Do Solar Tides affect Earth's Orbit?

[Buzz Bloom]

Solar tides on the Earth are currently approximately 1/2 as strong as lunar tides. The lunar tides affect the Earths rotation as well as the moon's orbit. Therefore it seems plausible that the solar tides would affect the sun's rotation and the Earth's orbit. Can someone confirm for me that this is so? If so, can someone cite any reference that would provide a value of this effect?

 

[scottdave]

As any 2 masses will exert gravitational force on one another, all the celestial bodies will affect each other's motion. It is the amount of force, which is greatly reduced for great distances, since there is distance² in the denominator.

 

[Drakkith]

The lunar tides affect the Earths rotation as well as the moon's orbit.

True. The Moon's orbit is affected because the tides on Earth transfer angular momentum to the Moon, gradually increasing its orbital radius.

Therefore it seems plausible that the solar tides would affect the sun's rotation and the Earth's orbit.

Possibly. Any effect would probably be very tiny since the Earth is so far away from the Sun though. I've never read anything about this, so I can't give any references.

 

[mfb]

The effects of tidal gravity generally scale with the negative third power of the distance. The Sun is 420 times more distant than the Moon, so the effects are a factor 70 million weaker. For tides on Earth the Sun is still relevant because it has 27 million times the mass of the Moon, but if you consider the effect of Earth's tides on other objects there is no such factor.

 

[Buzz Bloom]

Hi @Drakkith and @mfb:

The Moon's orbit is affected because the tides on Earth transfer angular momentum to the Moon

I don't mean to be pedantic, but Drakkith's kind of incomplete description of the tidal mechanism is generally what is presented in the material I have been to find and read, e.g., Wikipedia.

https://en.wikipedia.org/wiki/Tidal_acceleration
The gravitational torque between the Moon and the tidal bulge of Earth causes the Moon to be constantly promoted to a slightly higher orbit and Earth to be decelerated in its rotation.​

I have not been able to find a detailed description of how the tide causes these effects. I present below my guess about this, and if anyone finds an error, I would much appreciate being corrected. I intentionally here omit any discussion of (a) a mechanism preceding the permanent formation of oceans and (b) solar tides.
1. The moon's tidal force creates a bulge in the oceans both facing towards and away from the moon.
2. As the Earth rotates, these bulges hit the coastlines of continents. When this happens a force is applied to the coastlines. The effective direction of this force depends on the amount of slope of the coastline being hit. It the coastline were a vertical cliff, the force would be tangential to the surface of the Earth. If the coastline were a very shallow slope, a large component of the force would be directed towards the center of the Earth.
3. Over the course of a sidereal day, the sum of all these forces would be a day's vector with a component tangent to the equator and a radial component towards the center of the Earth.
4. The tangent component applies a torque to the Earth's rotation that slows the rotation.
5. The entire day's vector also applies a force to the Earth with one component (a) in the plane of the moon's orbit which points away from the moon, one component (b) in the plain of the moon's orbit normal to component (a), and one component (c) normal to the orbital plane. Component (a) acts to force the Earth further from the moon, which does two things: (i) the average distance between Earth and moon increases, and (ii) The Earth's and moon's orbits around their common center of gravity are changed. The combination of (b) and (c) primarily act to change the orbits, which may also have an effect changing the distance between Earth and moon. Technically (and perhaps pedantically) speaking, the tides do not move the moon away from the Earth, but moves the Earth away form the moon.
6. If the moon continues to keep one face toward the Earth, then there is also a reduction in the moon's angular momentum from its rotation about it's axis.
7. I do not see a clear explanatory reason for a day's force to subtract from the angular momentum of the Earth the exact angular momentum which is gained by the changes to the Earth's and moon's orbit relative to their common center of gravity, and to the moon's rotational angular momentum. Let the sum of all the day's forces over the course of a lunar sidereal cycle be called a month's force. The month's force will have a component along the tangent of the Earth's orbit around the sun. It might be either forward or backward relative the Earth orbital motion. In either case, it would change the Earth's orbit and the orbit's angular momentum. Therefore the total angular momentum of the entire sun, Earth, moon system is preserved.

The Sun is 420 times more distant than the Moon, so the effects are a factor 70 million weaker. For tides on Earth the Sun is still relevant because it has 27 million times the mass of the Moon

I think your numbers are close but a little off, mfb.
https://en.wikipedia.org/wiki/Moon
https://en.wikipedia.org/wiki/Sun

D_sun = Distance Earth to sun = 1.496×10⁸ km
D_moon = Distance Earth to moon = 3.84399⁵ km
R_dist = D_sun / D_moon = 389.17
R_dist³ = 5.8942627x10⁷

M_sun = 1.9855×10³⁰ kg
M_moon = 7.342×10⁸ kg
R_mass = M_sun / M_moon = 2.7043×10⁷

R_tide = R_dist³ / R_mass = 2.18

Your value for R_tide = 70 / 27 = 2.59.

In either case, the solar tide today, although smaller, is currently a substantial fraction compared to the lunar tide. When the moon was much closer to the Earth the solar tide would would have been a much smaller fraction.

Regards,
Buzz

 

[Drakkith]

7. I do not see a clear explanatory reason for a day's force to subtract from the angular momentum of the Earth the exact angular momentum which is gained by the changes to the Earth's and moon's orbit relative to their common center of gravity, and to the moon's rotational angular momentum.

From wiki's article on tidal acceleration:

The mass of the Moon is sufficiently large, and it is sufficiently close, to raise tides in the matter of Earth. In particular, the water of the oceans bulges out towards and away from the Moon. The average tidal bulge is synchronized with the Moon's orbit, and Earth rotates under this tidal bulge in just over a day. However, Earth's rotation drags the position of the tidal bulge ahead of the position directly under the Moon. As a consequence, there exists a substantial amount of mass in the bulge that is offset from the line through the centers of Earth and the Moon. Because of this offset, a portion of the gravitational pull between Earth's tidal bulges and the Moon is not parallel to the Earth–Moon line, i.e. there exists a torque between Earth and the Moon. This boosts the Moon in its orbit, and slows the rotation of Earth.

2. As the Earth rotates, these bulges hit the coastlines of continents. When this happens a force is applied to the coastlines. The effective direction of this force depends on the amount of slope of the coastline being hit. It the coastline were a vertical cliff, the force would be tangential to the surface of the Earth. If the coastline were a very shallow slope, a large component of the force would be directed towards the center of the Earth.

I don't think this is correct. None of the force should be directed towards the center of the Earth except the weight of the water. The force of the water on the Earth, discounting its weight, should all be tangential to the Earth's axis of rotation. A shallow coastline just means the water will move up the coast (high tide) and the tangential force will be spread out across a large area instead of all being applied at the base of a cliff. That's my understanding at least.

 

[Buzz Bloom]

Because of this offset, a portion of the gravitational pull between Earth's tidal bulges and the Moon is not parallel to the Earth–Moon line, i.e. there exists a torque between Earth and the Moon.

Hi Drakkith:

Thank you much for your post. I read your Wiki quote, but I confess I do not understand this logic all. Unfortunately, the article does not reference any authoritative source which might explain this better.

What does it mean that there is a "torque between Earth and the Moon"? I can imagine that there are two interpretations.
(1) The bulge on the side of the Earth towards the moon pulls the moon with a component in the direction of the moon's orbit. The would increase the velocity of the moon and add to it's orbital angular momentum. That is OK, but the bulge on the opposite side of the Earth would offset most of this. I would be delighted to see a source with the math calculating that this effect would approximately result in the known rate of increase in the distance between the Earth and the moon. My intuition tells me that this calculation has never been made and published.
(2) Some unexplained mechanism causes the bulge to impart a gravitational torque on the Earth's crust. As the bulge approaches a coastline it will gravitationally attract the higher terrain it is approaching. However, relative to the center of the Earth, the bulge is not moving (much), but the coast line is moving towards the bulge as the Earth rotates. Therefore the gravitational force from the bulge will pull the coastline towards the bulge in the same direction the coastline is moving. This force as a torque will INCREASE the angular momentum of the earth, not decrease it. Consequently, I can make no sense of this. The concept of my post has the bulge pushing against the coastline, and imparting a torque that decreases the rotational angular momentum.

Regards,
Buzz

 

[Drakkith]

(1) The bulge on the side of the Earth towards the moon pulls the moon with a component in the direction of the moon's orbit. The would increase the velocity of the moon and add to it's orbital angular momentum. That is OK, but the bulge on the opposite side of the Earth would offset most of this.

It does to some extent, but the bulge on the opposite side is further away from the Moon and therefore doesn't contribute as much torque as the nearer bulge.

I would be delighted to see a source with the math calculating that this effect would approximately result in the known rate of increase in the distance between the Earth and the moon. My intuition tells me that this calculation has never been made and published.

I have no idea why you would think that this hasn't been done before, but here's one paper doing exactly that: http://science.sciencemag.org/content/sci/273/5271/100.full.pdf

 

[Buzz Bloom]

Hi @Drakkith:

Your link to the Science article is exactly what I have been searching for unsuccessfully for some time. I do not have ready online access to Science, but my town librarian has been helpful before in getting me access to specific old articles.

Since you did not comment on my (2) discussion, I am assuming you did not find any errors in it needing correction.

Regards,
Buzz

 

[Drakkith]

Your link to the Science article is exactly what I have been searching for unsuccessfully for some time. I do not have ready online access to Science, but my town librarian has been helpful before in getting me access to specific old articles.

Dang it, I keep falling into the trap of finding articles while on a campus computer, where many journals are open to me. My apologies.

Since you did not comment on my (2) discussion, I am assuming you did not find any errors in it needing correction.

It doesn't look correct. Most noticeably because the force of attraction from the bulge on the coastline is extremely weak (due to the very low mass of the coastline) and also because the Earth can't exert a torque on itself. That's my understanding of the situation at least.

 

[Vanadium 50]

Since you did not comment on my (2) discussion, I am assuming you did not find any errors in it needing correction.

That's not how it works.

 

[Buzz Bloom]

That's not how it works.

Hi Vanadium 50:

I suppose that since you have now told me that, I will be now be able to figure out from the insight you have given me exactly how it actually does work.

Regards,
Buzz

 

[Vanadium 50]

The PF membership does not have an obligation to respond to any message, including ones of the "prove me wrong" part. Silence does not mean acceptance. You are not entitled to demand a response.

 

[Buzz Bloom]

That's not how it works.

You are not entitled to demand a response.

Hi Vanadium:

I apologize for misunderstanding your post #11. I thought the antecedent of "it" was the mechanism under discussion regarding the tidal affects on the Earth's rotation and the moon's orbit. You post #13 makes it clear now that "it" refers to the protocols of the PF.

Regards,
Buzz

 

[mfb]

I was lazy and took the current distance to the Moon instead of the mean distance. To be precise we would need the mean inverse cubed distance, and no one here wants to calculate that.

The bulge moves with respect to the surface, that induces friction. The bulge will move with the Earth's rotation until the gravitational pull from the Moon cancels the friction. As an extremely simplified picture, it looks like this (from here). Friction is a force counterclockwise for the tides and clockwise (slowing the Earth) for the rest of Earth. The gravitational attraction of the Moon is a force clockwise for the tides and counterclockwise (adding angular momentum) for the Moon.
No coastlines involved in this simplified picture that gives the main features of the Earth/Moon tide system.

Coastlines make everything messy. Instead of two nice bulges you get this:

Colors are the height of tides, lines are places of equal phase, places where lines meet don't have tides. The simple model from above would have the white lines going from pole to pole with nothing else. As you can see it is much more complicated, but the basic idea is still the same.

My intuition tells me that this calculation has never been made and published.

You underestimate the amount of effort going into science by a huge factor if you really think this could be possible.

 

[snorkack]

A simple question about quantification:

Suppose that no torques operate on Sun-Earth system and therefore the total angular momentum of Sun-Earth system were conserved

Further suppose that Sun were a deep ball of gas, unlike the shallow seas of Earth, and therefore rotated without friction.
The two exchangeable angular momenta should then be just Earth rotation and Earth revolution.
Furthermore, due to Third Law of Kepler, the angular momentum of Earth increases if Earth spirals out. The speed indeed slows, but the lever arm of the speed increases more than the slowing of the speed.

Therefore, as solar tides slow down Earth rotation, they cause Earth to climb to a higher orbit.

How much would Earth sidereal year stretch, in constant former hours, if due to Solar tides the length of the day were to expand from 24 hours to 25?
How much would Earth sidereal year stretch, in constant former hours, if due to Solar tides the day would extend forever (due to tidal lock with revolution)?

 

[mfb]

Even a pure ball of gas has internal friction. If you replace the Sun by superfluid helium things might look differently.

Earth's orbital angular momentum is 2.66*10⁴⁰ Js, its angular momentum from the rotation is 5.84*10³³ Js. The Earth will slow down, but the effect on the orbit is completely negligible.

 

[snorkack]

Earth's orbital angular momentum is 2.66*10⁴⁰ Js, its angular momentum from the rotation is 5.84*10³³ Js. The Earth will slow down, but the effect on the orbit is completely negligible.

Increase of Earth orbital angular momentum by a fraction of 10^-7 would include Earth orbital speed decreasing by a fraction of 10^-7, lever arm increasing by a fraction of 2*10^-7 and year increasing by a fraction of 3*10^-7.
The whole Earth angular momentum from rotation is about a fraction of 2.2*10^-7 of angular momentum from revolution, not exactly in the same direction. So the Earth orbital angular momentum can be increased by a fraction of 2*10^-7, which would increase the year length by fraction of 6*10^-7.
A year is about 3*10⁷ s, so stopping Earth rotation would stretch year by about 18 seconds.

Slowing Earth rotation to hundreds of days, like Venus, would have a similar effect.

How on Earth did Venus wind up with retrograde rotation? That's what we need to figure out to venusform Earth...

 

[Buzz Bloom]

Hi @scottdave, @Drakkith, @mfb, @snorkack:

I much appreciate your participation in this thread. I believe the thread has helped me with much of my confusion. In particular the discussion about the mechanism by which the lunar tides effect the Earth's rotation and the moon's orbit helped me to correct some of my wrong assumptions. However, some confusion about this remains, so I started another thread specifically to explore this issue.

https://www.physicsforums.com/threads/tidal-effects-on-Earth's-rotation-and-moons-orbit.933441/​

Regards,
Buzz