How much do surfaces tilt due to tidal forces?

[Adrian B]

I've read that tides deform the Earth's crust by about 40cm. When I try to visualize the tidal bulge approaching me and then receding away from me, it seems like the local surface under my feet would tilt slightly one way as the bulge approaches, then level out, and then tilt slightly the other way as the bulge recedes. Similar to a surfboard as a wave passes underneath it.

Is this picture correct? If "yes" does anybody have a rough figure for the maximum tilt one would "experience" on Earth due to this? Arc-seconds? Micro arc-seconds?

 

[andrewkirk]

Say the deformation is to approximately an ellipse. Then the ellipse has a ratio of semi-major to semi-minor axes $\frac{1+\epsilon}{1-\epsilon}$ where $\epsilon$ is 40cm / 6371km = $6\times 10^{-8}$. This can be modeled by the ellipse:

$$\frac{x^2}{(1+\epsilon)^2}+\frac{y^2}{(1-\epsilon)^2}=1$$

The tilt is the angle between the tangent to the ellipse and the line from the point on the ellipse to the origin, which is the centre of mass. That tilt is zero at the x and y intercepts. So it seems reasonable to guess that maximum tilt might be near angles of 45 degrees to the axes. There the angle of the line to COM is 45 degrees. The gradient of the ellipse is:

$$\frac{d}{dx}\left[(1-\epsilon)\sqrt{1-\frac{x^2}{(1+\epsilon)^2}}\right]=-\frac{(1-\epsilon)x}{\sqrt{1-\frac{x^2}{(1+\epsilon)^2}}}$$

$x$ is approximately $\frac{1}{\sqrt{2}}$ at that point, so this gives the gradient as $1-1.3\times 10^{-7}$. Taking the arctan gives an angle that differs from 45 degrees by $3\times 10^{-6}$. Multiplying that by $60^2$ gives about 0.013 seconds of arc.

E&OE

 

[Baluncore]

Multiplying that by 60^2 gives about 0.013 seconds of arc.

I agree.

An ellipse can be approximated by a circle of constant radius plus a significant 2'nd harmonic sinewave. One cycle of that 2'nd harmonic covers about 20 Mm, as that is half the Earth circumference, ignoring the flattening. (Remember that Napoleon declared the distance from the North Pole to the Equator, through Paris to be 10Mm, don't you just love the metric system).
If the vertical peak to peak Earth Tide amplitude is 0.4m, half that is the sinewave amplitude = 0.2 m
So scale the 20 Mm by 2π to get 3183100 m. The maximum slope of Sine is at zero = Cos(0) = 1.
Maximum slope of surface is therefore 0.2 in 3183100 = 6.283e-8.
Atan(6.283e-8) = 3.6e–6 deg = 0.01296 arcsec

 

[Adrian B]

Thanks folks!