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This question could apply to stars orbiting a black hole, or planets orbiting a star, or moons orbiting a planet. The example numbers don't matter, just the concept.

Lets say that a star has a gravitational time dilation factor of 2 of the level of an observer on Earth (i.e. time runs twice as slow as an observer on earth.) At a certain altitude above the star the time dilation factor is 1.5. This should mean that anything at that altitude should also be running at that same time rate of 1.5. But a planet, which would have its own time dilation factor of 1.5 if it was out in distant space, is orbiting the star at that altitude.

So what is the total time dilation on the planet to the observer on earth? Is it 1.5 x 1.5 = 2.25? Or does the planet just retain its own factor of 1.5? Or something else?

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As an approximation you can add the gravitational potentials and therefore multiply the time dilation factors. This will be a good approximation in weak fields, but factors of $1.5$ or $2$ are not weak fields! So for such large amounts of time dilation, multiplying them together is a very rough approximation; really it is only good for some sort of qualitative guide. Indeed I think that the factor 2 is about as large as it can get for a star (any larger and the star must collapse to a black hole).

For an exact treatment one would need to solve the field equation, taking into account its non-linearity. This is non-trivial and would require numerical methods.

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  • $\begingroup$ Thanks. So let me move to more common amounts, like 1.1 for the star (with 1.01 for the altitude above the star) and 1.01 for the planet. So the planet in orbit would end up at 1.01 x 1.01 = 1.02 $\endgroup$ – foolishmuse Aug 31 '20 at 20:07
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    $\begingroup$ How good of an estimate multiplying time dilations would produce depends not on their absolute values but on how characteristic length/curvature scales for each of the bodies compare to each other. For instance, for extreme mass ratio binaries multiplying time dilations would work well even up to Schwarzschild radii (of both bodies). $\endgroup$ – A.V.S. Sep 1 '20 at 14:56
  • $\begingroup$ Additionally, time dilation of an observer on a planet would also depend on the state of motion of that planet (in OP the planet is “orbiting”). And even for weak field approximation the terms $v^2/2$ and $\Phi$ would be of comparable magnitude (virial theorem guarantees that). So even in weak field your answer would not work. (-1). $\endgroup$ – A.V.S. Sep 1 '20 at 18:05
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    $\begingroup$ @A.V.S. That time dilation depends on state of motion is not denied by my answer but merely omitted in the interests of brevity, since this is obvious to anyone with a minimal knowledge of these things. One can often make a separation into "gravitational" and "Doppler" contributions. $\endgroup$ – Andrew Steane Sep 2 '20 at 7:59
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Yes, I think you are right, and the factors should be multiplied by each other. This must work correctly for, at least, the first order of approximation. The time dilation depends on the gravitational potentials, which, in your example, the gravitational potentials of the star and the planet should be added up together. This means, to some extent, that the time dilation factors are multiplied by each other.

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  • $\begingroup$ @G.Smith If the fields are weak, the sum of the $g$-potentials are approximately equal to the multiplication of dilation factors: $\sqrt{1-a-b}\approx \sqrt{1-a}\sqrt{1-b}$ $\endgroup$ – Mohammad Javanshiry Aug 31 '20 at 18:16
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    $\begingroup$ Agreed. I was wrong. But note that a time dilation of 1.5 is not caused by a weak field. $\endgroup$ – G. Smith Aug 31 '20 at 18:18

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