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I'm been researching this question but haven't been able to find a conclusive answer. As I understand it, there are two forms of tidal energy: tidal potential energy and tidal current energy. Tidal potential energy is concerned with the difference in height between the high and low tides, and tidal energy uses the actual kinetic energy of currents to generate power.

In both cases, the ability to generate power from the tides is from the influence of the Moon and the Sun's gravity on the tides. As I understand it, because the Moon is much closer to the Earth it is roughly 2.2 times stronger gravitational force on the Earth's tide than the Sun, however the Sun does contribute. The Sun also contributes in an additive manner to the Moon's influence during "spring tides" but is in opposition during "neap tides."

However, opposing waves do not "cancel out" the energy, it just moves it. As seen from this link: https://skullsinthestars.com/2010/04/07/wave-interference-where-does-the-energy-go/

So I'm curious what percent of tidal power can be contributed to the Sun. I don't need an exact percentage, I'm just curious if it is high enough to not be negligible.

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Click here for reference. Then as far as tidal potential energy goes in terms of displacements...

$$\frac{\Delta {{r}_{sun}}}{\Delta {{r}_{moon}}}\simeq \frac{{{m}_{sun}}}{{{m}_{moon}}}\frac{{{R}_{moon}}^{3}}{{{R}_{sun}}^{3}}\simeq \frac{1.99\times {{10}^{30}}}{7.35\times {{10}^{22}}}\frac{{{\left( 3.84\times {{10}^{8}} \right)}^{3}}}{{{\left( 1.5\times {{10}^{11}} \right)}^{3}}}\simeq 4.29\times {{10}^{30-22+24-33}}\\\simeq 4.29\times {{10}^{-1}}\simeq 43\%$$

UPDATE: (based on comments) enter image description here Consider two masses ${{m}_{1}}$ and ${{m}_{2}}$ whose co-ordinates are measured relative to the centre of a third mass, M, via distances ${{r}_{1}}$ and ${{r}_{2}}$. Point E lies on the surface of mass M (assumed to be spherical) and for now is assumed to also be in the plane of all three masses for convenience. Let $\angle EO{{R}_{1}}=\theta $and $\angle {{R}_{2}}O{{R}_{1}}=\phi $. We have using the same approximations as before $${{V}_{T}}={{V}_{m1}}+{{V}_{m2}}=-\frac{G{{m}_{1}}{{r}^{2}}}{2R_{1}^{3}}\left( 3\cos \left( \theta \right)-1 \right)-\frac{G{{m}_{2}}{{r}^{2}}}{2R_{2}^{3}}\left( 3\cos \left( \theta -\phi \right)-1 \right),0\le \phi ,\theta \le \frac{\pi }{2}$$ Tidal displacement due to both moon, ${{m}_{1}}$and sun, ${{m}_{2}}$ as a first approximation can be given by $$\Delta r=\frac{{{r}^{4}}}{2M}\left\{ \frac{{{m}_{1}}}{R_{1}^{3}}\left( 3\cos \left( \theta \right)-1 \right)+\frac{{{m}_{2}}}{R_{2}^{3}}\left( 3\cos \left( \theta -\phi \right)-1 \right) \right\}$$ From which we have then $$\Delta {{r}_{spring}}=\frac{{{r}^{4}}}{2M}\left\{ \frac{{{m}_{1}}}{R_{1}^{3}}+\frac{{{m}_{2}}}{R_{2}^{3}} \right\}\left( 3\cos \left( \theta \right)-1 \right)$$ and $$\Delta {{r}_{neap}}=\frac{{{r}^{4}}}{2M}\left\{ \frac{{{m}_{1}}}{R_{1}^{3}}\left( 3\cos \left( \theta \right)-1 \right)+\frac{{{m}_{2}}}{R_{2}^{3}}\left( 3\sin \left( \theta \right)-1 \right) \right\}$$ Q: Are there positions on the earth where the tidal displacement for both a spring and neap tide are the same? Equating the two displacements we find $\cos \left( \theta \right)=\sin \left( \theta \right)\Rightarrow \theta =\frac{\pi }{4}$ independent of mass. More generally we might ask when the system is not in a neap-tide, then where will the displacements become equal to that when it is in a neap-tide. Then we are considering $$\sin \left( \theta \right)=\cos \left( \theta -\phi \right)=\sin \left( \tfrac{1}{2}\pi -\theta +\phi \right)\Rightarrow \theta =\tfrac{1}{4}\pi +\tfrac{1}{2}\phi $$ Which defines a function $\theta \left( \phi \right)$ describing a point on the earth which has a displacement equivalent to that at neap tide. Note: $\theta \left( 0 \right)=\arctan \left( 1 \right)=\frac{\pi }{4}$ which was our previous result. So the point moves linearly with respect to angular separation, starting at $\pi /4$ in a spring tide and converging upon itself at a neap tide (which we logically expect). Note that this analysis delivers a single point where rather I think it would be a locus of points created at the intersection of two spheroids. We have the prolate spheroidal bulge due to the moon, which we hold fixed, and then add to that a tilted spheroid from the sun which as it moves towards its position for a neap-tide, becomes an oblate-spheroid. As it tilts it creates a locus of intersections on the surface where tidal displacements are equivalent to that at a neap-tide. Also there are other solutions here i haven't considered (due to the periodicity of the functions) in which a more careful analysis will bring to the fore. However i'd suggest moving directly to a more general coordinate set and try and capture the full locus.

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    $\begingroup$ interesting factoid I've just learned: since the Sun and Moon have the same apparent size, the ratio of their tidal forces is just the ratio of their densities. $\endgroup$
    – Roger Wood
    Apr 27 at 6:12
  • $\begingroup$ Thank you for the answer. Also, as this chapter points out, the range for the tidal force provided by the sun alone ranges from 36% to 57% depending on the position of the sun. link Is there any legitimacy to the claim that tidal energy that is provided during "spring tides" (which the Sun has additive effects with the moon for higher tide) is cancelled out with "neap tides" in which the Sun's forces are in opposition to the Moon's? $\endgroup$
    – user699279
    Apr 27 at 18:49
  • $\begingroup$ @user699279 Ahh...i think i see where the 36-57% is coming from, since the orbits are elliptical and so the R's change slightly. However i don't understand what you mean by energy being 'cancelled'. $\endgroup$ Apr 27 at 23:09
  • $\begingroup$ @mathstackuser12 Sorry, I should clarify. I mean to say, that because generating power from the tides is either concerned with the difference between the high and low tides in terms of height, or the kinetic power of the currents themselves, then, for the period of time where the Sun is opposing the Moon's effect on the tides, does this imply that the Sun restricts the ability for the tidal energy to be generated. Or in other words, is it correct to say that the Sun helps the generation of power during "spring tides" and opposes it during "neap tides". And if so, is this net effect positive? $\endgroup$
    – user699279
    Apr 28 at 1:05
  • $\begingroup$ @user699279 to me there is no restriction of energy per se, but rather when the sun is in such a position relative to the moon its contribution to the tidal energy is such that tidal 'bulge' due to the sun prohibits the bulge due to the moon. the displacements are at 90 degrees to each other so for any finite amount of liquid the resulting tidal displacement inline with the moon must certainly be less. By cancel do you mean the displacement due to the sun cancels that due to the moon? if so i don't think it can since in magnitude it is always less. $\endgroup$ Apr 28 at 1:26

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