Relativistic time dilation on Mars compared to Earth?

Question

What is the time dilation in Mars, compared to earth? Can we accurately calculate it? What information is needed to do these calculations?

Answer 1

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:

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

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

3. 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.

Answer 2

Let $G$ be the constant of gravity ($6.67\times 10^{-11} m3/(kgs^2)$), $C$ the speed of light ($2.99792\times 10^8 m/s$), $M_S$ the mass of the sun ($1.9891\times 10^{30} kg$), $r_E$ the mean radius of the orbit of Earth ($1.496\times 10^{11} m$), $r_M$ the mean radius of the orbit of Mars ($2.2799\times 10^{11} m$), $ρ_E$ the mean radius of the Earth ($6.37\times 10^6 m$), and $ρ_M$ the mean radius of Mars ($3.3895\times 10^6 m$).

You need to calculate what the radius of a body would have to be to form a black hole (Schwarzschild radius) before you can calculate the effect of its gravity on time dilation.

Schwarzschild radius of Sun $$R_S=(2GM_S)/C^2 =2.95237\times 10^3 m$$

Time dilation from Sun’s gravity $$Δt_S=1-\sqrt{(1-R/r_o )}$$ Earth: $9.86755\times 10^{-9}$ Mars: $6.47477\times 10^{-9}$
Difference: $3.39278\times 10^{-9}$

Orbital velocity $$v_p=\sqrt{(GM_S)/r_o }$$ Earth: 2.97801E+4 m/sec Mars: 2.41231E+4 m/sec

Time dilation from orbital velocity $$Δt_v=1-\sqrt{1-(v_p/C)^2} $$ Earth: $4.93378\times 10^{-9}$ Mars: $3.23739\times 10^{-9}$ Difference: $1.69639\times 10^{-9}$

Schwarzschild radius of planet $$R_P=(2GM_p)/C^2$$ Earth: $8.86409\times 10^{-3}$ Mars: $9.52445\times 10^{-4}$

Time dilation from planet’s gravity $$Δt_p=1-\sqrt{1-R_p/ρ_p}$$ Earth: $6.95769\times 10^{-10}$ Mars: $1.40499\times 10^{-10}$ Difference: $5.55269\times 10^{-10}$

Time dilation from planet's rotation is insignificant and varies with latitude, so I ignored it here.

I get about $5.64444\times 10^{-9}$ or about 1.78 seconds a year or 3.35 seconds a Mars year.

If you more properly use the dilations and not the deltas, you should multiply the gravitational and velocity sums. It results in the same answer to a reasonable precision.