I What is the angular velocity of a satellite?

[binis]

TL;DR Summary

Angular speed ω is by definition the runned angle dθ per time dt elapsed: ω=dθ/dt. If the time elapsed in the center of the earth is dt, the dilated time elapsed on satellite is dt′. What is the satellite's angular speed? Is it dθ/dt or dθ/dt′?

Angular velocity ω is by definition the runned angle dθ per time dt elapsed: ω=dθ/dt. If the time elapsed in the center of the Earth is dt, the dilated time elapsed on satellite is dt′. What is the satellite's angular velocity? Is it dθ/dt or dθ/dt′?

 

[haushofer]

As you are obliged to ask in relativity theory: according to whom?

 

[PeroK]

What is the satellite's angular velocity? Is it dθ/dt or dθ/dt′?

Why not both? Both might be useful quantities.

 

[Dale]

Summary:: Angular speed ω is by definition the runned angle dθ per time dt elapsed: ω=dθ/dt. If the time elapsed in the center of the Earth is dt, the dilated time elapsed on satellite is dt′. What is the satellite's angular speed? Is it dθ/dt or dθ/dt′?

If the time elapsed in the center of the Earth is dt, the dilated time elapsed on satellite is dt′. What is the satellite's angular velocity? Is it dθ/dt or dθ/dt′?

That is a relative quantity. That means that its value depends on the coordinate system chosen. Roughly speaking, if you choose the coordinate system $(t,r,\theta,\phi)$ then the angular velocity will be $d\theta/dt$. If you choose the coordinate system $(t’,r,\theta,\phi)$ then the angular velocity will be $d\theta/dt’$. Either choice is valid.

Any time you have a relative quantity you must specify the coordinate system you are using. Otherwise it does not have a unique definite value.

 

[binis]

Any time you have a relative quantity you must specify the coordinate system you are using.

Let's stay in the local (ECI) frame.I think I am getting it wrong,in this frame,there is not dt',there is not time dilation?Where the dt' is appear?

 

[Dale]

Let's stay in the local (ECI) frame.I think I am getting it wrong,in this frame,there is not dt',there is not time dilation?Where the dt' is appear?

Ok, so in your OP you listed $dt$ as the time at the center of the Earth and $dt’$ as the time at the satellite. But the time coordinate in the ECI is not either of those. The ECI uses time as at the surface of the Earth for hypothetical clocks that are not fixed to the earth, but stay in place as the Earth rotates under it. So if we use $dT$ to denote the time in the ECI $(T,r,\theta,\phi)$ coordinates then the angular velocity in the ECI would be $d\theta/dT$

 

[PeterDonis]

The ECI uses time as at the surface of the Earth for hypothetical clocks that are not fixed to the earth, but stay in place as the Earth rotates under it.

Actually, I don't think even this is quite correct. As I understand it, the ECI uses hypothetical points in space not fixed to the earth, but staying in place as the Earth rotates underneath them, to fix the simultaneity convention and the directions of the spatial axes. But the clock rate in the ECI is the rate of clocks at rest on the geoid of the rotating Earth, which will not be the same (it will be slightly slower) as the clock rate of actual "inertial" clocks sitting at rest in the ECI at the altitude of the surface of the Earth. (I put "inertial" in quotes because of course such hypothetical "at rest in the ECI" clocks would have to have nonzero proper acceleration in order to maintain altitude, and their clock rates would be affected by the Earth's gravitational potential.)

 

[Dale]

Actually, I don't think even this is quite correct.

You could be correct. I am going from memory. So then a clock attached to the Earth at the level of the geoid would not be at rest in the ECI, but its proper time would match the ECI coordinate time.

 

[PeterDonis]

a clock attached to the Earth at the level of the geoid would not be at rest in the ECI, but its proper time would match the ECI coordinate time.

That is my understanding, yes. Unfortunately I can't find any good sources online at the moment for this; none of the references I can find talk about the time coordinate at all, they only talk about the spatial coordinates.

 

[Motore]

Possibly relevant:
https://www.itu.int/dms_pubrec/itu-r/rec/tf/R-REC-TF.2018-0-201208-W!PDF-E.pdf

 

[vanhees71]

For GPS I found this a good reference:

http://www.nbmg.unr.edu/staff/pdfs/Blewitt_Encyclopedia_of_Geodesy.html

 

[binis]

That is a relative quantity. That means that its value depends on the coordinate system chosen. Roughly speaking, if you choose the coordinate system $(t,r,\theta,\phi)$ then the angular velocity will be $d\theta/dt$. If you choose the coordinate system $(t’,r,\theta,\phi)$ then the angular velocity will be $d\theta/dt’$. Either choice is valid.

But only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.Relative velocity ω is dθ/dt = dθ/dt' => dt=dt' => no time dilation ?

 

[jbriggs444]

But only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.Relative velocity ω is dθ/dt = dθ/dt' => dt=dt' => no time dilation ?

If we are working within the framework of special relativity, the frame in which the satellite is at rest is not inertial. That matters.

When using a non-inertial frame of reference, things get... tricky. You can't always make the coordinates line up.

 

[Dale]

But only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.Relative velocity ω is dθ/dt = dθ/dt' => dt=dt' => no time dilation ?

No, this assertion fails. First, it fails because angular velocity is not the same as velocity so it doesn’t follow the same rules. Second, the physically meaningful angular velocity is invariant not relative, and is measured by local gyroscopes. Third, there is no requirement that relative rotation be reciprocal. Fourth, when spacetime is curved even relative linear velocity is not physically meaningful for spatially separated objects like in this scenario.

If I had to pick one reason as the most important I would pick the second. The idea that only relative angular velocity is physically meaningful is very wrong. Angular velocity is not a relative quantity, it is invariant and is easily measured locally with a gyroscope without reference to any other object. That is an essential and fundamental difference between angular velocity and linear velocity.

 

[jbriggs444]

Angular velocity is not a relative quantity, it is invariant and is easily measured locally with a gyroscope without reference to any other object.

I have one small quibble with this.

Whether an object is rotating or not is an invariant fact of the matter. It either rotates or does not. All inertial frames will agree. However, the rate at which an object rotates (and whether it rotates rigidly) is not invariant. Not all inertial frames will agree about the rate of rotation of an unaccelerated object about its center of mass.

For instance, the rotation rate of the second hand on an analog wristwatch cannot be a relativistic invariant if time dilation is a thing.

 

[etotheipi]

Wait, I thought @Dale refers to the endomorphism $\mathbf{\Omega} = \mathbf{\Omega}_{\text{FW}} + \mathbf{\Omega}_{\text{rot}}$ where $\mathbf{\Omega}_{\text{FW}}$ depends only on invariant 4-acceleration $\mathbf{a}$ of the observer and $\mathbf{\Omega}_{\text{rot}}(\mathbf{v}) = \boldsymbol{\omega} \times_{\mathbf{u}} \mathbf{v}$ depends only on the invariant 4-rotation $\boldsymbol{\omega}$ of the local frame, i.e. such that $d\mathbf{v}/d\tau = \mathbf{\Omega}(\mathbf{v})$. Also $\mathbf{a}$ and $\boldsymbol{\omega}$ can both be determined by the observer himself with local experiments.

 

[Dale]

However, the rate at which an object rotates (and whether it rotates rigidly) is not invariant. Not all inertial frames will agree about the rate of rotation of an unaccelerated object about its center of mass.

This is exactly analogous to coordinate acceleration and proper acceleration. What you are talking about above is the coordinate rotation. What I am talking about is the “proper” rotation. That is invariant because it is the rotation measured by gyroscopes carried by the object. All frames must agree on the outcome of such a measurement.

By the way, does anyone know the correct term for “proper” angular velocity? I don’t think that I have seen such a term used before.

 

[etotheipi]

By the way, does anyone know the correct term for “proper” angular velocity? I don’t think that I have seen such a term used before.

I think the formalism is this, although I hope someone will correct/improve: given a vector field $\mathbf{v}(\tau) = v^{\mu} \mathbf{e}_{\mu}(\tau)$ fixed with respect to an observer $\mathscr{O}$ [i.e. $v^{\mu}$ are independent of $\tau$], then one may define an endomorphism $\mathbf{\Omega}$ by$$\frac{d\mathbf{v}}{d\tau} = \mathbf{\Omega}(\mathbf{v}) = \underbrace{c(\mathbf{a} \cdot \mathbf{v})\mathbf{u} - c(\mathbf{u} \cdot \mathbf{v})\mathbf{a}}_{\mathbf{\Omega}_{\mathrm{FW}}(\mathbf{v})} + \underbrace{\boldsymbol{\epsilon}(\mathbf{u}, \boldsymbol{\omega}, \mathbf{v}, \, \boldsymbol{\cdot} \, )}_{\mathbf{\Omega}_{\mathrm{rot}}(\mathbf{v})}$$where $\mathbf{u} = d\mathscr{O}/d\tau$ is the 4-velocity of $\mathscr{O}$, $\mathbf{a}$ is the 4-acceleration of $\mathscr{O}$ and $\boldsymbol{\omega}$ is the 4-rotation of $\mathscr{O}$'s local frame $\{ \mathbf{e}_{\mu} \}$.

 

[Dale]

I think the formalism is this, although I hope someone will correct/improve: given a vector field $\mathbf{v}(\tau) = v^{\mu} \mathbf{e}_{\mu}(\tau)$ fixed with respect to an observer $\mathscr{O}$ [i.e. $v^{\mu}$ are independent of $\tau$], then one may define an endomorphism $\mathbf{\Omega}$ by$$\frac{d\mathbf{v}}{d\tau} = \mathbf{\Omega}(\mathbf{v}) = \underbrace{c(\mathbf{a} \cdot \mathbf{v})\mathbf{u} - c(\mathbf{u} \cdot \mathbf{v})\mathbf{a}}_{\mathbf{\Omega}_{\mathrm{FW}}(\mathbf{v})} + \underbrace{\boldsymbol{\epsilon}(\mathbf{u}, \boldsymbol{\omega}, \mathbf{v}, \, \boldsymbol{\cdot} \, )}_{\mathbf{\Omega}_{\mathrm{rot}}(\mathbf{v})}$$where $\mathbf{u} = d\mathscr{L}_{\mathscr{O}}/d\tau$ is the 4-velocity of $\mathscr{O}$, $\mathbf{a}$ is the 4-acceleration of $\mathscr{O}$ and $\boldsymbol{\omega}$ is the 4-rotation of $\mathscr{O}$'s local frame $\{ \mathbf{e}_{\mu} \}$.

Yes. Mathematically, that seems correct (@vanhees71 do you spot any problems I missed), but is there a specific term for that? I am not aware of it.

 

[Nugatory]

But only relative velocity is physically meaningful. We can choose to interpret either satellite or Earth to be at rest.Relative velocity ω is dθ/dt = dθ/dt' => dt=dt' => no time dilation ?

You could make the same argument for two clocks moving inertially (that is, in a straight line with no rotation, angular velocity, centripetal force, or any other things that make circular motion more tricky). It might be worth working out for yourself why the argument fails in that case before taking on the more complicated orbiting satellite case.

 

[vanhees71]

That looks plausible to me, but I must say I've never thought about "angular velocity" in relativity. If I understand it right, what @etotheipi defines as $\boldsymbol{\omega}$ quantifies the additional change of a vector along an observer's timelike worldline with time which goes beyond the Fermi-Walker transport, and this precisely describes the rotation of this vector relative to a Fermi-Walker (i.e., rotation-free) transported tetrad of the observer, and that's of course just describing an "angular velocity" in close analogy to how you define the angular velocity in Newtonian rigid-body dynamics.

 

[weirdoguy]

But only relative velocity is physically meaningful.

You've already been told, in the previous thread you were participating in, that this is not true.

 

[binis]

the physically meaningful angular velocity is invariant not relative, and is measured by local gyroscopes. Angular velocity is not a relative quantity, it is invariant and is easily measured locally with a gyroscope without reference to any other object.

An observer on the moon's surface sees the Earth revolve around the moon.

 

[Dale]

An observer on the moon's surface sees the Earth revolve around the moon.

So what? That is not the earth’s physically meaningful angular velocity.

Please address the points I raised before proceeding.

 

[PeterDonis]

does anyone know the correct term for “proper” angular velocity?

The relevant invariant is the vorticity of the congruence of worldlines describing the object.

 

[Dale]

The relevant invariant is the vorticity of the congruence of worldlines describing the object.

Of course. I cannot believe I never made the mental connection between vorticity and what I was mentally calling proper rotation. It is obvious in retrospect.

 

[PeterDonis]

Of course.

Note, though, that this sense of "rotation" corresponds to "spin", not "orbiting". For example, in the case of the Earth, it describes the Earth's rotation about its axis, but not its orbiting about the Sun. In the case of the Moon, this can get confusing because its (sidereal) rotation about the Earth has the same angular velocity as its rotation about its axis, because of tidal locking. But the vorticity of the congruence of worldlines that describes the Moon still only describes the latter, not the former.

 

[PeterDonis]

In the case of the Moon, this can get confusing because its (sidereal) rotation about the Earth has the same angular velocity as its rotation about its axis, because of tidal locking.

I should probably also point out that there are other relativistic effects--Thomas precession, de Sitter precession, and Lense-Thirring precession--that affect the relationship between "spin" and "orbital" rotation. These are all very small for bodies like the Moon, Earth, and Sun, so they are usually ignored.

 

[Dale]

Note, though, that this sense of "rotation" corresponds to "spin", not "orbiting".

Yes, but that is intended since it is the only invariant angular velocity. Orbital angular velocity is not invariant.

In the case of the Moon, this can get confusing because its (sidereal) rotation about the Earth has the same angular velocity as its rotation about its axis, because of tidal locking.

That brings up a good point. Even in Newtonian physics with no time dilation the concept of relative angular velocity is ambiguous. If you have a planet which spins once per day and a moon which spins once per week and orbits once per month, then what is the relative angular velocity?

 

[PeterDonis]

Orbital angular velocity is not invariant.

In the general case, no, because there is no general invariant notion of "coming back to the same place" or "going through 360 degrees of angle" or any other concept that could be used to define what one "orbit" means.

There are some special cases where I think an invariant notion of "one orbit" could be defined, but they're very special--I think one would need the spacetime to be stationary and spherically symmetric, or at least axially symmetric with the plane of the orbit being an "equatorial plane" of the axial symmetry. And any such system could at most have one "primary" and one "satellite"--a system like the Solar System, where we have satellites orbiting planets orbiting the Sun, would not meet the requirements.

 

[etotheipi]

Even in Newtonian physics with no time dilation the concept of relative angular velocity is ambiguous. If you have a planet which spins once per day and a moon which spins once per week and orbits once per month, then what is the relative angular velocity?

It's a good point but I wouldn't say there's any real ambiguity; in rigid body dynamics, the angular velocity vector between any two frames $Oxyz$ with basis $\{\mathbf{e}_i \}$ and $P\zeta \eta \xi$ with basis $\{ \tilde{\mathbf{e}}_i \}$ is understood to be the vector $\boldsymbol{\omega}$ such that$$\frac{d\tilde{\mathbf{e}}_i}{dt} \big|_{Oxyz} = \boldsymbol{\omega} \times \tilde{\mathbf{e}}_i$$where $\dot{\mathbf{u}}|_{Oxyz} := \dot{u}^i \mathbf{e}_i$. This object $\boldsymbol{\omega}$ characterises only the changing relative orientation of these two frames, and says nothing about their relative translation.

[Sometimes people refer to an "orbital angular velocity" of a particle with respect to a point, but this really ought to be understood as the angular velocity of a particular rotating frame in which the origin and that particle are at rest, with respect to the other 'fixed' frame under consideration.]

 

[PeterDonis]

I wouldn't say there's any real ambiguity

Sure there is; the quantity you describe is frame-dependent. For example, say we want the angular velocity of the Moon around the Earth. For the "Earth frame", do we use the sidereal frame (basis vectors remain fixed with respect to the distant stars), the solar frame (basis vectors remain fixed with respect to the Earth-Sun line), or the Earth-centered frame (basis vectors remain fixed with respect to the rotating Earth)? And even those don't exhaust the possibilities. Look up how many different kinds of lunar months there are.

 

[etotheipi]

Sure there is; the quantity you describe is frame-dependent. For example, say we want the angular velocity of the Moon around the Earth. For the "Earth frame", do we use the sidereal frame (basis vectors remain fixed with respect to the distant stars), the solar frame (basis vectors remain fixed with respect to the Earth-Sun line), or the Earth-centered frame (basis vectors remain fixed with respect to the rotating Earth)? And even those don't exhaust the possibilities. Look up how many different kinds of lunar months there are.

Right, but the point I was making is that an angular velocity vector is defined between two specified frames, which must be explicitly stated when you define it. So the angular velocity vector between an Earth-fixed basis and a moon-fixed basis is $\boldsymbol{\omega}_1$, the angular velocity vector between a sidereal-fixed basis and a moon-fixed basis is another angular velocity vector $\boldsymbol{\omega}_2$, et cetera., and these are not equal vectors!

That's what I mean by "no ambiguity", i.e. each is relating one specified basis to a different specified basis.

 

[PeterDonis]

an angular velocity vector is defined between two specified frames

Yes, which just pushes the ambiguity into the definition of the frames. It still doesn't give you a single answer to the title question of this thread. It just makes clearer where the ambiguity is; it doesn't get rid of it.

 

[etotheipi]

Yes, which just pushes the ambiguity into the definition of the frames. It still doesn't give you a single answer to the title question of this thread. It just makes clearer where the ambiguity is; it doesn't get rid of it.

Ah okay, I guess there was a little bit of ambiguity about what was being referred to as being ambiguous

 

[vanhees71]

I think one must be a bit careful with the notion of "vorticity". In fluid dynamics you have on the one hand a kinematic vorticity tensor related to the displacement vectors of the congruence given by the worldlines of the fluid elements and the vorticity tensor (or better circulation tensor to distinguish it from the kinematic vorticity tensor), which is a two form defined from the enthalpy current, $w^{\mu}=h u^{\mu}$, where $h$ is the proper enthalpy density (i.e., the enthalpy density as measured in the local fluid-rest frame). The vorticity tensor then is defined by
$$\Omega_{\mu \nu}=\nabla_{\nu} w_{\mu}-\nabla_{\nu} w_{\mu}=\partial_{\nu} w_{\mu}-\partial_{\mu} w_{\nu}.$$
That's proportional to the kinematic vorticity tensor only for "dust", i.e., a pressureless fluid which is in free fall and the worldlines are thus geodesics.

Of course these quantities are in some sense frame dependent, i.e., it depends on the congruence of worldlines defined by the fluid. The distinct local frames where the physics of the tensor quantities is most easily interpreted are of course the local rest frames of the fluid, which thus is characterized by scalar quantities measured in the local rest frames of the fluid.

A very nice reference about fluid dynamics in GR is

L. Rezzolla and O. Zanotti, Relativistic hydrodynamics, Oxford University Press, Oxford (2013),
https://dx.doi.org/10.1093/acprof:oso/9780198528906.001.0001

 

[PeterDonis]

I think one must be a bit careful with the notion of "vorticity".

The notion I am referring to in this thread, which is the one arising from the kinematic decomposition of a congruence, is what you are calling "kinematic vorticity".

 

[binis]

So what?

The observer is watching his wristwatch running faster than a clock on the earth.Real?

 

[Ibix]

The observer is watching his wristwatch running faster than a clock on the earth.Real?

Sure.

I think your problem here is that you are trying to treat the Earth and Moon as symmetric and interchangeable. They aren't. So you should not be surprised that there are physical consequences to this lack of symmetry. Furthermore, you're trying to treat a GR problem using SR, which I suspect is causing headaches for those of us who know how that doesn't work.

Let's consider a pure SR case - a clock on the end of a piece of string swung in a circle around my head. I can easily adopt polar coordinates centered on me, and measure the orbital period of the clock. The clock could also adopt polar coordinates (for clarity, with the zero angle pointing in the same direction as mine) centered on itself. In this system it could measure the orbital period of my head, and would come out with a value that is lower than my value by a factor of $\gamma$.

You should not be surprised that the answers are not symmetric. One calculation is performed against an inertial frame and one against a non-inertial frame. The usual "everything's symmetric" rule of relativity only applies between inertial frames.

 

[Dale]

The observer is watching his wristwatch running faster than a clock on the earth.Real?

Yes, this has been experimentally demonstrated countless times with different types of experiments. This is how nature works.

 

[binis]

Yes, this has been experimentally demonstrated

Moon is the natural satellite of the earth. In the same way an observer on an artificial satellite is watching his watch running faster than a clock on the earth.True?

 

[binis]

Conclusively,time is running faster on the moon than the Earth but slower on some satellites. What about the ISS?

 

[Ibix]

Moon is the natural satellite of the earth. In the same way an observer on an artificial satellite is watching his watch running faster than a clock on the earth.True?

It depends on the altitude of the satellite because gravitational effects also affect clock rates. There is an altitude at which the two rate differences cancel, and which is faster depends on whether the satellite is above or below this altitude. That's why I suggested you consider a pure SR problem.

 

[Ibix]

In #42, @binis said: Conclusively,time is running faster on the moon than the Earth but slower on the satellites.

No. Time does not run fast or slow. There are just different paths through spacetime with different elapsed times along them. Thus if you find a way to compare tick rates of clocks (for example, standing beside one and watching the other through a telescope) you may find that they tick at different rates. But clocks in the same location may tick at different rates compared to each other, and different observers and different methods of observing and interpreting results may lead to different conclusions about which clock is ticking faster.

That said, there are symmetries in Schwarzschild spacetime that allow you to define a global time coordinate with a sensible relation to physically measurable times. In this spacetime, clocks at rest in this coordinate system will all regard other such "at rest" clocks as ticking slower than them if the other clock is below them and faster if they are above them. This is sometimes stated in popsci sources as "time runs slower near masses", but this is wildly unhelpful because it ignores all the caveats in my previous paragraph.

Just think of clocks as measuring the interval along a worldline, like an odometer measures distance along a spatial path. Use flocks of clocks to define a global notion of time (different flocks produce different such notions, of course). Compare clock rates. Don't think in terms of time running fast or slow.[/i][/i]

 

[Dale]

Moon is the natural satellite of the earth. In the same way an observer on an artificial satellite is watching his watch running faster than a clock on the earth.True?

Conclusively,time is running faster on the moon than the Earth but slower on some satellites. What about the ISS?

See this figure

At an altitude of about 3000 km is where the switch occurs. Above that altitude the time on a satellite runs faster than the time on Earth (gravitational time dilation is more important than kinematic time dilation). Below that altitude time on a satellite runs slower (kinematic time dilation is more important than gravitational). The ISS is below 3000 km and the moon is above 3000 km

 

[binis]

So the angular velocity vector between an Earth-fixed basis and a moon-fixed basis is $\boldsymbol{\omega}_1$

what is the magnitube of this vector?

 

[etotheipi]

I suppose you're asking about classical mechanics here? The two bases are related by a unique, orthogonal, time-dependent matrix $(\mathcal{R}_{ij})$ such that $\tilde{\mathbf{e}}_i = \mathcal{R}_{ij} \mathbf{e}_j$. Then take the time-derivative with respect to the frame $Oxyz$,$$\frac{\mathrm{d}\tilde{\mathbf{e}}_i}{\mathrm{d}t}\big|_{Oxyz} = \frac{\mathrm{d} \mathcal{R}_{ij}}{\mathrm{d}t} \mathbf{e}_j = \frac{\mathrm{d} \mathcal{R}_{ij}}{\mathrm{d}t} \mathcal{R}^{-1}_{jk} \tilde{\mathbf{e}}_k := \omega_{ik} \tilde{\mathbf{e}}_k$$and because $\mathcal{R}_{ij}$ is orthogonal we have $\mathcal{R}^{-1}_{jk} = \mathcal{R}_{kj}$ thus the tensor $\omega_{ik} := \dot{\mathcal{R}}_{ij} \mathcal{R}_{kj}$. You can easily check that $\omega_{ij}$ is antisymmetric (do it!), thus is reducible to a single index object via $\omega_i := \frac{1}{2} \epsilon_{ijk} \omega_{jk}$ which define the components of a vector $\boldsymbol{\omega}$. You can just take the Euclidean norm $||\boldsymbol{\omega}|| = \sqrt{\sum \omega_i^2}$ for the magnitude.

 

[Dale]

I suppose you're asking about classical mechanics here?

He is talking about general relativity and specifically two spatially separated observers in a curved spacetime.

 

[A.T.]

The ISS is below 3000 km and the moon is above 3000 km

One should add, that the Moon has non-negligible mass and thus it's own gravitational time dilation. So the time dilation graph for near Earth orbits of small objects doesn't strictly apply to the Moon.

 

[Dale]

One should add, that the Moon has non-negligible mass and thus it's own gravitational time dilation. So the time dilation graph for near Earth orbits of small objects doesn't strictly apply to the Moon.

Yes, that is a good point. I was neglecting the lunar gravitational time dilation (pure laziness, no justification).