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We know satellite is moving , so there should be time dilation due to motion, also we know due to Earth's Gravitational field there should be time dilation, what should I do, just calculate both of them and add?

Chandra Prakash
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  • see the answers to https://physics.stackexchange.com/questions/1061/why-does-gps-depend-on-relativity – mike stone Aug 05 '18 at 15:40
  • The two should be multiplied, but it comes out practically identically to adding them: rate due to motion is $K(1+r_v)$ where $K$ is the rest frame rate of the clock and $r_v$ is the rate shift factor due to motion. Clock rate after gravitational time dilation is $K(1+r_g)$ where $r_g$ is the rate shift factor due to difference in gravitational potential. Taking both together gives a rate of $K((1+r_v)(1+r_g) = K(1+r_v+r_g +r_v+r_g)$, which is extremely close to $k(1+r_v+r_g)$ because $r_v$ and $r_g$ are both very small. – S. McGrew Aug 05 '18 at 15:58

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Yes, just add. From GR the satellite clock goes faster, but then SR slows it down a bit, though it's still faster than on the ground.

Taylor and Wheeler's book "Exploring Black Holes" has a short chapter on GPS doing just this. The first edition (which I read) is out of print, but there is an online second edition, which I didn't read, but which I see does have a chapter on GPS.

Job Stancil
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You should not really separate special relativistic time dilation and general relativistic time dilation into two separate effects and you should read John Rennie's excellent answer to Is gravitational time dilation different from other forms of time dilation?

As an approximation you can use the expression for time dilation of an object in a circular orbit in the Schwarzschild metric. $$ d\tau_S = \left(1 - \frac{3GM}{c^2 r_S}\right)^{1/2}\ dt,$$ where $d\tau$ represents a tick of the clock in its reference frame (the proper time), $dt$ is the tick length as observed by a distant observer in asymptotically flat space-time and $r$ is the orbital radius. This formula assumes a spherically symmetric gravitational potential and that the satellite orbits outside the Earth!

For an observer on the surface of the Earth, then they are not in orbit, so the above equation does not apply. In fact to a good approximation they are equivalent to a stationary observer in the Schwarzschild metric at distance $r$ from the centre. This approximation becomes very good for an observer at the north or south poles and is roughly true at the equator because the speed of an observer at the equator is about 15 times less than that of a satellite in orbit just above the Earth's surface. $$d \tau_{\rm O} \simeq \left(1 - \frac{2GM}{c^2r_O}\right)^{1/2} \ dt,$$ where $d\tau_O$ would be the ticks measured at the surface of the Earth and $r_O$ is the Earth radius.

If we divide one equation by the other, you can get the expression for the relative time-dilation. $$ \frac{d\tau_O}{d\tau_S} = \left( \frac{1 - 2GM/c^2r_O}{1- 3GM/c^2r_S}\right)^{1/2}$$

ProfRob
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