# Would Special Relativity Predict Time Dilation of a Geostationary Satellite Compared to an Observer on Earth?

Consider a geostationary satellite at some arbitrary distance above the Earth's equator and consider a person directly below it, standing on the Earth's equator. Both satellite and person exist on the same radial line.

General relativity tells us that the observer on the Earth will experience time dilation due to being further inside the gravitational well. This will slow the person's clock relative to the satellites clock.

However, would it be correct to say that, since the satellite has greater linear velocity (it's further out and rotating with the same angular velocity as the person on Earth), we should be taking into account special relativity, which would cause a time dilation of the satellites clock relative to the person's clock? Therefore there would be two competing time dilations, that of the satellites clock and that of the person's clock, and the actual time difference between the clocks must take both effects into account.

I've had counter suggestions, saying that since both observers are stationary relative to each other in the rotating frame, special relativity doesn't play a role when comparing the satellites clock to that of the person.

Yes, both special relativity and general relativity have to be taken into account. The total time dilation is given by $$\frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}},$$ where $\text{d}\tau$ is the time measured by a moving clock at radius $r$, and $\text{d}t$ is the coordinate time measured by a hypothetical stationary clock infinitely far from the gravitational field.

For a person standing on the equator, we have $r_\text{eq}=6378\,\text{km}$ and $v_\text{eq}=0.465\,\text{km/s}$, and for a geostationary satellite $r_\text{s} = 42164\,\text{km}$ and $v_\text{s} = 3.074\,\text{km/s}$. This enables you to calculate $\text{d}\tau_\text{eq}/\text{d}t$ and $\text{d}\tau_\text{s}/\text{d}t$, and finally the ratio $\text{d}\tau_\text{eq}/\text{d}\tau_\text{s}$.

• Pulsar: "both special relativity and general relativity have to be taken into account." -- That's certainly preferable instead of saying "SR and/or GR predict". However, your formulation gives the impression as if SR should be taken into account in addition to GR. "The total time dilation is given by [...]" -- I wonder (especially) about the meaning of $v$ in your formula. (The OP wrote of "linear velocity" ... this question makes my doubt more precise). "This enables you to calculate [...]" -- Consider giving a value of $\frac{G~M}{c^2}$. Commented Apr 7, 2015 at 21:32
$$\frac{\text{d}\tau}{\text{d}t} = \sqrt{ 1 - \frac{2GM}{rc^2} - \frac{v^2}{c^2}},$$
$$\frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}},$$