ssope said:
No it means I'm struggling to arrive at the correct answer and that my pride wouldn't let me post that in my previous post. If I had the math I'd show my work.
I would not worry about your pride too much. It would take a rocket scientist to come up with an exact answer. For our purposes we are are going to have to make a lot of simplifications to come to an approximate answer. Initially I propose we assume the following simplifications with maybe more to be added later:
1) The Earth, Moon and Sun are perfect spheres.
2) The Earth's orbit is exactly on the equatorial plane of the sun.
3) The Earth's axis is exactly orthogonal to the Sun's equatorial plane. (i.e. the equator of the Earth always lies on the equatorial plane of the Sun.
4) The Moon's orbit is exactly on the equatorial plane of the Earth and Sun.
5) All orbits are perfectly circular.
6) We tidy up the Solar system by removing all the other planets and orbiting junk.
7) The mass of the satellite is insignificant compared to the mass of the moon.
(We might have to remove the Sun later.)
To start you off here are some equations you might need.
The basic time dilation factor is:
\tau = t \sqrt{(1-2GM/(Rc^2)}\sqrt{(1-{v_L}^2/c^2)}
where:
\tau is the proper time interval measured by a clock on the satellite.
t is the time interval according to a clock at infinity.
R is the distance the satellite is from the gravitational body.
v_L is the "local velocity" according to an observer at R that is stationary with respect to the distant stars.
Note that the above time dilation equation is based on the Schwarzschild metric which assumes a non rotating gravitational body which is not true for the Earth and Moon and really we should be using the Kerr metric but that is more complicated and for the rotation velocities of the massive bodies we are studying the Schwarzschild metric is a good enough approximation.
Note that all commonly know GR metrics assume a single gravitational body in otherwise empty space so they can not be directly applied here. However time dilation is a function of gravitational potential and I think it is safe to assume that gravitational potentials are simply additive.
Now we come to the problem of determining the velocity of the satellite. Calculating the orbital velocity for a given radius is not too difficult. However you should bear in mind that this orbital velocity in an equatorial orbit around the Moon (orbit type 1) subtracts from the velocity of the Moon around the Earth when the satellite is nearest the Earth in a prograde (anticlockwise looking from the North pole) orbit and adds to the velocity in a retrograde orbit. The gravitational potential will also vary periodically as the satellites distance from the Earth varies in an equatorial orbit. We can remove the periodic velocity variation by considering an orbit around the Moon that remains on a plane than intersects the centres of the both the Moon and the Earth which is at right angles to the equatorial plane (orbit type 2). Strictly speaking the direction is constantly changing but the magnitude of the velocity is constant and for time dilation we only need the magnitude (or speed). Unfortunately the variation in gravitational potential remains in this orbit. We could also consider an orbit around the moon that is orthogonal to a line joining the Moon and the Earth (orbit type 3) such that the satellite is always visible from the Earth and appears to roll like wheel. This removes the periodic variation of the gravitational potential but the periodic variation in velocity remains. Without doing the calculations I imagine the periodic variations in the time dilation are much smaller than the magnitude of the average time dilation. Anyway, you should specify what orbit type you have in mind.
We can also note that calculations for the time dilation of GPS satellite clocks can be calculated to a very good accuracy by ignoring the gravitational influence of the Sun, Moon etc and the velocity variations due to the orbit velocity of the GPS satellite adding or subtracting from the Earth's orbital velocity around the Sun.
You might want to consider changing the question to "How does the clock rate of clock on the surface of the Moon compare to the clock rate on the surface of the Earth?" as that might be slightly simpler.
You also need to specify where on the surface of the Earth the clock is situated. The rate of a clock at the equator of the Earth is different form one situated at one of the poles.
To calculate orbital velocities you need the following equation:
{v_{L}} = \sqrt{\frac{GM}{R(1-2GM/(R_{o}c^2)}}
Where:
R is the orbital radius of the satellite.
Ro is the location of the "stationary observer".
For an observer at infinity (Ro = infinity) the equation reduces to the familiar Newtonian equation:
{v_{L}} = \sqrt{\frac{GM}{R}}
For R = Ro = 3GM/c^2 the result is:
{v_{L}} = \sqrt{\frac{GMc^2}{(3GM(1-2GM/(3GM)}} = \sqrt{\frac{GMc^2}{(3GM-2GM)}} = c
which is the photon orbit where particles need to travel at the speed of light to orbit.
That should give you a good start. For now it would be useful if you would specify the type of orbit (1,2 or 3) and whether you mean clockwise or anticlockwise as viewed from high above the North pole or from the surface of the Earth's equator.
