# Does Earth experience any significant, measurable time dilation at perihelion?

Is there any measurable time dilation when Earth reaches perihelion? Can we measure such a phenomena relative to the motion of the outer planets?

• Possible duplicate of Since the Earth is moving through space, do we experience time dilation?
– JMac
Mar 28, 2017 at 11:44
• -1. No research effort. Mar 29, 2017 at 4:12
• @sammygerbil: that's a bit harsh. It's not a trivial problem. How many of the site members could do the calculation I describe? Mar 29, 2017 at 5:41
• @JohnRennie : I was referring to effort to look for readily-available information, such as on wikipedia or this site. Mar 29, 2017 at 6:35
• @TomDales Probably 2 things I should point out with regards to your response to sammy gerbil. 1) I don't think a discussion between the users of an internet Q&A site is acceptable as a reference in a university paper (unless you're analyzing the dynamics of scientific discussion more than the content itself). And 2) You're not likely to find us responding in the form of conversation; this site is more for Q&A and less for forum-like discussions, PLUS this is fairly well-understood material with one answer and little room for debate. Still, good question
– Jim
Mar 29, 2017 at 12:38

We can calculate the time dilation for an object moving in the Sun's gravitational field using the Schwarzschild metric. Strictly speaking this is an approximation since the Sun is rotating and not spherical, but it will give us a pretty good answer.

The Schwarzschild metric is (writing it in terms of the proper time):

$$c^2d\tau^2 = \left( 1 - \frac{r_s}{r}\right) c^2dt^2 - \frac{dr^2}{1 - r_s/r} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 \tag{1}$$

where $r_s$ is the Schwarzschild radius of the Sun:

$$r_s = \frac{2GM_{Sun}}{c^2}$$

At the perihelion and aphelion the motion is tangential so the radial velocity is zero and therefore $dr=0$. Also we will arrange our coordinates so that all motion is in the equatorial plane so $\theta=\pi/2$ and $d\theta=0$. Substituting these values into equation (1) we find the metric simplifies considerably to:

$$c^2d\tau^2 = \left( 1 - \frac{r_s}{r}\right) c^2dt^2 - r^2 d\phi^2 \tag{2}$$

If the tangential velocity is $v$ then the angle $d\phi$ moved in a time $dt$ is just:

$$d\phi = \omega dt = \frac{v}{r}dt$$

and we substitute this into equation (2) to get:

$$c^2d\tau^2 = \left( 1 - \frac{r_s}{r}\right) c^2dt^2 - r^2 \left(\frac{v}{r}\right)^2dt^2$$

which we rearrange to give us the equation for the time dilation:

$$\frac{d\tau}{dt} = \sqrt{1 - \frac{r_s}{r} - \frac{v^2}{c^2}} \tag{3}$$

According to NASA's fact sheet the values of $v$ and $r$ at perihelion and aphelion are:

\begin{align} r_p &= 1.4709 \times 10^{11} \,m \\ v_p &= 30290 \,m/s \\ r_a &= 1.5210 \times 10^{11} \,m\\ v_a &= 29190 \,m/s \end{align}

And the Schwarzschild radius of the Sun is $r_s \approx 2953$ m. Putting these figures into our equation (3) gives us:

\begin{align} d\tau/dt \,\text{perihelion} &= 0.99999998486 \\ d\tau/dt \,\text{aphelion} &= 0.99999998555 \end{align}

we can make these numbers a bit more digestible by expressing them as time lost per day e.g. how many seconds a day do clocks on the Earth run slower as a result of the time dilation. If we do this we find:

\begin{align} \text{perihelion loss} &= 1.308 \,\text{ms/day} \\ \text{aphelion loss} &= 1.248 \,\text{ms/day} \end{align}

And the difference between the two is about $60\mu$s per day. So clocks run about $60\mu$s per day more slowly at perihelion than they do at aphelion.

This is easily measurable in principle since atomic clocks have the accuracy to measure shifts this small. However there are practical difficulties. The time dilation is measured relative to a stationary observer outside the Sun's gravitational influence, and we can't easily put an atomic clock somewhere outside the orbit of Pluto to make the comparison. We could put a satellite in an exactly circular orbit at the average orbital radius of the Earth, and in that case our clocks would run about about $30\mu$s per day faster than the satellite clock at aphelion and the same amount slower at perihelion.

A quick footnote:

Count Iblis points out that pulsars make an excellent clock outside the gravitational influence of the Sun, and we can measure pulsar frequencies with enough accuracy to detect the $60\mu$s per day change between perihelion and aphelion. If anyone has a reference for this feel free to edit it into this answer.

• @Jim: good point. I've extended my answer accordingly and I agree it is an improvement. Mar 29, 2017 at 5:39
• We have good pulsar timing methods, so this effect is visible. I've read that the much smaller monthly periodic effect due to the Earth orbiting the Earth- Moon barycenter is also visible from pulsar timing data. Mar 29, 2017 at 7:53
• @JohnRennie Hooray for mutual cooperation towards the betterment of content for all users! Also, thanks for double checking my math. I found approximately $5.96\times10^{-5}s$ for the difference between a perihelion and aphelion day and (I must have been smoking something) my head went "Yeah, with a -5, that must be $6\mu s$". No. What's interesting is your $60\mu s$ is in coordinate time and mine was in perihelion time. Not surprising, given $d\tau/dt$, just interesting
– Jim
Mar 29, 2017 at 12:31
• You should be getting a difference of about 3.4 milliseconds per day rather than 60 microseconds per day. Mar 30, 2017 at 16:59
• @DavidHammen: $60\mu$s is the difference in the time dilation at perihelion and aphelion. I get $1.25$ms per day for the time dilation at aphelion and $1.3$ms per day at perihelion. Mar 30, 2017 at 17:18

The GPS depends on corrections to the timing from General and Special relativity because the satellites are in a smaller gravitational field and they are running with a high enough velocity.

Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly (see the Special Relativity lecture). Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion .

Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away (see the Black Holes lecture). As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.

The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)! This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and 38 microseconds is 38,000 nanoseconds. If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day! The whole system would be utterly worthless for navigation in a very short time.

The Time dilation - Earth & Jupiter , should tell you that corresponding differences will be found between the perihelion and aphelion of the earth.