Effect of time dilation on a satellite

[PeroK]

Correct. As Earth's gravitational field is weak you can treat both time dilation effects as independent. Add the velocity/height relation for an orbit and you can get a closed formula. Here is a graph. At 3000 km clocks run at the same speed as on the ground (for circular orbits). Below they run slower, above they run higher.

I thought I'd seen the calculation once that the ISS was high enough. Apparently not. I'll remember that. Thanks.

 

[Janus]

Is there a calculation for that? I would have thought that in the limit of a very large orbit, the gravitational effect would dominate. And, in the limit of a very low orbit, velocity based time dilation would dominate.

For a clock in a circular orbit , the time dilation is
$$ t_0 = t_f \sqrt{1- \frac{3GM}{r c^2}}$$
when r is the orbital radius
Ignoring effects due to the rotation of the Earth, time dilation at the Earth's surface is
$$ t_0 = t_f \sqrt{1- \frac{2GM}{r_e c^2}}$$

with r_e being the radius of the Earth

$$ t_f \sqrt{1- \frac{3GM}{r c^2}}= t_f \sqrt{1- \frac{2GM}{r_e c^2}}$$
when
$$ r = \frac{3}{2}r_e $$

or the altitude of the orbit is 1/2 Earth radius above the surface of the Earth.
Below that, clocks run slow compared to surface clocks, above it, they run faster.

 

[member 656954]

but that theme has, perhaps, been over-done

Pretty much any theme has been overdone Only the telling differentiates your iteration from the others. Which is where the challenge of writing arises from, of course, so go with your idea, @some bloke, and make the telling the best it can be. That's always worth reading, no matter how common (or uncommon) the scenario is.