The Office of Science and Technology Policy, part of the government of the United States of America, issued this statement outlining the need for a Lunar Standard Time (LST) standard. They state:

For example, to an observer on the Moon, an Earth-based clock will appear to lose on average 58.7 microseconds per Earth-day with additional periodic variations.

I take that to mean a moon-based clock would require a -1 leap second (repeating a second) every 46.6 years in order to stay synchronized with an Earth-based clock to within 1 second.


Where does this 58.7 $\mu$s / day slip come from? The moon's relative velocity? It's weaker gravitational field? If both are factors, what are their relative contributions?

Bonus question:

What are the 'periodic variations' mentioned in the statement, how big and frequent are they? E.g. what is the largest periodic perturbation to the mean 58.7$\mu$s / day dilation?

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    $\begingroup$ To be pedantic, a second on the Moon is by definition the same as a second on Earth. The time differences reflect different durations, not different units of time. $\endgroup$ Commented Apr 4 at 21:36
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    $\begingroup$ A related current thread on our sister site: space.stackexchange.com/q/65783/38535 $\endgroup$
    – PM 2Ring
    Commented Apr 5 at 1:33
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    $\begingroup$ @MarcoOcram, that's like saying the length of a yard is always a yard, even when you measure with a rubber yardstick that stretches. $\endgroup$ Commented Apr 5 at 15:17
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    $\begingroup$ @MichaelHall. No, you misunderstand the subtlety of it. $\endgroup$ Commented Apr 5 at 15:57
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    $\begingroup$ @MichaelHall no it isn't, it's like saying a yard is a yard, even if you use your (non-stretchy) yardstick to measure a big stretchy thing. $\endgroup$
    – N. Virgo
    Commented Apr 6 at 9:45

4 Answers 4


The time dilation is mainly gravitational and, to a first approximation, calculated by the Schwarzschild metric, which is $$c^2\mathrm{d}\tau^2 = -\left(1-\frac{r_S}{r}\right)c^2\mathrm{d}t^2 + \left(1-\frac{r_S}{r}\right)^{-1}\mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2$$ where $r_S = 2GM/c^2$ is the Schwarzschild radius, $\mathrm{d}\tau$ is the time of the observer and $\mathrm{d}t$ is the time at infinity (i.e. the time in the absence of gravity). So we see that $$\mathrm{d}\tau \approx \left(1-\frac{1}{2}\frac{r_S}{r}\right)\mathrm{d}t = \left(1+\frac{\Phi}{c^2}\right)\mathrm{d}t$$ where $\Phi$ is the gravitational potential. This means that the fractional time dilation between two points is, to first order, proportional to the difference in their gravitational potential. (This result can also be obtained using linearized gravity.)

Plugging in the numbers for the Earth-Moon system, the potential is -62.58 MJ/kg at Earth's surface and -3.86 MJ/kg at the Moon's surface. This works out to a value of about 56.5 microseconds per day faster for the Moon.

The orbital speed of the Moon contributes, through special relativity, a time dilation of 0.5 microseconds per day, resulting in a net total of around 56 microseconds per day. This is consistent with the result here.

The largest fluctuation comes from the variation in the Moon's distance to the Sun as it orbits Earth, which contributes a difference of 4.5 MJ/kg between the closest and farthest points. This is followed by the eccentricity of the Moon's orbit causing the Earth-Moon distance itself to vary.

  • $\begingroup$ Some Schwarzschild radii, in metres, using data from ssd.jpl.nasa.gov/astro_par.html Sun 2.95325008e+3 Earth 8.87005594e-3 Moon 1.09102017e-4 $\endgroup$
    – PM 2Ring
    Commented Apr 5 at 2:18
  • $\begingroup$ Thank you for your time answering. To be more clear though, the relativistic dilation opposes the gravitational dilation? I.e. 56.5 - 0.5 = 56? $\endgroup$
    – cms
    Commented Apr 8 at 0:07
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    $\begingroup$ @cms Yes, using the (approximately) inertial frame of the Earth, the Moon's clock is 56.5 microseconds faster due to it being higher up in Earth's gravity, but its orbital velocity makes it 0.5 microseconds slower due to special relativistic time dilation (which tells us that moving clocks run slower). $\endgroup$ Commented Apr 8 at 10:29

It's basically due to gravitational time dilation. Ratio of proper times of Moon vs Earth is : $$\tag 1\large \frac{t_{moon}}{t_{earth}} = \frac {{\sqrt {1-{\frac {2Gm}{rc^{2}}}}}}{{\sqrt {1-{\frac {2GM}{Rc^{2}}}}}} ,$$

where $M,m$ are Earth and Moon masses and $R,r$ - Earth and Moon radii. Substituting celestial objects parameters we get this ratio to be $\approx 1.000~000~000~665$. That's how much faster time on the Moon lapses compared to Earth.

So according to (1) formula while on Earth passes 24 hours, on the Moon passes $1.000 000 000 665 × 24~\text{hours} \approx 1\text{ day} + 57~\mu s$.

Hence the difference of Moon-Earth proper times within full Earth day is $57\mu s$.

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    $\begingroup$ This only considers the two bodies in isolation and ignores the important fact that both their gravitational fields are present simultaneously. $\endgroup$ Commented Apr 4 at 23:16
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    $\begingroup$ Well, your answer based on both planets gravitational potentials, gives gravitational time dilation-based speedup in the Moon as $56.5\mu s$ which rounded to the nearest integer gives exactly my solution. So, my approximation is good enough, because it's relative error is only about $0.9\%$ with respect to including distant field effects (as in your answer). Sure, every correction matters. $\endgroup$ Commented Apr 5 at 6:06
  • $\begingroup$ Yeah, I just thought I'd point it out since it's still a significant correction (about a microsecond) to the vacuum isolated calculation. $\endgroup$ Commented Apr 5 at 9:32

The vast majority of the difference comes from the significantly less mass of the moon compared to the Earth. So gravitational time dilation.
To a MUCH lesser extent there will be kinetic time dilation from (as you say) the relative movement of the moon compared to the Earth.

The small variances will come as the moon moves in it's elliptical orbit around the Earth, and to a lesser extent around the Sun, slipping deeper into and out of the gravitational time dilation of each.


Main Question

The main correction is due to the gravitational effects of the Earth clock being in the potential well of the Earth, followed by the effects of the Moon clock being in the potential well of the Moon and of the Earth, and the effects of the relative speeds of the clocks are only felt in the third significant figure. I obtain an overall correction of 56.0 $\mu$s/day and cannot reproduce the statement value of 58.7 $\mu$s/year unless ignoring the fact that the Moon clock is in the potential well of the Moon.


An answer can be provided in terms of the Schwarzschild metric interval, which says $$c^2 d\tau^2 = \left(1 - \frac{r_s}{r}\right)c^2dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1} dr^2 - r^2d\theta^2 - r^2\sin^2\theta d\phi^2$$ in the usual Schwarzschild-Droste coordinates, where $d\tau$ is a proper time interval and $r_s$ is the Schwarzschild radius.

If we set the Moon's orbital plane to be fixed at $\theta=\pi/2$ ($d\theta=0$) and assume $r_s/d_m \ll 1$, then an object in a Moon-like orbit around the Earth, at the distance of the Moon, $d_m$ will satisfy $$ \frac{d\tau_{m,o}}{dt} = \left[1 - \frac{r_{s,e}}{d_m} - \frac{v_m^2}{c^2} \right]^{1/2}\ ,$$ where $v_m$ is the orbital speed of the Moon and $r_{s,e}$ is the Schwarzschild radius of the Earth.

However, a clock on the Moon's surface will run slower than this because of the Moon's gravitational potential that can also be represented with the Schwarzschild metric (assuming the gravitational influence of the Earth can be neglected at the surface of the Moon - i.e. $M_m/r_m^2 \gg M_e/d_m^2$). $$\frac{d\tau_{m,s}}{d\tau_{m,o}} = \left[1 - \frac{r_{s,m}}{r_m}\right]^{1/2}\ ,$$ where $r_{s,m}$ is the Schwarzschild radius of the Moon and we assume the Moon's surface moves at the same velocity as something in orbit around the Earth at the distance of the Moon.

Finally, a clock on the Earth's surface has its own proper time given by $$\frac{d\tau_{e,s}}{dt} = \left[1 - \frac{r_{s,e}}{r_e} - \frac{v_e^2}{c^2} \right]^{1/2}\ , $$ where $v_e$ is the speed of the clock on the Earth's surface with respect to the centre of the Earth.

Putting this together, the ratio of proper time intervals for a clock on the surface of the Moon and a clock on the surface of the Earth is $$\frac{d\tau_{m,s}}{d\tau_{e,s}} = \frac{d\tau_{m,s}}{d\tau_{m,o}} \frac{d\tau_{m,o}}{dt} \frac{dt}{d\tau_{e,s}} = \left[1 - \frac{r_{s,m}}{r_m}\right]^{1/2} \left[1 - \frac{r_{s,e}}{d_m} - \frac{v_m^2}{c^2} \right]^{1/2} \left[1 - \frac{r_{s,e}}{r_e} - \frac{v_e^2}{c^2} \right]^{-1/2}\ .$$

All the terms $r_{s,m}/r_m$, $r_{s,e}/d_m$, $r_{s,e}/r_e$, $v_m^2/c^2$ and $v_e^2/c^2$ are $\ll 1$, so binomial expansions can be employed, keeping only terms at this order. $$\frac{d\tau_{m,s}}{d\tau_{e,s}} = 1 - \frac{1}{2}\left[ \frac{r_{s,m}}{r_m} + \frac{r_{s,e}}{d_m} + \frac{v_m^2}{c^2} - \frac{r_{s,e}}{r_e} - \frac{v_e^2}{c^2}\right]\ . $$

We can now put in numbers and explicitly write down the numerical values of each of the correction terms to see which are the most important. I have used $r_{s,e} = 8.869\times 10^{-3}$ m, $r_{s,m} = 1.09\times 10^{-4}$ m, then assuming clocks on the equators of the Moon and Earth, so $r_e = 6.378 \times 10^6$ m, $r_m = 1.74\times 10^6$ m. For $v_e$, I assume $0.465\times 10^3$ m/s, for a mean orbital speed I assume $v_m = 1.02\times 10^3$ m/s and for an average Earth-Moon distance, $d_m = 3.85 \times 10^8$ m. $$\frac{d\tau_{m,s}}{d\tau_{e,s}} = 1 - \frac{1}{2}\left[ 6.26\times 10^{-11} + 2.30\times 10^{-11} + 1.16\times 10^{-11} - 1.391\times 10^{-9} - 2.40\times 10^{-12}\right] $$

Thus the most important term is due to a clock on the Earth in the deep gravitational potential well of the Earth; the next most important is the clock on the Moon being in the potential well of the Moon. Then, that the clock on the Moon is still in the potential well of the Earth and the speed terms make contributions to the third significant figure.

In summary $$\frac{d\tau_{m,s}}{d\tau_{e,s}} = 1 - 6.48\times 10^{-10}\ .$$ Multiplying the correction term by the number of seconds in 24 hours, we have that a clock on the Moon runs faster by 56.0 $\mu$s/day.

Bonus Question

There will then be periodic terms on the period of the Moon's orbit around the Earth. These will: (i) cause $d_m$ to change by about $\pm 5$%, which, from the contribution of the second term, leads to variations of up to $\pm 0.9$ $\mu$s/day. (ii) There is also a contribution from the Earth and Moon sitting at different points in the Sun's gravitational potential by up to (approximately) the distance between the Earth and the Moon. This leads to a change in the relative time dilation of up to $$ \Delta \left(\frac{d\tau_{m,s}}{d\tau_{e,s}}\right) \simeq \frac{1}{2} \left[ \frac{r_{s,\odot}}{d_e} - \frac{r_{s,\odot}}{d_e + d_m}\right] \simeq \frac{d_m r_{s,\odot}}{2d_e^2}\ ,$$ where $d_e$ is the Earth-Sun distance and $r_{s,\odot}$ is the Schwarzschild radius of the Sun. Using $d_e = 1.50\times 10^{11}$ m and $r_{s,\odot}=2.96\times 10^3$ m yields a periodic correction of up to $\pm 2.2$ $\mu$s/day.


I cannot reproduce the statement value of $58.7$ $\mu$s/day. I note however, that this is what I get if I ignore the correction for the clock being on the surface of the Moon... i.e. I think that the figure of $58.7$ $\mu$s/day is the correction for a clock in orbit around the Earth at the distance of the Moon, not a clock on the Moon's surface.

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    $\begingroup$ Interesting. The values I got were 60.14 for Earth's gravity at Earth's surface and 2.71 for Moon's gravity at the Moon's surface. The Moon still being in Earth's gravitational field is another -1 (making it 59.14), and the Moon's orbital velocity is 0.5. Earth's rotational velocity is much smaller than all of this (maximum of 0.1 at equator), that's why I didn't include it as it's even smaller than the Sun's correction. Therefore I couldn't get the proposed value of 58.7. $\endgroup$ Commented Apr 6 at 13:10
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    $\begingroup$ @VincentThacker Exactly. The statement figure appears to have ignored the 2.7 $\mu$s/day correction due to the clock being on the Moon's surface. It is instead the correction for a clock in orbit at the distance of the Moon. $\endgroup$
    – ProfRob
    Commented Apr 6 at 13:23

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