How do you calculate tidal amplitudes from the fluid Love numbers? In my course on planetary physics I saw an approximate expression for the displacement from equilibrium tide:

$$\Big|\frac{V_T}{g}\Big| \approx \frac{3}{4}\frac{M_\odot R_p^4}{M_p a^3}$$

where $V_T$ is the tide generating potential, $g$ is the gravity, $M_p$ is the mass of the planet, $R_p$ the radius and $a$ is the semi-major axis of the orbit. The displacement would then be $\xi = h\frac{V_T}{g}$. This does not make a distinction between radial amplitudes at the equator or at the poles. I'm asking because I was reading this article. I was hoping the radial amplitude would correspond to my formula above (given to me without reference or derivation), but with a Love number h = 0.77 I get $\xi = 0.64$m instead of $\xi = 1.93$m.

The tidal potential in my notes is $$V_T(r) = -\frac{GM}{a}\Big(\frac{r}{a}\Big)^2(1 + 3e\cos(M))\Big[-\frac{1}{2}P_2^0(\cos\theta) + \frac{1}{4}P_2^2(\cos\theta)(\cos(2\lambda) + 4e\sin(M)\sin(2\lambda)\Big] $$ $e$ is the eccentricity, the $P$ are Legendre polynomials: $P_2^0 = \frac{1}{2}(3\cos^2\theta - 1)$ and $P_2^2 = 3\sin^3\theta$. $\lambda$ is the longitude measured with respect to the long axis of the planet. I played around with this formula a bit, but I never find an amplitude of 1.93. It could be that this is just not what is meant by radial "amplitude".

The fact that no method of calculation is mentioned in the article makes me think it's either really obvious or really well known.


1 Answer 1


Contrary to what you are saying, the expression $\xi = h\frac{V_T}{g}$ does make a distinction between radial amplitudes at the equator or at the poles. This distinction is ensured by $V_T$ being a function of the coordinates $\theta$ and $\lambda$.


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