# Symmetry in gravitational time dilation

In special relativity, when two observers move in respect to each other, each observes the other's clock tick slower.

An observer floating in space far from gravitational fields, observes earth clock ticking slower; at what rate is the clock of the floating observer ticking as observed from earth?

$$\frac{t_E}{t_S} = \sqrt{1 - \frac{2GM}{c^2r_E}}$$
where $r_E$ is the radius of the Earth. So the clock on Earth runs slow as observed from space.
$$\frac{t_S}{t_E} = \frac{1}{\sqrt{1 - \frac{2GM}{c^2r_E}}}$$