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Calculating gravitational time dilation

  1. Mar 11, 2010 #1
    I'm trying to calculate the time difference for something orbiting very high above the moon in comparison to the surface of the Earth.
  2. jcsd
  3. Mar 11, 2010 #2


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  4. Mar 11, 2010 #3


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    Like I asked before:
    And if you want to include velocity, kev calculated in post #8 of this thread that for a clock in a circular orbit, its time dilation (relative to a clock at infinity I think) would just be a product of velocity-based time dilation and gravitational time dilation.
  5. Mar 16, 2010 #4
    I must also be a product of time-dilation based on gravity because if on the surface of the moon its 1/6th earth gravity, it must be far more orbiting high above the moon. I do understand that velocity based time-dilation would indeed come into play, and yes I would like to accurately calculate for both.
  6. Mar 16, 2010 #5


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    What must be far more? Gravitational time dilation is less when you're farther from a source of gravity, not more.
    As kev showed on that thread I linked to, the total time dilation for an object in a circular orbit is just the velocity-based time dilation (velocity measured relative to the center of the body it's orbiting) multiplied by the gravitational time dilation.
  7. Mar 16, 2010 #6
    If you are interested an equation given by Dr Greg in this thread https://www.physicsforums.com/showthread.php?t=383884&page=3 arrives at the same conclusion from a different direction. (See posts#9 and #35 of that thread).
  8. Mar 16, 2010 #7
    Does anyone here want, better, can anyone here to calculate the time difference for two clocks. One clocking orbiting the moon, the other clock on the surface? I need to double check my math. It doesn't matter what altitude you use, as long as you show your work I'll be able to cross check it with mine.
  9. Mar 16, 2010 #8
    Can we assume a moon in otherwise empty space, because if we have to take the gravity of the Earth and Sun into account then things get really complicated.

    Do you want answers for a rotating moon moon or a static moon?

    [EDIT] I have re-read your original post and I now think you want to compare the rate of a clock orbiting the moon, with the rate of a clock on the surface of the Earth? Is that right?
    Last edited: Mar 17, 2010
  10. Mar 16, 2010 #9


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    What about the second question I asked?
    Also, when you say you want to double check your math, does that mean you've already calculated the gravitational and velocity-based time dilation for each clock? If so what'd you get?
  11. Mar 17, 2010 #10
    Yes you are correct
  12. Mar 17, 2010 #11
    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.
  13. Mar 17, 2010 #12
    I wonder if putting one of the clocks in a Geosynchronous orbit above the moon to cut back on velocity based time-dilation when compared to a clock on the surface of the earth.

    [Sorry about the triple post, but the forum won't let me delete posts, I thought they'd be automerged]
  14. Mar 17, 2010 #13

    Jonathan Scott

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    Time dilation is a tiny effect. It is close to zero everywhere except close to massive objects. At distance r from spherical mass m the fraction by which clocks are slowed compared with their rate a long way away is simply Gm/rc2, equal to the Newtonian potential difference divided by c2. (This is accurate provided that the resulting fraction is small compared with 1, which applies for example to all gravitational potentials within the solar system). Another way of putting this is that a clock at that location will run (1-Gm/rc2) times more slowly than a distant one in empty space.

    You can plug in the mass and radius of the earth (for example using the calculator built into Google search) to find out what this is at the surface of the earth.

    If you want to also take account of the time dilation due to velocity in a circular orbit, you can make use of the fact that for such an orbit v2/r = Gm/r2 where r is the radius of the orbit and m is the mass about which it is orbiting. If you multiply both sides by r/c2 you find that v2/c2 is equal to Gm/rc2, the time dilation fraction.

    From the usual Special Relativity formula, the time dilation due to the speed is given by sqrt(1-v2/c2), which for non-relativistic speeds is equal to 1-1/2(v2/c2), where the last part corresponds to the kinetic energy per unit energy. That's half as much as the gravitational time dilation.

    If you want an accurate time dilation for something in orbit around the moon compared with the surface of the earth, you also have to take into account the time dilation in the vicinity of the moon due to the earth, which is the value of Gm/rc2 when m is the mass of the earth and r is the radius of the moon's orbit around the earth. That is, time dilation at orbiting object is made up of time dilation relative to empty space caused by the moon plus time dilation caused by the orbital speed plus time dilation of the zone around the moon caused by the earth plus the time dilation due to the speed at which the moon moves around the earth. In general, you'd also need to think about how the orbital speeds add and subtract depending on the orientation of the orbits.

    However, the difference in time dilation is dominated by the time dilation at the surface of the earth, with the other factors making only a little difference, so as an approximate estimate you can ignore everything else!
  15. Mar 17, 2010 #14
    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:

    [tex] \tau = t \sqrt{(1-2GM/(Rc^2)}\sqrt{(1-{v_L}^2/c^2)}[/tex]


    [itex]\tau[/itex] 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.
    [itex]v_L[/itex] 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:

    [tex] {v_{L}} = \sqrt{\frac{GM}{R(1-2GM/(R_{o}c^2)}}[/tex]


    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:

    [tex] {v_{L}} = \sqrt{\frac{GM}{R}}[/tex]

    For R = Ro = 3GM/c^2 the result is:

    [tex] {v_{L}} = \sqrt{\frac{GMc^2}{(3GM(1-2GM/(3GM)}} = \sqrt{\frac{GMc^2}{(3GM-2GM)}} = c [/tex]

    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.
  16. Mar 2, 2011 #15
    I have a question, but if it has already been covered above then please forgive me for being lazy and just scanning rather than reading.

    I am standing at sea level next to a 1000m tall skyscraper. By my foot is a clock.

    At the top of the skyscraper is another clock.

    I am not moving relative to either clock.

    As I understand it these two clocks do not tick at the same rate. The one at the top of the skyscraper ticks faster than the clock by my foot.

    I watch the clock by my foot for 1 day. How much time passes on the clock at the top of the skyscraper?
  17. Mar 2, 2011 #16


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    Do you want to calculate the gravitational time dilation alone (idealizing the Earth as a nonrotating sphere) or do you want to include the time dilation due to different velocities in an Earth-centered coordinate system?
  18. Mar 3, 2011 #17
    I want to calculate the gravitational time dilation from my perspective of standing at the foot of the skyscraper. Neither clock is moving relative to me so I would assume that there is no velocity component and wondering why I would need to idealise the Earth as a non-rotating sphere?
  19. Mar 6, 2011 #18
    If the Earth is idealised as non rotating and spherical then you calculate the relative rate of the two clocks using:

    t_{top} = t_{base} \frac{\sqrt{1-\frac{2GM}{(R+h)c^2}}}{\sqrt{1-\frac{2GM}{(R)c^2}}}

    where R is the radius of the Earth and h is the height of the tower.

    You need to assume a non rotating Earth, because if the Earth is rotating then the top of the tower is travelling through space faster than the base, even though they appear to be stationary relative to each other to an observer at the base. Imagine you have a 6ft piece of string in your hand with a toy plane attached to the far end. Another plane is attached to the string about 1ft away from your hand. The planes do not have engines but the propellers free wheel in the wind and indicate how fast the planes travel through the air. When you whirl the planes around you the propeller of the plane furthest from you spins the fastest because that plane travels through more air per unit time than the inner plane, even though both planes are stationary relative to you at the centre. They are are not stationary relative to the air and are travelling at different speeds through the air.

    If the Earth is rotating (which it usually is) then you need to allow for the different speeds of the top and base of the tower and you need to use this formula:

    t_{top} = t_{base} \frac{ \sqrt{1-\frac{(v_{top})^2}{c^2}} \sqrt{1-\frac{2GM}{(R+h)c^2}} }{ \sqrt{1-\frac{(v_{base})^2}{c^2}} \sqrt{1-\frac{2GM}{(R)c^2}} }

    where [itex]v_{base}[/itex] is the velocity of the base of the tower through space and [itex]v_{top}[/itex] is the speed of the top of the tower through space.

    The above equations are based on the Schwarzschild metric which assumes a non rotating spherical gravitational body, but they are a reasonable approximation because of the low rotational velocity of the Earth. For more accurate results you should use the Kerr metric which assumes a rotating body. Where you are on the Earth also affects the velocity. A person on the Equator is travelling faster through space than a person at the poles.* For these reasons idealising the Earth as non-rotating and spherical, as Jesse suggested, considerably simplifies things, but at a cost of some accuracy.

    A very similar question was asked recently here https://www.physicsforums.com/showthread.php?t=474350 and you can find more info and some calculations using real data there.

    * It is interesting to note that while a clock at sea level at the Equator is travelling through space faster than a clock at sea level at one of the poles, the clock at the Equator is at a greater radius radius from the centre due to the non-spherical shape of the Earth and the velocity effect cancels out the gravitational effect exactly and all clocks at sea level anywhere on the Earth tick at the same rate relative to each other.

    P.S. While it is tempting to think you are stationary while standing at the base of the tower on the surface of the Earth, all you have do is hold a gyroscope in your hand to reveal that is not true.
    Last edited: Mar 6, 2011
  20. Mar 6, 2011 #19
    Excellent. Thanks, yuiop. That clarifies the situation for me. I guess the mistake I was making is that I was confusing special relativity where zero relative velocity means the same reference frame with general relativity where you can only be in the same reference frame if you are at the same radius. If you have a different radius then you must be in a different reference frame.

    Just one further question. If the tower is at the North Pole, then there would be no difference in velocity between the top and bottom of the tower, so do I then forget about the velocity part of the equation?
  21. Mar 6, 2011 #20


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