Effect of gravity on time dilation.

adrian53300

Can anyone tell me What is the relationship between the orbital radius of a satellite and the time gained per second due to its weaker gravity at its altitude?
In relation to that how much time does the International Space station lose per-second due to weaker gravity at its altitude?
Also What about a GPS? How much time does it gain if it is on Earth?

I have read some articles but i have yet to found the exact value. Very interested in the values. Please help. Thanks

Bill_K

The apparent clock rate of a satellite will be affected firstly by its orbital motion. From Newtonian mechanics and assuming a circular orbit, mv²/r = GMm/r², implying the orbital velocity is v²/c² = GM/c²r and γ = 1/√(1-v²/c²) ≈ 1 + GM/2c²r.

According to Wikipedia the Schwarzschild radius of the Earth is 2GM/c² = 8.9mm. The radius of the Earth is r = 6378 km, so for a surface-skimming satellite, γ ≈ 1 + 3.5 x 10^-10.

Secondly, the effect of the gravitational potential. ds² = (1-2GM/c²r) dt², so ds/dt = √(1-2GM/c²r) ≈ 1 - GM/c²r. Larger by a factor of two, but what matters here is that you must take the difference between the gravitational potential at the surface vs the potential at the orbit.

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See attached.

Attached Files

The Schwarzschild solution to the General Theory of Relativity gives the following equation for .doc (29.5 KB, 8 views)