Tidal effect of Sun on Earth

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Summary:

How does the tidal effect of the Sun on the Earth affect the Earth's spin, the Moon's orbit about the Earth, and the Earth's orbit about the Sun?
It's well known that the tidal effect of the Moon on the Earth causes the Earth's rotation to slow down and the radius of the Moon's orbit to increase over time. However, the Sun also exerts a tidal effect on the Earth, which should also contribute to slowing down the Earth's spin. This raises two questions:

(1) Does the Sun's tidal effect, by changing the total tidal bulge on the Earth, change how the Moon's orbital radius increases over time?

(2) Does the Sun's tidal effect change the Earth's orbit about the Sun? If so, how?

Please note that I have put this thread in Classical Physics because I do not want to complicate the analysis with relativistic effects; I want to stick to Newtonian gravity and Newtonian mechanics. (Relativistic effects should be extremely small for this scenario in any case.)
 
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What's the ratio of the radius of the earth's orbit about the sun compared to the radius of the moon's about the earth? And of the gradients?
 
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PeterDonis
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What's the ratio of the radius of the earth's orbit about the sun compared to the radius of the moon's about the earth? And of the gradients?
In SI units (meters for length, kilograms for mass, seconds for time, other units composed of those):

Moon's orbit: ##R_M = 3.844 \times 10^8##

Earth's orbit: ##R_E = 1.496 \times 10^{11}##

Ratio of Earth about Sun to Moon about Earth: ##389.2##

Moon's radius: ##r_M = 1.737 \times 10^6##

Earth's radius: ##r_E = 6.378 \times 10^6##

Gravitational constant ##G##: ##6.67 \times 10^{-11}##

Mass of Moon: ##M_M = 7.348 \times 10^{22}##

Mass of Sun: ##M_S = 1.989 \times 10^{30}##

Tidal gradient due to Moon across Earth: ##\frac{G M_M}{R_M^3} r_E = 5.503 \times 10^{-7}##

Tidal gradient due to Sun across Earth: ##\frac{G M_S}{R_E^3} r_E = 2.527 \times 10^{-7}##

Ratio of Sun gradient to Moon gradient: ##0.4592##
 
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......... and that gives a good reason for the monthly occurrence of two lots of spring tides and two lots of neap tides. The Sun's contribution is significant but less than the Moon's.
 
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Summary:: How does the tidal effect of the Sun on the Earth affect the Earth's spin, the Moon's orbit about the Earth, and the Earth's orbit about the Sun?

It's well known that the tidal effect of the Moon on the Earth causes the Earth's rotation to slow down and the radius of the Moon's orbit to increase over time. However, the Sun also exerts a tidal effect on the Earth, which should also contribute to slowing down the Earth's spin. This raises two questions:

(1) Does the Sun's tidal effect, by changing the total tidal bulge on the Earth, change how the Moon's orbital radius increases over time?

(2) Does the Sun's tidal effect change the Earth's orbit about the Sun? If so, how?
Question (1) seems more complicated for analysis. I'd guess there is small influence.
Question (2) yes. And there is mutual tidal effect. Just like Sun causes tides on Earth, system Earth-Moon causes small tides on the Sun. Earth and Sun are decreasing their rotation rates and the orbit should slowly grow due to loss of angular momenta
 
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Question (1) seems more complicated for analysis. I'd guess there is small influence.
A quantitative analysis would be fairly complicated, yes. But I think we can gain some important points just from a qualitative analysis

First: the Sun's tidal effect combines with the Moon's to cause a net tidal bulge on the Earth. This, in itself, does not change any rotation rates.

However, the bulge exerts a torque on the Moon, and the Moon therefore exerts and equal and opposite torque on the bulge. This torque is what causes the Earth's rotation to slow (and the corresponding torque on the Moon is what causes its orbital radius to increase--the combination of the two torques transfers angular momentum from the Earth's spin to the Moon's orbit).

Second: why does the bulge exert a torque on the Moon? Because the bulge is not aligned with the Moon--more precisely, it is not directly along the line from the Earth's center to the Moon's center. It is "pulled" out of that line by the Earth's rotation. The bulge will be ahead of the Moon's position in the sky by some angle (which I believe is called ##\alpha## in the literature) which will be the outcome of a balance between the torque on the bulge exerted by the Moon and the "pull" on the bulge exerted by the Earth's spin.

Now for the key point: where is the Sun in the sky relative to the bulge? To first order, i.e., neglecting any effects of the Sun itself on the size of the bulge, the Sun moves in the sky relative to the Moon, so it will also move in the sky relative to the bulge. So the Sun's torque on the bulge will not be constant; it will vary over the course of a month as the relative positions of the Sun and Moon in the sky go through a complete cycle. When the Sun is behind the bulge, as the Moon is, its torque will increase the slowing of the Earth's spin; when the Sun is ahead of the bulge, its torque will decrease the slowing of the Earth's spin. Over the course of a month, to first order, these effects will cancel, so the average rate of slowdown of the Earth's spin will be the rate due to the Moon. This will then also be the average rate of increase of the Moon's orbit.

So far I have not taken into account higher order effects. The most obvious one is that the Earth itself orbits the Sun, but I am not sure at this point what, if any, change that makes in the above analysis.

Question (2) yes.
Are you sure? If the analysis above is correct, then to first order, the Sun's torque on the bulge cancels over the course of a month, so there will be no net exchange of angular momentum between the Earth's spin and the Sun-Earth orbit.

system Earth-Moon causes small tides on the Sun
Yes, but they are indeed very small; I think small enough that they can be ignored in this discussion. However, I have not done the explicit calculation; if you think these tides are large enough to matter, doing the math to check that would help.
 
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Now for the key point: where is the Sun in the sky relative to the bulge? To first order, i.e., neglecting any effects of the Sun itself on the size of the bulge, the Sun moves in the sky relative to the Moon, so it will also move in the sky relative to the bulge. So the Sun's torque on the bulge will not be constant; it will vary over the course of a month as the relative positions of the Sun and Moon in the sky go through a complete cycle. When the Sun is behind the bulge, as the Moon is, its torque will increase the slowing of the Earth's spin; when the Sun is ahead of the bulge, its torque will decrease the slowing of the Earth's spin. Over the course of a month, to first order, these effects will cancel, so the average rate of slowdown of the Earth's spin will be the rate due to the Moon. This will then also be the average rate of increase of the Moon's orbit.

So far I have not taken into account higher order effects. The most obvious one is that the Earth itself orbits the Sun, but I am not sure at this point what, if any, change that makes in the above analysis.
Earth's orbit is slightly elliptical so Sun has a small effect on the bulge size. Off course, tidal effects of this kind influences way much less and accumulate slower than the Moon's tidal effects, but they exists.

Yes, but they are indeed very small; I think small enough that they can be ignored in this discussion. However, I have not done the explicit calculation; if you think these tides are large enough to matter, doing the math to check that would help.
I believe they are small enough. Effects of solar wind and presence of other planets might be more important for the possible change of Earth's orbit.
 
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PeterDonis
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Earth's orbit is slightly elliptical
Yes, and more importantly, the period of course is not an integral number of months, so yes, this will introduce a small higher order effect.
 

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