Time Dilation Experiment I am currently thinking of doing an experiment for science fair involving time dilation which will use the equations for both time dilation based off of speed and for gravitational dilation to determine if there is a speed and orbit altitude a satellite can maintain to nullify the effects of time dilation.
From the sources I have looked at, satellites gain about an extra 1.7-1.9 seconds every century because they are not effected by gravitational time dilation as much as us on earth. What I need a little help with is finding the equation for gravitational time dilation.
I have found out how to do it for velocity, which was pretty straightforward with the Lorentz equation and finding gamma. Is there a different formula for an orbiting object as opposed to an object on the surface.
Any response would be appreciated and thanks for taking the time to read this.
Original question: https://www.reddit.com/r/Physics/comments/2mic2l/time_dilation_experiment/
 A: The time dilation factor (relative to a stationary observer at infinity) for a moving clock at a constant radius is:
$$\frac{d \tau}{dt} = \sqrt{1-\frac{2GM}{c^2 r} - \frac{v^2}{c^2}}$$
where G is the gravitational constant, M is the mass of Earth, r is the radius from the center of the Earth, and v is the velocity of the clock. So, a stationary observer sitting on the surface of Earth would experience a factor:
$$\sqrt{1-\frac{2GM}{c^2 r_\oplus}}$$
where $r_\oplus$ is the radius of the Earth. What we are then looking for is a stable orbit where the amount of time dilation experienced by the satellite is exactly equal to the amount experienced by someone stationary on Earth. In other words:
$$\sqrt{1-\frac{2GM}{c^2 r_\oplus}} = \sqrt{1-\frac{2GM}{c^2 r} - \frac{v^2}{c^2}}$$
This can be simplified a bit:
$$\frac{2GM}{c^2 r_\oplus} =\frac{2GM}{c^2 r} + \frac{v^2}{c^2}$$
Additionally, we know from ordinary Newtonian mechanics that the orbital velocity of a test particle is given by:
$$v^2=\frac{GM}{r}$$
So plugging that in to our equation above we get:
$$\frac{2GM}{c^2 r_\oplus} =\frac{2GM}{c^2 r} + \frac{GM}{c^2r}$$
Simplifying:
$$r= \frac{3}{2} r_\oplus$$
In other words, a satellite orbiting at a height of $\frac{1}{2} r_\oplus$ above the Earth's surface will age at the same rate as someone stationary on Earth at either the North or South pole. You can do exactly the same calculation for someone not at the poles, but their velocity due to Earth's rotation will depend on their distance from the equator.
A: The Lorentz Transformation of Time Dilation formula is rather simple:
$$
T=\gamma T_0 = \frac{T_0}{\sqrt{1-\frac{v^2}{c^2}}}
$$
Note that $v$ is the velocity relative to earth. Let's just assume that they are moving at a speed $100m/s$ slower than Earth. We would plug in the equation for, say, 150 years and receive:
$$
T = \frac{150}{\sqrt{1-\frac{100^2}{c^2}}} = \frac{150}{\sqrt{1-\frac{10000}{8.98755179\times 10^{16}}}} = \frac{150}{\sqrt{0.99999999999988873499442721987363}}
\\
T = \frac{150}{0.99999999999994436749721360838933} = 150.00000000000834487541795920585 \text{yrs}
$$
This leads to $00000000000834487541795920585 \times 1year$. Assuming $3.15569\times10^7$ seconds per year, this would mean $\approx .000263s$ must be added per 150 years. I can only assume, since the atomic clocks are offset more than that, that they are moving less than $-100m/s$ relative to Earth (perhaps on the order of km/s slower), but hopefully this helped you with the formulation of the problem :)
