When you look at a solar eclipse, you see that

*The apparent diameters of the sun and the moon are
almost exactly equal.*

It follows that

*If the sun were as dense as the moon, the solar
tidal force would be almost exactly equal to the
lunar tidal force.*

Here is why:

Let *r* be the radius of the moon, and *R* the radius of the sun;
let *d* be the distance of the moon, and *D* the distance of the sun;
let *rho* be the density of the moon, and *Rho*
the density of the sun.

The tidal force of the moon separating two points of mass 1,
lined up with the
moon and at distance *d*
and *d*+1 (*d* much larger than 1) is approximately
the *derivative* of the gravitational field strength of
the moon (mass = *m*) at distance *d*:

d/d*x* [*G* *m* / *x*^2] evaluated at *d*,
i.e. -2 *G* *m* / *d*^3.

Writing *m* in terms of *r* and *rho*:
*m*=(4/3)*pi* *r*^3 *rho*
(mass = volume times density) gives the moon's tidal force
as

-2 (4/3) *G* *pi* *rho* *r*^3 / *d*^3 =
(-8/3) *G* *pi* *rho* (*r* / *d*)^3.

Doing the same calculation for the solar tidal force gives
the same expression, but with *Rho*, *R*, *D* instead.

Now the equality of the apparent diameters of the moon and the sun
means that *r* / *d* and *R* / *D* are almost exactly
equal. So if *rho* and *Rho* were equal, the
tidal forces would be almost exactly the same!

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