower than people on Earth? I could work this out for myself I suppose, but someone in the forum probably has the answer at hand. I hope.
  2. jcsd
  3. Nov 28, 2014 #2


    Staff: Mentor

    Yes, but that doesn't mean we'll necessarily give it without making you do any work. ;)

    You are right about the general tendency of the two effects; however, you should also consider that, for an object in orbit, the two are related, because the velocity is a function of the altitude. So you can actually combine them into a single formula that only depends on altitude, which I will write as:

    \frac{d\tau}{dt} = \sqrt{1 - \frac{3 G M}{c^2 r}}

    where ##r## is the orbital radius, measured from the center of the Earth (I am assuming a circular orbit for simplicity), ##G## is Newton's gravitational constant, ##M## is the Earth's mass, ##c## is the speed of light, and ##d\tau / dt## gives the object's time dilation relative to an idealized observer at rest "at infinity" (i.e., far enough away from the Earth that its gravity is negligible, and at rest relative to the Earth so there is no velocity effect).

    For an observer at rest on the Earth's surface, things are a little more complicated because the observer's velocity is determined by the Earth's rotation speed. The formula that applies to such an observer therefore has to include both altitude and velocity separately:

    \frac{d\tau}{dt} = \sqrt{1 - \frac{2 G M}{c^2 R} - \frac{v^2}{c^2}}

    where ##R## is the Earth's equatorial radius and ##v## is the velocity of an observer on the Earth's equator (relative to the hypothetical observer at rest "at infinity"). (Again, I assume an observer on the equator for simplicity.)

    If you work out the numbers using the above formulas, you should find that, for objects in orbit about the Earth, there is a radius ##r## at which ##d\tau / dt## is exactly the same as it is for an observer on the Earth's equator. For orbits below that radius, the SR effect predominates and orbiting clocks run slow compared to clocks on the Earth's surface (their ##d\tau / dt## is smaller). For orbits above that radius, the GR effect predominates and orbiting clocks run fast compared to clocks on the Earth's surface (their ##d\tau / dt## is larger).

    Note, btw, that all these formulas are only valid in weak gravitational fields, like the Earth's (or even the Sun's). In strong fields the formulas get more complicated.
  4. Dec 3, 2014 #3
    Thanks, Peter, for the excellent reply.
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