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

CC BY-SA 4.0
2023-03-01T00:54:09.823Z
2023-03-01T00:54:09.823Z
Michael_1812
<p>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:</p>
<p><span class="math-container">$$\Big|\frac{V_T}{g}\Big| \approx \frac{3}{4}\frac{M_\odot R_p^4}{M_p a^3}$$</span></p>
<p>where <span class="math-container">$V_T$</span> is the tide generating potential, <span class="math-container">$g$</span> is the gravity, <span class="math-container">$M_p$</span> is the mass of the planet, <span class="math-container">$R_p$</span> the radius and <span class="math-container">$a$</span> is the semi-major axis of the orbit. The displacement would then be <span class="math-container">$\xi = h\frac{V_T}{g}$</span>. This does not make a distinction between radial amplitudes at the equator or at the poles. I'm asking because I was reading <a href="https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/2018JE005569" rel="noreferrer">this article</a>. 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 <span class="math-container">$\xi = 0.64$</span>m instead of <span class="math-container">$\xi = 1.93$</span>m.</p>
<p>The tidal potential in my notes is
<span class="math-container">$$V_T(r) = -\frac{GM}{a}\Big(\frac{r}{a}\Big)^2(1 + 3e\cos(M))\Big[-\frac{1}{2}P_2^0(\cos\theta) + \frac{1}{4}P_2^2(\cos\theta)(\cos(2\lambda) + 4e\sin(M)\sin(2\lambda)\Big] $$</span>
<span class="math-container">$e$</span> is the eccentricity, the <span class="math-container">$P$</span> are Legendre polynomials: <span class="math-container">$P_2^0 = \frac{1}{2}(3\cos^2\theta - 1)$</span> and <span class="math-container">$P_2^2 = 3\sin^3\theta$</span>. <span class="math-container">$\lambda$</span> 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 &quot;amplitude&quot;.</p>
<p>The fact that no method of calculation is mentioned in the article makes me think it's either really obvious or really well known.</p>

<planet><gravity><tidal-forces>
Calculating the radial tidal amplitude on a planet from the fluid Love numbers
theWrongAlice
# Calculating the radial tidal amplitude on a planet from the fluid Love numbers

(Asked by: username_0 on 2021-05-13)

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](https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/2018JE005569). 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.

**Tags:** planet, gravity, tidal-forces

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## 1 answers

### Answer by username_1 on 2023-03-01. Score: 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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