## Why looking at a solar eclipse should tell you that the solar and lunar tides are of comparable strength.

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/dx [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!

Back to Main Tide page.