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This question is inspired by the (now looking at it, improper use of, my bad) comments section here in Physics SE. I'm not sure I could explain better than our short discussion does below:

No, the sun is not extremely dense. The conditions at the center of the sun are quite extreme in terms of pressure and temperature, and the density is quite impressive if one takes into account that the sun is made up mainly of hydrogen (plasma) which under "normal" conditions has a very low density, but in absolute terms the sun has a low density. Indeed the moon has about three times the density of the sun, a fact that, given the curious coincidence that sun and moon disks have almost exactly the same size as seen from the earth, explains why the moon has more tidal effect than the sun has. – Marc van Leeuwen

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@MarcvanLeeuwen isn't the increased tidal effect more a result of gravity's strength being inversely proportional to the square of distance between two objects? – TCooper

@TCooper No, since if that were true the sun would easily beat the moon, as it is $p$ times as far away, $p$ times the linear size, and so $p^3$ times as voluminous, for some large factor $p$ of proportionality (I guess around 300, thought I'd need to check the numbers). However tidal effect is proportional to the derivative of the vector-valued gravitational strength (or more properly acceleration), which follows an inverse cube law. That precisely cancels the volume effect, so only density remains. – Marc van Leeuwen

That's very interesting. Frankly I don't follow and will have to read up, so thank you for sharing. I was just referencing some old grade school knowledge... I always understood gravity as a relationship of mass and distance. Not accounting for volume or density at all (highly simplified) though I did find this: https://oceanservice.noaa.gov/education/tutorial_tides/tides02_cause.html Is it simply a case of the noaa needing to update their website? – TCooper

So when it comes to measuring the gravitational effect of the sun and moon on our tides, is the higher density of the moon, or the further distance of the sun, a larger contributing factor to the moon having a greater effect?

TCooper
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We get pretty close to answering all your questions here:

I always understood gravity as a relationship of mass and distance. Not accounting for volume or density at all …

Yes, gravity depends on mass and distance. Mass, in turn, depends on volume and density. In fact, mass is the product of volume and density.

As has been pointed out by other users and by the NOAA page you referenced, tidal forces are inversely proportional to the cube of the distance. They are also directly proportional to the mass and therefore directly proportional to the density.

According to the NOAA page, the effect of distance is a factor of about 59 million.

The densities of the Sun and Moon are about $1.4 \text{ } \text{g} \cdot\text{cm}^-3$ and $3.3 \text{ } \text{g} \cdot \text{cm}^-3$, respectively. This means the effect of density is a factor of less than 3.

Clearly, the larger factor, by a huge margin, is distance.

These numbers look a bit silly (59 million vs 3), because they ignore volume. Like Marc van Leeuwen’s comment explained, the distance and volume effects cancel out, leaving just the density effect – a factor of a little over 2.

Brian Drake
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The tidal effects is related to the difference between gravitional field on 2 opposite points of the surface of the earth.

As the field is proportional to $\frac{M}{r^2}$, the difference is proportional to its derivative $\frac{M}{r^3}$. (Because the earth diameter is small compared to the distance to moon or sun).

$$F_s \propto \frac{M_{sun}}{r_{sun}^3} $$ $$F_m \propto \frac{M_{moon}}{r_{moon}^3} $$

The relation between that figures is about 0,45. That is: the solar effect on tides of sun is less than half the effect of the moon.

$M = \rho \frac{4}{3}\pi R^3$, where $R$ is the radius of a sphere where the density is being measured. As the density of the moon is bigger than the sun, the tidal effects in the surface of each of those bodies is bigger in the case of the moon.

But in our case of tides on earth, if we apply (for the sun for example):

$$F_s \propto \frac{M_{sun}}{r_{sun}^3} = \frac{\frac{4}{3}\pi \rho_{av} r_{sun}^3}{r_{sun}^3} = \frac{4}{3}\pi \rho_{av}$$

$\rho_{av}$ is not here the density of the sun, but only its mass divided by the volume of a sphere with a radius equal to the distance between sun-earth. (Imagine the sun in its future as a red giant - the effect in our tides doesn't change).

The same can be calculated for the moon.

That reasoning is only to show the effect of density on tides. It is real, but we must use that modified concept of average density, not the density of the bodies properly.

Claudio Saspinski
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