# Does tidal lock = no rotation

• tony1grendel

#### tony1grendel

So my co-workers and I have had a dispute after hearing a podcast on NPR about Earth's rotation.

One question was about what would happen if one side of the Earth always faced the sun (tidal locked) and another question was about what would happen if the Earth stopped rotating.

From what I've gathered tidal lock does not mean no rotation.

But if the Earth stopped rotating would a probable outcome be a tidal lock?

Is it possible for a body in orbit to never rotate?

Lastly, my co-worker is convinced the moon on the left, which always has the same side pointed to the center, is not rotating.

To me it seems the one on the right is not rotating. But ,like I asked before, is the right side even possible?

Tidal lock means the rotation and orbit are synchronous in a 1:1 relationship. That is, it takes the same amount of time to complete one rotation and one orbit.
The moon on the left is tidally-locked and rotating once per orbit, the moon on the right is not tidally-locked and is not rotating.

The situation on the right is possible, but is not stable. All systems of pairs of orbiting bodies will tend to synchronise in a full tidal lock over time, or crash into one another if certain initial conditions are met.

The situation on the right is possible, but is not stable. All systems of pairs of orbiting bodies will tend to synchronise in a full tidal lock over time, or crash into one another if certain initial conditions are met.
Venus is not tidally locked to Sun, and is very slowly rotating backwards.
Under the tidal influence of Sun, is the rotation of Venus slowing, and will the rotation of Venus eventually go to full stop before a slow forward rotation starts?

Venus is not tidally locked to Sun, and is very slowly rotating backwards.
Under the tidal influence of Sun, is the rotation of Venus slowing, and will the rotation of Venus eventually go to full stop before a slow forward rotation starts?
Yes. I don't see that as being a different set-up. The rotation is in the opposite direction, and so is the displacement of the tidal bulge and the resultant torques.

Lastly, my co-worker is convinced the moon on the left, which always has the same side pointed to the center, is not rotating.

To me it seems the one on the right is not rotating. But ,like I asked before, is the right side even possible?

Your co-worker is wrong. The one of the left is absolutely rotating and is doing so at a rate of 1 rotation per orbit.

I suppose that a planet can be tidally locked yet still rotating - if the pole was pointed at the sun.

I suppose that a planet can be tidally locked yet still rotating - if the pole was pointed at the sun.

It doesn't need to be pointed at the Sun. A tidally locked body is still rotating at a rate of once per orbit.

Dotini
It doesn't need to be pointed at the Sun. A tidally locked body is still rotating at a rate of once per orbit.

Question: If a small planet was tidally locked - in extremely close orbit to its star - with the north pole pointed at the star, would a loose object at the south pole feel heavier than another similar object at the north pole?

Tidal locking doesn't work that way. The spin north pole would have the same direction as the orbit north pole.

But from what you asked about, it's the substellar point and its antipode. To lowest order, they have the same size of acceleration of gravity, though in opposite directions.

That's a counterintuitive feature of tides. The Moon and the Sun each make 2 evenly-spaced tides a day, and not just 1 tide, as one would naively expect. That's because the ocean away from the Earth is left behind as the Moon and the Sun pull on the Earth. Remember that the acceleration of gravity is independent of mass and composition. That's the Equivalence Principle, and there are some strong upper limits to violations of it.

Expanding to the next order in (planet radius) / (distance from its star), one does find an effect, but its size is about (tide size) * (that factor), and that is teeny teeny teeny tiny except for a star in a close binary-star system or something similar for planets.

Question: If a small planet was tidally locked - in extremely close orbit to its star - with the north pole pointed at the star, would a loose object at the south pole feel heavier than another similar object at the north pole?

I think you have to determine the difference, between acceleration due to gravity from the planet and star...

and the acceleration required to maintain circular motion...

the difference would be supplied by the surface of the planet pushing up against the bottom of the object, which would be experienced as weight

D = distance from CMstar to CMplanet
R = radius of planet
M = mass of planet
M = mass of star

at sub stellar point, gravities pull in opposite directions, but the required centripetal force is lower

F/m = -GM/(D-R)2 + GM/R2

a = -( D - R ) w2

at ant stellar point, gravities pull in concert, but more centripetal force is required

F/m = -GM/(D+R)2 - GM/R2

a = -( D + R ) w2

note that W is the orbital frequency

D w2 = GM/D2

to lowest order in

e = R/D​

and defining

g = GM/R2
g = GM/D2

SS point

F/m = -g ( 1 + 2e ) + g
a = -g ( 1 - e )​

AS point

F/m = -g ( 1 - 2e ) - g
a = -g ( 1 + e )​

SS point

F/m = -g ( 1 + 2e ) + g - W
=
a = -g ( 1 - e )​

AS point

F/m = -g ( 1 - 2e ) - g + W'
=
a = -g ( 1 + e )​

W = g - 3 e g

W' = g - 3 e g

so to lowest order, objects would have the same weight at both poles

to second order,

SS point

F/m = -g ( 1 + 2e - e2 ) + g - W
=
a = -g ( 1 - e )​

AS point

F/m = -g ( 1 - 2e - e2) - g + W'
=
a = -g ( 1 + e )​
W = g - 3 e g + e2 g

W' = g - 3 e g - e2 g

so to next lowest order, objects would have higher weight at the SS point

In explaining the above graphic, you say the moon on the left has a typical 1:1 sidereal spin rate:

"Tidal lock means the rotation and orbit are synchronous in a 1:1 relationship. That is, it takes the same amount of time to complete one rotation and one orbit. The moon on the left is tidally-locked and rotating once per orbit, the moon on the right is not tidally-locked and is not rotating."

You also say that the moon on the right is possible, but not stable. By "not stable," I assume you mean that the image on the right still has one remaining clockwise (CW) polar axial rotation (since that moon's surface is still moving in relation to its primary's surface), thus it appears that the moon on the right has a 0:1 sidereal spin rate:

"The situation on the right is possible, but is not stable. All systems of pairs of orbiting bodies will tend to synchronise in a full tidal lock over time, or crash into one another if certain initial conditions are met."

Though not common, an astronomical body with a clockwise (CW) rotation can also be spun down to 1:1, as (CW) rotating Venus proves, so is Venus the situation you had in mind by saying the graphic on the right is possible, but not stable and wouldn't last for very long?

If so, I agree that a 0:1 sidereal spin rate wouldn't be stable for very long, since once Venus spins down to that 0:1 rate (1 CW polar axial rotation per each CCW orbit), Venus at that point will still have one last CW polar axial rotation left to lose before the planet is fully despun at 1:1, so Venus would relatively quickly pass thru that 0:1 sidereal spin rate.

Correct?

That's almost what I meant.

The lack of stability simply means that in the presence of tidal forces, a non-rotating (in an inertial frame) moon will inevitably end up as rotating. I.e., it will have some angular acceleration around its rotational axis. Conversely, a full (1:1) tidal lock means no angular acceleration of the moon, so the rotation remains constant over time.

A 0:1 rotation is then not merely 'not stable for very long', but not stable at all - at no point in the evolution of Venus' rotation prior to reaching a 1:1 lock (shouldn't happen before the Sun engulfs it) will the angular acceleration equal zero.
This is assuming a simplified, circular orbit-case with no outside influences.

I assume you mean that the image on the right still has one remaining clockwise (CW) polar axial rotation (since that moon's surface is still moving in relation to its primary's surface), thus it appears that the moon on the right has a 0:1 sidereal spin rate:
once Venus spins down to that 0:1 rate (1 CW polar axial rotation per each CCW orbit)
These two bits are confusing, though. I think you're using a rotating frame here, as otherwise the underlined parts make no sense. This is not what is normally used to describe synchronous motion. Whenever you see a rotation : orbit ratio, it means rotation in an inertial frame of reference.

Welcome to PF, by the way :)

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That's almost what I meant.

The lack of stability simply means that in the presence of tidal forces, a non-rotating (in an inertial frame) moon will inevitably end up as rotating. I.e., it will have some angular acceleration around its rotational axis. Conversely, a full (1:1) tidal lock means no angular acceleration of the moon, so the rotation remains constant over time.

I think I understand your viewpoint. What's confusing is that the sidereal versus the center-point perspectives always differ by one apparent rotation.

E.g., from the sidereal perspective Earth has 366.25 CCW rotations, each 23 hours, 56 minutes and a few seconds long. But from the center-point perspective, Earth has one less rotation, 365.25 CCW rotations, each 24 hours long, which is one less apparent rotation than from the sidereal perspective.

From the sidereal perspective, Venus currently has .92 CW rotations, but since the sidereal and the center-point perspectives always differ by one apparent rotation, Venus actually has nearly two full solar days per orbit (1.92 CW polar axial rotations per orbit).

Today, since Venus has slightly less than one apparent sidereal rotation (only .92 sidereal rotations), it appears that Venus has recently passed thru a 1:1 sidereal spin rate as that planet is in the process of being despun – how else could Venus currently have a .92:1 sidereal spin rate if Venus didn't recently pass thru a 1:1 rate?

If so, then Venus is now slowing to that mythical 0:1 sidereal spin rate, before finally stopping at a final (and 2nd) 1:1 sidereal spin rate as it becomes tidally locked.

For example:
If our moon spun down from the CWW direction, then from our center-point perspective, our moon spun down like this (starting randomly with 4 polar axial rotations, but there may have been more):

CCW: 4 ~ 3 ~ 2 ~ 1 ~ 0

If our moon spun down from the CW direction (which it may have done), then from our center-point perspective, our moon spun down like this (the same):

CW: 4 ~ 3 ~ 2 ~ 1 ~ 0

No matter which direction our moon spun down from, from Earth's center-point perspective, our moon spun down to ZERO polar axial rotations.

From from the sidereal (outside) perspective, our moon *APPEARS* to have spun down like this from the CCW direction:

CCW: 4 ~ 3 ~ 2 ~ 1

Using the ROTATION:ORBIT convention, for a CCW spinning moon, you would write the above like this, with our moon stopping at one (1) *apparent* rotation per each 360˚ orbit of the moon:

CCW: 4:1 ~ 3:1 ~ 2:1 ~ 1:1

From the center-point perspective (i.e., from the sun), Venus would appear to spin down similarly to our moon if our moon had spun down from the CW direction:

CW: 4 ~ 3 ~ 2 ~ 1 ~ 0

From the above center-point perspective, currently at 1.92 CW polar axial rotations, Venus recently had 2 full CW rotations, and is currently heading towards 1 CW rotation, before stopping at zero, thereafter with one face pointing inwards towards the sun in a tidal lock.

Using the ROTATION:ORBIT convention again, you would write Venus' above spin-down like this, with Venus stopping at one (1) sidereal rotation, per each 360˚ CCW orbit around the sun:

CW: 4:1 ~ 3:1 ~ 2:1 ~ 1:1 ~ 0:1 ~ 1:1

Remember, Venus currently has nearly two full CW polar axial rotations left (1.92), so Venus must have already passed thru one 1:1 sidereal spin rate. How else could Venus bleed off nearly two more CW rotations and get to that final 1:1 rate, if Venus doesn't need to pass thru a 0:1 sidereal spin rate, as depicted on the right side in the above graphic?

A 0:1 rotation is then not merely 'not stable for very long', but not stable at all - at no point in the evolution of Venus' rotation prior to reaching a 1:1 lock (shouldn't happen before the Sun engulfs it) will the angular acceleration equal zero.
This is assuming a simplified, circular orbit-case with no outside influences.

In my opinion, as Venus passes thru that transitory 0:1 sidereal spin rate, Venus will still have one (1) full CW polar axial rotation left to lose, so as Venus passes thru that transitory 0:1 rate, Venus would still have angular momentum around its polar axis at that point.

I.e., the 0:1 spin rate is an optical illusion. Look at the moon on the right in the upper image, and while many people see a zero-rotating moon, I see a moon with one (1) CW polar axial rotation as it orbits in the opposite CCW direction, and two 360˚ spins in the opposite direction, around two different axes (one axis its barycenter, and the 2nd axis around its polar axis), these two 360˚ spins in opposite directions merely cancel each other out ...MOMENTARILY, since a 0:1 sidereal spin rate is transitory.

As for Venus being engulfed by our sun going red giant before Venus becomes tidally locked, Venus only has 1.92 polar axial rotations left to lose, and Magellan's cloud piercing radar found that Venus had lost 6 minutes of rotation during the 16 years Magallan observed Venus back in the 1990s:

https://tallbloke.wordpress.com/201...-to-discover-that-venus-spin-is-slowing-down/

Welcome to PF, by the way :)

Thank you.
:)

I think I understand your viewpoint. What's confusing is that the sidereal versus the center-point perspectives always differ by one apparent rotation.

E.g., from the sidereal perspective Earth has 366.25 CCW rotations, each 23 hours, 56 minutes and a few seconds long. But from the center-point perspective, Earth has one less rotation, 365.25 CCW rotations, each 24 hours long, which is one less apparent rotation than from the sidereal perspective.
This one rotation of difference is exactly the same difference between the 0:1 and 1:1 rotation : orbit relations.
It would probably be less confusing if you dropped this switching of reference frames. Stick to the inertial frame (i.e. the one used for the sidereal rather than the tropical year).

From the sidereal perspective, Venus currently has .92 CW rotations, but since the sidereal and the center-point perspectives always differ by one apparent rotation, Venus actually has nearly two full solar days per orbit (1.92 CW polar axial rotations per orbit).

Today, since Venus has slightly less than one apparent sidereal rotation (only .92 sidereal rotations), it appears that Venus has recently passed thru a 1:1 sidereal spin rate as that planet is in the process of being despun – how else could Venus currently have a .92:1 sidereal spin rate if Venus didn't recently pass thru a 1:1 rate?
If Venus has 0.92 clockwise rotations per year, then it has -0.92 counter-clockwise rotations per year. As a convention, the counter-clockwise direction is the direction used for positive values in orbital and rotational periods. So, current rotation : orbit relation for Venus is -0.92:1, and it had passed -1:1 rather than 1:1.

edit: here's me not checking the facts. Venus' rotational period is longer than the orbital period, so it's definitely not -0.92:1. but -1.08:1.

Using the ROTATION:ORBIT convention again, you would write Venus' above spin-down like this, with Venus stopping at one (1) sidereal rotation, per each 360˚ CCW orbit around the sun:

CW: 4:1 ~ 3:1 ~ 2:1 ~ 1:1 ~ 0:1 ~ 1:1
So this is wrong for the aforementioned reason.

In my opinion, as Venus passes thru that transitory 0:1 sidereal spin rate, Venus will still have one (1) full CW polar axial rotation left to lose, so as Venus passes thru that transitory 0:1 rate, Venus would still have angular momentum around its polar axis at that point.
No. When its rotation in an inertial frame of reference is 0, it necessarily means that its rotational angular momentum is also 0. As tidal interactions continue to accelerate the planet, it gains rotational angular momentum until it settles in a stable 1:1 synchronous rotation.

I.e., the 0:1 spin rate is an optical illusion. Look at the moon on the right in the upper image, and while many people see a zero-rotating moon, I see a moon with one (1) CW polar axial rotation as it orbits in the opposite CCW direction, and two 360˚ spins in the opposite direction, around two different axes (one axis its barycenter, and the 2nd axis around its polar axis), these two 360˚ spins in opposite directions merely cancel each other out ...MOMENTARILY, since a 0:1 sidereal spin rate is transitory.
The moon on the right rotates only in a non-inertial, rotating reference frame.
There's nothing wrong in using a non-inertial frame to describe motion, as long as you clearly state if you're using it and are consistent about it. The default for most applications in astrodynamics is the inertial frame.
It's strongly advisable not to switch frames as you analyse motion to avoid confusion and wrong results - for example, while the moon on the right rotates in any rotating reference frame, and in particular rotates at 1 rotation per orbit in a frame rotating with the angular velocity equal to orbital angular velocity of the moon, in the same frame the moon is not orbiting - it is hovering in one place above the observer, supported by a pseudo-force, and has no orbital angular momentum as a result.

I think the bottom line here is: stick to the inertial frames, or at least don't switch frames, and watch the signs. Otherwise you seem to get it right.

By the by, I wasn't aware of the discrepancy in Venus' rotation period. That's interesting. I wouldn't jump to conclusions, though.

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This one rotation of difference is exactly the same difference between the 0:1 and 1:1 rotation : orbit relations.
It would probably be less confusing if you dropped this switching of reference frames. Stick to the inertial frame (i.e. the one used for the sidereal rather than the tropical year).

In order to fully understand the sidereal perspective, then it's imperative to compare the sidereal (OUTSIDE) perspective with the center-point perspective (such as from Earth or from the sun, as the case may be).

Let's review your claim that there's only a difference of one (1) rotation between -0:1 and +1:1 (NOTE - as you suggested, I'm now adding a neg "-" sign for retrograde CW orbits, and a "+" sign for regular CCW prograde orbits).

Clearly, the sidereal (view from the stars) and center-point perspectives (view from Earth, or other primary), they always differ by ONE (1) apparent rotation:

Synodic Motion

The word synodic derives from the Greek word for meeting or assembly. It is motion relative to a conjunction or alignment of sorts. A synodic or solar day is the time it takes the sun to successively pass the meridian (astronomical noon). A mean solar day is 24 hours (the “mean” is there to average over the effect of the analemma). The Earth has to rotate more than 360° for the sun to come back to “noon”.

A synodic year is the time it takes for a planet-sun alignment to reoccur. For the case of the sun, it is the time it takes the sun to come to the same place on the ecliptic (equinox to equinox) and is called a Tropical Year. A tropical year is 365.242 mean solar days (366.242 sidereal days. It is just over 20 minutes shorter than a sidereal year (again, the effect of precession).

http://astro.unl.edu/naap/motion3/sidereal_synodic.html

If Earth has 366.242 sidereal days, then we need to subtract ONE (1) to determine Earth's actual CCW polar axial rotations, as explained here:

Sidereal Time

Another way to see this difference is to notice that, relative to the stars, the Sun appears to move around the Earth once per year. Therefore, there is one fewer solar day per year than there are sidereal days. This makes a sidereal day approximately 365.24 ⁄ 366.24 times the length of the 24-hour solar day, giving approximately 23 hours, 56 minutes, 4.1 seconds (86,164.1 seconds).

Venus rotates retrograde with a sidereal day lasting about 243.0 earth-days, or about 1.08 times its orbital period of 224.7 earth-days; hence by the retrograde formula its solar day is about 116.8 earth-days, and it has about 1.9 solar days per orbital period.

https://en.wikipedia.org/wiki/Sidereal_time

The reason that the sidereal perspective always has one (1) more apparent 360° rotation per CCW orbit, is because typically from the sidereal perspective the orbiting body's polar axial rotations (if any) are aggregated in with the orbiting body's 360° orbit.

Conversely, for astronomical bodies with a retrograde CW rotation, such as Venus, you instead need to ADD one (1) to their sidereal rotation rate to determine the actual number of their polar axial rotations:

"All the planets of the Solar System orbit the Sun in a counter-clockwise direction as viewed from above the Sun’s north pole. Most planets also rotate on their axis in a counter-clockwise direction, but Venus rotates clockwise (called “retrograde” rotation) once every 243 Earth days—the slowest rotation period of any planet. To an observer on the surface of Venus, the Sun would rise in the west and set in the east.

"A Venusian sidereal day thus lasts longer than a Venusian year (243 versus 224.7 Earth days). Because of the retrograde rotation, the length of a solar day on Venus is significantly shorter than the sidereal day, at 116.75 Earth days (making the Venusian solar day shorter than Mercury’s 176 Earth days); one Venusian year is about 1.92 Venusian (solar) days long."

http://blog.world-mysteries.com/science/puzzles-of-the-moon/

Venus has 1.92 actual CW polar axial rotations per each orbit, which is slightly less than two full solar days. However, Venus only has -0.92 sidereal rotations (224.7 divided by 243 = 0.92). The reason why you need to ADD one (1) for CW rotating bodies, is the same reason for subtracting one (1) for CCW bodies, because the sidereal perspective likewise aggregates together a retrograde body's CW polar axial rotations with the orbiting body's single CCW 360° orbit.

An astronomical body with a prograde 360° CCW orbit with a retrograde 360° CW rotation, these two 360° spins in opposite directions (around two different axes) merely cancel each other out, which is why you need to add one (1) to Venus' -0.92 sidereal rotations to determine Venus' actual 1.92 CW polar axial rotations – it's as simple as that.

Can we now agree that the sidereal (outside) perspective differs from the center-point perspective, by one (1) APPARENT rotation?

As the below orrery spins, if you were to view it from above, the moon in this orrery would behave exactly like the moon does (on the left) in the above graphic that the poster had asked about:

Let's compare the original poster's above graphic with the orrery:

In normal operation the orrery's faux-moon would match the moon on the left. However, if you reached out and forced the orrery's faux-moon to point at one wall as you cranked that orrery around, then the faux-moon would necessarily be forced to spin CW around its supporting metal spindle (360° one time), as the moon on the right is doing. If you reached out and spun the faux-moon CW 0.92 times per revolution, then that is what Venus is now doing, -0.92 from the sidereal perspective, but 1.92 360° CW rotations from the sun's center-point perspective.

NOTE – When viewing either the orrery or any graphic such as the graphic above, you are necessarily placed in the outside sidereal position, so you need to either add one (1) or subtract one (1) to determine the moon's actual polar axial rotations in these models.

The moon's surface on the right is clearly moving in relation to the planet in the center (and would thus still be experiencing tidal braking forces). Imagine yourself standing on the planet on the right, and clearly you would observe the moon on the right making one 360° CW polar axial rotation per each CCW orbit, so while from the sidereal perspective the moon appears to be pointing in one direction and not rotating, ADD one (1) to that sidereal zero rotation rate, and the moon is actually making one (1) CW polar axial rotation per orbit.

That many people see a zero-rotating moon is just an optical illusion from the sidereal perspective since that moon is actually making TWO 360° spins in opposite directions.

When you're dealing with bodies in motion there are many possible optical illusions arising from the nature of the orbits. E.g., from Earth's perspective, there are times when Mars appears to be traveling backwards compared to the "fixed stars”:
Mars Retrograde Happens Every Two Years:

http://mars.nasa.gov/allaboutmars/nightsky/retrograde/

If Venus has 0.92 clockwise rotations per year, then it has -0.92 counter-clockwise rotations per year. As a convention, the counter-clockwise direction is the direction used for positive values in orbital and rotational periods. So, current rotation : orbit relation for Venus is -0.92:1, and it had passed -1:1 rather than 1:1.

edit: here's me not checking the facts. Venus' rotational period is longer than the orbital period, so it's definitely not -0.92:1. but -1.08:1.

Your 1.08 rotation claim is wrong since 1.08 is what you get by dividing Venus' 225 day orbit into Venus' 'APPARENT' sidereal rotation rate of 243 days:

Venus sluggishly rotates on its axis once every 243 Earth days, while it orbits the Sun every 225 days - its day is longer than its year! Besides that, Venus rotates retrograde, or "backwards," spinning in the opposite direction of its orbit around the Sun. From its surface, the Sun would seem to rise in the west and set in the east.

https://solarsystem.nasa.gov/planets/profile.cfm?Object=Venus

243 days divided by 225 days = 1.08
However, Venus' 225 day orbit divided by its 243 day apparent sidereal rotation rate, instead gives the correct answer of 0.92 sidereal rotations per orbit.

1.08 is instead how many times Venus needs to rotate around its polar axis per each sidereal day in relation to a Venusian year (i.e., Venus' sidereal day is a tad longer than a Venusian year):

Sidereal Time

Another way to see this difference is to notice that, relative to the stars, the Sun appears to move around the Earth once per year. Therefore, there is one fewer solar day per year than there are sidereal days. This makes a sidereal day approximately 365.24 ⁄ 366.24 times the length of the 24-hour solar day, giving approximately 23 hours, 56 minutes, 4.1 seconds (86,164.1 seconds).

Venus rotates retrograde with a sidereal day lasting about 243.0 earth-days, or about 1.08 times its orbital period of 224.7 earth-days; hence by the retrograde formula its solar day is about 116.8 earth-days, and it has about 1.9 solar days per orbital period.

https://en.wikipedia.org/wiki/Sidereal_time

Clearly, Venus currently has 1.92 ACTUAL 360° CW polar axial rotations per complete orbit, (Venus has nearly two (2) solar days per orbit), but if you prefer, we can use sidereal days, which would be one (1) less, or -0.92 sidereal days per orbit.
We still have the same dilemma, if Venus currently has -0.92 CW sidereal rotations per orbit (-0.92:1), then Venus must have recently passed thru -1:1 to now be at -0.92:1.

Since Venus also clearly has nearly two (2) actual CW polar axial rotations per orbit (and nearly two solar days remaining), that means Venus has also passed thru a 2 full solar day rotation to wind down to only 1.92 solar days that it has today, which means that Venus has nearly two full polar axial rotations left to lose before being fully despun at your sidereal +1:1 tidally locked rate.

I already explained, from the SIDEREAL perspective, how it would be possible for Venus to lose nearly two (2) actual CW polar axial rotations and go from its current -0.92:1 sidereal rotation rate to a tidally locked rotation rate of +1:1:

CW: -4:1 ~ -3:1 ~ -2:1 ~ -1:1 ~ -0:1 ~ +1:1

If you can explain how Venus can bleed off nearly two full CW polar axial rotations, starting at its current -0.92:1 rate without passing thru that zero sidereal rotation rate, then I'd like to see that done?

Also, if you believe that a tidally locked body has one (1) CCW rotation per orbit (once despun at your +1:1), then you may want to explain how Venus will not only lose nearly two CW polar axial rotations starting from -0.92:1, but then once Venus is fully despun at +1:1, please do explain how Venus will suddenly start rotating one (1) full time in the opposite CCW direction after Venus becomes tidally locked at +1:1?

The sidereal perspective has its practical uses, but as Venus proves, the sidereal perspective has its quirks, too.

No. When its rotation in an inertial frame of reference is 0, it necessarily means that its rotational angular momentum is also 0. As tidal interactions continue to accelerate the planet, it gains rotational angular momentum until it settles in a stable 1:1 synchronous rotation.

Are you claiming tidal interactions can accelerate a planet's polar axial rotation? If you really do mean its rate of rotation, then do you have a citation for that? Yes, tidal braking can increase an orbiting body's orbital speed, but we're talking about decelerating Venus' rotational speed from -0.92:1 down to +1:1. Increased orbital speed is just where the lost rotational energy is transferred to (along with some loss to heating), but that isn't what you appear to be claiming?

As far as I know, whether a planet or moon has either a prograde or retrograde axial spin, tidal braking forces would slow down either type of rotating body. Of course, with a retrograde CW rotating (and CCW orbiting) body, the primary's gravity would be pulling the planet's tidal bulge in the opposite direction of its orbital direction, which may decrease its orbital speed a tad, but not sure about that and I'm just throwing out the possibility? Even so, a planet the size of Venus likely wouldn't have its orbital speed affected too much as it loses only two more CW polar axial rotations.

The moon on the right rotates only in a non-inertial, rotating reference frame.

There's nothing wrong in using a non-inertial frame to describe motion, as long as you clearly state if you're using it and are consistent about it. The default for most applications in astrodynamics is the inertial frame.

Let's try to keep this simple and explain it in a way a layman can understand. By a "non-inertial frame" you mean from Earth's perspective.

It's strongly advisable not to switch frames as you analyse motion to avoid confusion and wrong results - for example, while the moon on the right rotates in any rotating reference frame, and in particular rotates at 1 rotation per orbit in a frame rotating with the angular velocity equal to orbital angular velocity of the moon, in the same frame the moon is not orbiting - it is hovering in one place above the observer, supported by a pseudo-force, and has no orbital angular momentum as a result.

The seemingly zero-rotating moon on the right in the above graphic would NOT be hovering in one place above an observer standing on the planet in the center, since such an observer would clearly see that moon both orbiting CCW as well as rotating around its polar axis CW, ONCE per each orbit. The only thing not stable about a seemingly zero-rotating moon (from the sidereal perspective), is that such a moon would still have one last CW rotation left to lose, so it would merely pass thru -0:1, then go to -0.01:1, then -0.02:1, then -0.03:1, then -0.04:1, etc, until stopping at a final +1:1 sidereal rate. As that moon passes thru that -0:1 rate, the background stars would slowly begin to move again in relation to its surface, and gradually increase their movement until the moon reached its final +1:1 rate.

It's only from the sidereal perspective that such a body would seemingly pass thru that transitional -0:1 rate as it was being despun. On planet Venus, its days will just get gradually longer and longer (going from 1.92 days per orbit to zero days), until Venus eventually runs out of steam and locks, at which point Venus will only show one face inwards towards the sun, just as our moon now only shows one face inwards towards Earth.

Unlike our own moon, Venus' backside will be in perpetual darkness one it locks to the sun.

Further, if you stood on that seemingly zero-rotating moon's surface on the right, then the stars wouldn't appear to move much (the stars would move a tad due to librations), but the Earth would rise in the West and set in the East each orbit. If you instead stood on the other moon's surface (the left moon in the graphic), as astronauts have already done, then Earth would hang nearly motionless in the sky, except for minor movements again due to librations, caused by the moon's elliptical orbit, as well as by the tilt of the moon's orbit in relation to Earth's orbital plane.

I think the bottom line here is: stick to the inertial frames, or at least don't switch frames, and watch the signs. Otherwise you seem to get it right..

I feel I have it exactly right, and it's also necessary to switch between reference frames to help explain which frame better represents the reality of the movements that we're discussing here, as I hope I'm now doing with you.
:)

By the by, I wasn't aware of the discrepancy in Venus' rotation period. That's interesting. I wouldn't jump to conclusions, though.

Despite being our closest planetary neighbor, since Venus has been shrouded in clouds, it was only relatively recently by using radar that Venus's axial rotation could even be studied, and there doesn't appear to be much published on the planet's unique rotation. In my opinion, to understand what our own moon is doing, then you need to understand what Venus is doing.

For example, as you try to harmonize how Venus can lose nearly two full polar axial rotations, and be despun down from its current -0.92:1 to a tidally locked +1:1, then it may help you to understand what your +1:1 final sidereal rate actually means. The tip to that I already posted above:

If Earth has 366.242 sidereal days, then we need to subtract ONE (1) to determine Earth's actual CCW polar axial rotations.

Likewise, if our moon has a sidereal spin rate of one (1), then subtract one (1) from our moon's apparent sidereal spins, just as you need to do for Earth's 366 sidereal spin total to find the actual polar axial rotations of both:

MOON: 1 – 1 = 0

EARTH: 366.25 – 1 = 365.25

If you consider that our fully despun +1:1 moon now has zero polar axial rotations, then everything makes sense as Venus must lose nearly two rotations to go from -0.92 down to +1:1 sidereal rotations:

CW: -4:1 ~ -3:1 ~ -2:1 ~ -1:1 ~ -0:1 ~ +1:1

I realize what I'm saying here may sound like heresy to you at first blush, but synchronous theory has been disputed for hundreds of years by many notable scientists and astronomers.

If you need further proof, then please try and explain our moon's longitudinal librations within your synchronous theory framework?

Most websites claim our moon has a steady CCW rotation rate that gets out of sync as our moon's orbital velocity varies as it passes thru its apogee and perigee. However, our moon's two maximum longitudinal librations actually occur midway between apogee and perigee!

That our moon's maximum longitudinal librations occur midway between apogee and perigee can be easily explained if you consider the possibility that our moon is today fully despun –– simply, midway between apogee and perigee, our fully despun moon is merely facing the empty focus of its elliptical orbit.

The sidereal perspective has its practical uses, but it's also not some 'God's Eye' view of reality!

Think about it.

Ken

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Back in the late 1880s synchronous rotation was thoroughly debated, and the majority of English astronomers back then eventually voted against the concept.

There are many arguments PRO & CON, but I like this one the best, especially his #5 reason below. Even though this letter was written in 1864, these were obviously very intelligent men. Sir John Herschel mentioned in this letter was one of the main proponents of synchronous rotation theory back then. There were actually two Sir Herschel astronomers back then, a father and a son, and they were both lunatic lunar rotators:

---snip---
THE MOON CONTROVERSY,

TO THE EDITOR OF THE ASTRONOMICAL REGISTER.

"Sir,—On reviewing the whole controversy respecting the moon's rotation, or non-rotation, on her own axis, it will be observed that mathematicians are ranged on one aide, and practical mechanics and engineers on the other, and that theory is at variance with practice. "Mathematicians have adopted a quasi definition of rotation, which leads to innumerable absurdities, e.g. that every atom of a spinning- top rotates on its own axis, that the ball on St. Paul's cathedral rotates on its own axis once in 23h, 56m,. etc, etc..

"They use the terms of "rotation" and "rotation on its own axis" as though they were synonymous in meaning, quite overlooking the distinction between axial and radial rotation. It would surely be quite possible for them to be more concise in the use of language without calling in question any single mathematical deduction or truth of any sort or kind. *It would be better to understand the term "rotation" in the sense used by mechanics, engineers, and men of plain common sense.*

"It is really astonishing how one writer after another, like Mr. Bird in the last No. of the Register, can persuade themselves that "turning round" necessarily involves "rotation," either axial or radial.

"Before closing the correspondence on this most interesting subject it may be as well to notice Sir John Herschel's illustration of axial rotation. He "plants a staff in the ground, and, grasping it in both hands, walks round it, keeping as close to it as possible, with his face always turned towards it, "when the unmistakable sensation of giddiness effectually satisfies him of the fact that he has rotated on his own axis!"

"Mr. Bertram Mifford, of Cheltenham, in a letter addressed to the Practical Mechanics' Journal, Dec. 1, 1859, thus effectually disposed of this most transparent fallacy by the following arguments;—

"1. Because giddiness may be produced in various other ways, by looking over a precipice, etc.

"2. Because (mechanically speaking) no man can walk round his own axis in any way, much less turn upon it when holding on to any fixed object, which must be external to his body. He may walk his axis round any object he pleases, but he cannot in any sense walk round his own axis.

"3. Because a man holding by a stick and revolving round it turns on the axis of the stick and not on his own axis, the axis of the stick in this case becoming the common axis of both man and stick.

"4. The alteration of the relative position of the objects round him is no proof that the man has turned on his own axis, he has merely dragged his axis after him, and has not turned upon it.

"5. "Keeping as close to it as possible" is certainly the nearest approach to turning on his own axis, but until he is impaled on the stick and turned on it he does not revolve (mechanically speaking) on his own axis.​
"I have the honour to be, very faithfully yours,

"Buckland: Nov. 12, 1864.

"ACADEMICUS."

John G-Wolbach Library, Harvard-Smithsonian Center for Aslrophysics • Provided by the NASA Astrophysics— The above letter (starting on page 19) and other old pro & con letters on the issue can be found here: http://articles.adsabs.harvard.edu//full/seri/AReg./0003//0000034.000.html

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Lastly, my co-worker is convinced the moon on the left, which always has the same side pointed to the center, is not rotating.

[PLAIN]http://upload.wikimedia.org/wikipedia/commons/5/56/Tidal_locking_of_the_Moon_with_the_Earth.gif[/QUOTE] [Broken]
The very short answer is: Your coworker is wrong.

Even from the point of view on Earth, the Moon is still observed to be rotating.

As the Moon moves across the sky, it must turn, otherwise you'd see one side, then the front, then the other side (like a car passing you, you see the front, then the side, then the back). The only way you can see the same face of the Moon from the Earth, is if the Moon is rotating to match your changing angle of observation.

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The very short answer is: Your coworker is wrong.

Unless, the co-worker is CORRECT and you're WRONG?

Even from the point of view on Earth, the Moon is still observed to be rotating.

The last I looked at our moon, it certainly didn't appear to be rotating when viewed from my backyard?

Where are you observing our moon from?

Anything in particular I should ingest before seeing this claimed rotational movement of our moon that you claim to see?
:)

As the Moon moves across the sky, it must turn, otherwise you'd see one side, then the front, then the other side (like a car passing you, you see the front, then the side, then the back). The only way you can see the same face of the Moon from the Earth, is if the Moon is rotating to match your changing angle of observation.

If you sit in the stands at a car race, then you will see both sides of each race car as they turn a lap. HOWEVER, if you stand in the infield, then you will only observe the left door of all the race cars.

Can't you see that this issue involves your viewing angle?

Are the race-cars revolving 360° around their center-mass as they complete each 360° lap (synchronous rotation), or are the cars merely TURNING about their center-mass as they complete a lap?

As the cars complete a lap, EXACTLY where is their turn axis?

A – is the axis around each car's center-mass; or

B – are the cars instead turning around an axis in the center of the racetrack; or

C – turning around BOTH axes at the same time?​

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Unless, the co-worker is CORRECT and you're WRONG?

He's not wrong. The Moon is rotating.

As the cars complete a lap, EXACTLY where is their turn axis?

A – is the axis around each car's center-mass; or

B – are the cars instead turning around an axis in the center of the racetrack; or

C – turning around BOTH axes at the same time?

The cars rotate about an axis through the middle of each car, halfway between the front and rear axles. They also rotate revolve around an axis passing through the center of this track. So the answer is C, both.

Are the race-cars revolving 360° around their center-mass as they complete each 360° lap (synchronous rotation), or are the cars merely TURNING about their center-mass as they complete a lap?

Turning requires a rotation.

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DaveC426913
I'd respond to KD, but Drakkith pretty much covered it.

There's really a very simple way of telling whether the moon is rotating or not. If you're on the surface of the Moon, do the stars, which represent a non-rotating coordinate system, move about the sky? The answer is yes. And they do so at a rate of one cycle per 27.321582 days. That's really all there is too it.

davenn and Bystander
There's really a very simple way of telling whether the moon is rotating or not. If you're on the surface of the Moon, do the stars, which represent a non-rotating coordinate system, move about the sky? The answer is yes. And they do so at a rate of one cycle per 27.321582 days.

Terrific, a simplistic RULE that seemingly solves everything!

However, we were discussing this graphic:

Consistent with your stated RULE, the image (our current Earth & moon) on the left does REVOLVE CCW around Earth with a sidereal rotation rate of 27.3, but is the moon in that graphic both REVOLVING around the Earth 360° and ROTATING around its center-mass (polar axis) 360°, as well?

Synchronous rotation theory holds that Earth's moon does indeed have two synchronous rotations, a 360° REVOLUTION around its barycenter, as well as a 2nd 360° ROTATION around its polar axis, and the issue here is, since both types of 360° spins would cause the stars to move while standing on such a moon, is it our moon's REVOLUTION or ROTATION, or both types of spin in conjunction which would cause the stars to move?

Can REVOLUTION & ROTATION in conjunction cause the starfield to stop (or at least, to slow down), as appears to be happening with the moon on the right?

Let's now examine the supposedly zero-rotating moon on the right. I see that moon as having two 360° spins, a 360° CCW REVOLUTION around its barycenter, as well as a 360° CW ROTATION around its polar axis. In relation to Earth in the center, from the outside perspective, that moon would not appear to have a rotation in relation to the Earth in the center, but from Earth's perspective, that moon would be rotating since its surface is moving.

If the surface of the moon on the right is still moving in relation to the Earth's surface, then tidal braking forces would still be slowing the moon on the right down – can you agree that the surface of the moon on the right is still turning when viewed from the center?

Clearly, if you were to stand on that Earth, then you would clearly see that moon both REVOLVING (orbiting) 360° CCW each orbit, as well as you would see the entire surface of that moon each orbit spinning 360° in the opposite CW direction.

If our own moon had spun down from the CW direction (as Venus is now doing), then our moon would have passed thru that same so-called zero-spin rate, which is a -0:1 sidereal spin rate. It is clearly possible for an astronomical body to pass thru a -0:1 in the process of being despun to a +1:1 sidereal spin rate. Actually, it would be impossible for a CW rotating body to be despun without passing thru that -0:1, as depicted on the right in the above graphic.

So, let's apply your RULE to that supposedly zero-rotating moon. If you were to stand on that supposedly zero-rotating moon's surface, then the stars would still move about its sky, but not in 27.3 days, but instead in 365 days since both the Earth and moon are together in orbit around the sun. 365 is not 27.3 days, but the stars would indeed still move while standing on that moon's surface!

If our moon was despun from the CW direction (as is now happening to Venus), then that -0:1 rate wouldn't have lasted for very long, perhaps only a matter of days, so the starfield would at most appear to slow down as our moon passed thru that -0:1 rate, before speeding up again. Of course, once our moon was fully tidally locked to Earth, since that happened many billions of years ago when our moon was much closer to Earth with a much faster orbital period, so the 27.3 day part of your rule wouldn't apply to our moon until relatively recently after our moon evolved into its current orbit, which continues to change to this day.

BTW – since no two lunar orbits are exactly the same, the "27.321582" day portion is merely an average (mean) sidereal rotation rate, so don't engrave that into stone.

That's really all there is too it.

Yep, it's surely as simple as that!
:)

Using your RULE, would you care to explain how CW rotating Venus, which now has a -0.92:1 sidereal rotation rate (but 1.92 solar days per orbit), how Venus can bleed off nearly two full CW polar axial spins to arrive at the classic +1:1 of a despun body?

As viewed from the sidereal perspective, my explanation is the only possible way to count Venus' spin down:

-1:1 ~ -0:1 ~ +1:1

Since Venus now has slightly less than a single CW sidereal rotation (-0.92:1), venus has recently passed thru that first -1:1 sidereal rotation rate, and is now heading towards that -0:1 rate before stopping at a final +1:1 sidereal spin rate. While passing thru that -0:1 sidereal spin rate, Venus will match that supposedly zero-spinning moon on the right side of the graphic under discussion here, so Venus would still have one final CW polar axial rotation left to lose before becoming tidally locked with the sun.

Of course, when viewing Venus' spin down from the center-point perspective, things get far less confusing.

The simple center-point is followed by the sidereal perspective, as Venus loses nearly 2 full CW polar axial rotations as that planet is despun – (in parenthesis is Venus' current spin rates from both perspectives):

~ 2 ~ (1.92) ~ 1 ~ 0

-1:1 ~ (-0.92) ~ -0:1 ~ +1:1

If someone here can logically explain how Venus can go from a sidereal spin rate of -1:1, without passing thru a transitory -0:1 sidereal spin rate, before finally stopping at a final +1:1 rate, then I'd appreciate if they would do it?

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Ken, I'm having a lot of trouble understanding what point you're trying to make in your posts. It's hard to parse this amount of text when you're not using standard terminology.

The issue of rotation is indeed simple - you can always measure whether you're in a rotating reference frame by simple experiments - like the Foucault pendulum. This is entirely unambiguous.

If that's not what this is about, then maybe you could distil it to a single question or an issue with previous answers that you're having here?

You asked earlier for a resource on tidal braking acting on rotation - any introductory book on astronomy should have a section on this process. Page 47 of the fourth edition of Astronomy and Astrophysics by Zeilik & Gregory is one place to find it.
A little bit of googling on tidal braking provides a lot of hits as well. This page has a short section around the bottom: http://www.das.uchile.cl/~mhamuy/moon.html
And this paper has some precise measurements:
http://onlinelibrary.wiley.com/doi/10.1029/JB093iB06p06216/abstract

Ken, I'm having a lot of trouble understanding what point you're trying to make in your posts. It's hard to parse this amount of text when you're not using standard terminology.

The issue of rotation is indeed simple - you can always measure whether you're in a rotating reference frame by simple experiments - like the Foucault pendulum. This is entirely unambiguous.

Sorry I was being ponderous, but you threw out that Venus had a -1.08 sidereal rotation rate instead of -0.92, and I had to figure that out and then explain where you were getting 1.08 from.

I also had to explain why the sidereal and center-point perspectives always differ by one (1) apparent rotation.

Hopefully, all those basics which I felt were necessary in order to analyze that supposedly zero-rotating graphic, are now behind us.

Not so sure the issue of determining true rotation is as simple as you claim since it is often a question of rotating or revolving in regards to which frame of reference, and around which spin axis?

If that's not what this is about, then maybe you could distil it to a single question or an issue with previous answers that you're having here?

Certainly, and I just did above, so here it is again (question in bold):

From the center-point venus currently has 1.92 CW polar axial rotations per orbit (a Synodic period of nearly two solar days per orbit), but from the sidereal perspective, Venus appears to have one less rotation, or -0.92 sidereal rotations per orbit.

The simple center-point perspective is followed by the sidereal perspective, as Venus loses nearly 2 full CW polar axial rotations as that planet is despun (in parenthesis is Venus' current spin rates from both perspectives):

~ 2 ~ (1.92) ~ 1 ~ 0

-1:1 ~ (-0.92) ~ -0:1 ~ +1:1

If someone here can logically explain how Venus can go from a sidereal spin rate of -1:1, without passing thru a transitory -0:1 sidereal spin rate, before finally stopping at a final +1:1 rate, then I'd appreciate if they would do it?

There's my simple question!

You asked earlier for a resource on tidal braking acting on rotation - any introductory book on astronomy should have a section on this process. Page 47 of the fourth edition of Astronomy and Astrophysics by Zeilik & Gregory is one place to find it.

A little bit of googling on tidal braking provides a lot of hits as well. This page has a short section around the bottom: http://www.das.uchile.cl/~mhamuy/moon.html

And this paper has some precise measurements:

http://onlinelibrary.wiley.com/doi/10.1029/JB093iB06p06216/abstract

Sorry to waste your time googling for tidal braking since I already know how that works. What I had asked you for was a citation which would support what you seemed to have claimed concerning tidal ACCELERATION:

No. When its rotation in an inertial frame of reference is 0, it necessarily means that its rotational angular momentum is also 0. As tidal interactions continue to accelerate the planet, it gains rotational angular momentum until it settles in a stable 1:1 synchronous rotation.

What I had asked was:

Are you claiming tidal interactions can accelerate a planet's polar axial rotation? If you really do mean its rate of rotation, then do you have a citation for that?

I doubt if you can find such a citation, so let's move on about that.

Tidal braking is acceleration. As in angular acceleration. The result of torques. The change of angular velocity. What's the issue here?

I still don't get where you got the 0.92 from. If you're using rotation : orbit notation, then it's as simple as dividing the period of rotation by the period of revolution normalized by the period of revolution:
(-245/224) : (224/224)

To your main question: Can you use standard terminology? What is 'centre-point', what is polar-axial rotation? Can you define those terms?

Apologies should you take my insistence on clarifying those terms as nagging. I simply did not want us to talk past each other.
But anyway, I read it again:
If someone here can logically explain how Venus can go from a sidereal spin rate of -1:1, without passing thru a transitory -0:1 sidereal spin rate, before finally stopping at a final +1:1 rate, then I'd appreciate if they would do it?
is it all about the boldened bit? But nobody said that this transition could not or should not happen, had they? Of course it has to happen - it's not like the planet can skip over some values of angular velocity in-between the initial and final value, and having no rotation is one such value.

However, we were discussing this graphic:

View attachment 88634
Actually, this graphic is leading us away from the OP's question.

Is our Moon rotating? Yes. As has been pointed out, several methods will demonstrate this, notably observation of background stars, and a pendulum.
This is what the image on the left is showing.

Discussion of the image on the right is - as seen by the amount of attention it is taking away from the central discussion - a straw man diversion. You can argue it all you want, it does not bear on the OP's question.

I'm going to request that further discussion, whether it be of hypothetical lunar rotations, or of Venus, be treated as off-topic.

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