# B Moon time versus Earth time...

1. Aug 17, 2016

### hsdrop

does the earth and the moon run on 2 different flows of time and if so what would be the difference ?? also would the face that we see of the moon be different that the part that faces away from the earth ???

2. Aug 17, 2016

### rootone

Time is time unless we are talking of objects moving near light speed relative to each other, so for the the Earth-Moon system, relativistic time dilation is insignificant.
A clock on the moon would run at the same rate as a clock on Earth, within any reasonable measure of accuracy.
Your second point. Moon orbiting satellites have shown that the side of the Moon facing away from Earth is more mountainous, less of flat lowlands.

Last edited: Aug 17, 2016
3. Aug 17, 2016

### hsdrop

i know that the affix of time dilation is insignificant with slower speed are no wait when compared to something moving near c but i was looking at the earth and the moon spinning around for say a billion years or more would it not add up then throw i would understand if the difference was only a day or so?? maybe?

4. Aug 17, 2016

### rootone

Probably would be more like a billion years (as seen from Earth) makes less then a second of difference to the clocks, but yes in principle they could drift slightly.

5. Aug 17, 2016

### Staff: Mentor

This question is too vague as you state it, since how we compare the "flow of time" depends on how we match up "the same moment of time" between the earth and the moon.

One possible more precise formulation would be: suppose we have two observers, one on the surface of the Earth and one on the surface of the Moon. And suppose that every time the Moon is directly overhead as seen by the observer on the Earth, the two observers record the readings on their respective clocks. (Here the passing of the Moon overhead is what defines "the same moment of time" for both observers.) How will the time intervals between two successive readings compare between the two clocks?

To answer this question, we need to take into account three contributions to "time dilation": the Earth's gravity, the Moon's gravity, and the motion of the observer relative to the center of gravity of the Earth-Moon system. The time dilation factor for each observer is then

$$\sqrt{1 - \frac{2GM_e}{c^2 r_e} - \frac{2GM_m}{c^2 r_m} - \frac{v^2}{c^2}}$$

where $M_e$ is the mass of the Earth and $r_e$ is the observer's distance from the Earth's center; $M_m$ is the mass of the Moon and $r_m$ is the observer's distance from the Moon's center; $v$ is the observer's velocity; $c$ is the speed of light; and $G$ is Newton's gravitational constant. Note that this is the time dilation factor relative to an observer at "infinity", i.e., one very far away from both the Earth and Moon, and at rest relative to the center of mass of the Earth-Moon system.

For parameter values, we use a velocity of 450 m/s for the Earth observer and 1022 m/s for the Moon observer (remember these are relative to the center of mass of the Earth-Moon system, so the main contribution for the Earth observer is from Earth's rotation and the main contribution for the Moon observer is the Moon's orbital velocity); we use an $r_e$ of the Earth's radius and an $r_m$ of the Moon's distance from Earth minus the Earth's radius, for the Earth observer; and we use an $r_e$ of the Moon's distance from Earth minus the Moon's radius, and an $r_m$ of the Moon's radius for the Moon observer.

Plugging in values, we obtain for the Earth observer a time dilation factor of

0.9999999993038333,

and for the Moon observer a time dilation factor of

0.9999999999506316.

The Moon observer's factor is closer to 1, so his clock will show more elapsed time between two successive readings (i.e., two successive overhead passages) than the Earth observer's clock.

From the above formula it should be clear that there would be some difference, because the distance $R_e$ will be different for the Moon observer; it will be the sum of the Moon's radius plus the Earth-Moon distance, instead of the difference between them. This gives a time dilation factor by the above formula of

0.9999999999507357.

It depends on what you consider "reasonable". The ratio of the two time dilation factors above differs from 1 by a few parts in $10^{10}$, and results in a difference in elapsed time of about 58 microseconds between successive overhead passages of the Moon (about 24 hours 50 minutes by Earth clocks). That is easily detectable with current technology.

The ratio of the two Moon observers' time dilation factors, however, differs from 1 by only a part in $10^{13}$ or so, and results in a difference in elapsed time of only about 9 nanoseconds between successive overhead passages of the Moon. That is detectable with current technology, but is much closer to the limit of accuracy of our best current atomic clocks.

6. Aug 17, 2016

### hsdrop

wow thank you so much for the math make me wonder that the farther and farther out we look as a hole how earth time would dilation in relationship to the solar system( other planets and moons around the sun) and milky way or even other galaxies

i will admit that it feels a bit counter intu vided to think of everything moving through time at different rates compared to everything else even throw the truth has been proven to be what it is thank you again

7. Aug 18, 2016

### Staff: Mentor

If we want to make these comparisons we have to take into account more gravity fields. For example, if we want to compare with an observer outside the solar system, we have to include the Sun's gravity (and possibly the gravity of other planets). If we want to compare with an observer outside our galaxy, we have to include the galaxy's gravity. That basically means adding more terms under the square root in the formula I gave, each one with a different $M$ (the sun's mass or the galaxy's mass) and a different $r$ (the distance from the center of the sun or the center of the galaxy). We also have to consider speeds relative to the appropriate center of mass (solar system or galaxy).

If we include a term for the sun's gravity and consider an observer on Earth, who will be 1 AU from the Sun and be moving at the Earth's orbital velocity (we can adjust if we like for the radius of the Earth and the velocity of the Earth's rotation), we get a time dilation factor of

0.9999999842799239,

which is already less than 1 by much more than the factor we calculated for the Earth observer just using the Earth's gravity and the Moon's. The factor for the observer on the Moon, including the Sun's gravity and using velocity relative to the center of mass of the solar system, is

0.999999984744275,

which is still closer to 1, so the Moon observer's clock will still be running faster than the Earth observer's clock.

I'll leave it to you to look up numbers for the galaxy and work out time dilation factors for that case.