Relativistic time dilation on Mars compared to Earth? What is the time dilation in Mars, compared to earth? Can we accurately calculate it? What information is needed to do these calculations?
 A: We can calculate the time dilation approximately using the weak field approximation. If the difference in the Newtonian gravitational potential between two points $A$ and $B$ is $\Delta\Phi$ then the weak field approximation tells us that the relative rate at which clocks at the two points tick is given by:
$$ \frac{\Delta t_A}{\Delta t_B} = \sqrt{1 - \frac{2\Delta\Phi_{AB}}{c^2}} \tag{1} $$
Let's be clear about the notation and sign conventions. Take your example of Earth and Mars as an example. $\Delta\Phi_{AB}$ is the change in the potential energy going from the Earth ($A$) to Mars ($B$), and since in going from the Earth to Mars means the potential energy becomes less negative that means $\Delta\Phi_{AB} \gt 0$. That means the right hand side of equation (1) is less than one, so the fraction $\Delta t_A/\Delta t_B$ is less than one. This means time ticks more slowly on Earth than it does on Mars.
To calculate $\Delta\Phi$ simply:


*

*calculate the (positive) potential energy change to leave the Earth's surface (staying at the same distance from the Sun)

*calculate the (positive) potential energy change to move from the Sun-Earth istance outwards to the Sun Mars distance

*calculate the (negative) potential energy change to descend to the surface of Mars (staying at the same distance from the Sun)
then add up the three potential energy changes to get the total $\Delta\Phi$ and plug it into equation (1). I'll leave this as an exercise for the reader.
Strictly speaking this only calculate the gravitational time dilation and ignores the time dilation due to the orbital velocity of the Earth and Mars. In the weak field limit you can simply multiply together the time dilation due to gravity and orbital velocity.
