IPhO 1996 · Theory problem 3

Mid-ocean tides in the Earth-Moon system

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In this problem we consider some gross features of the magnitude of mid-ocean tides on earth. We simplify the problem by making the following assumptions:

(i) The earth and the moon are considered to be an isolated system,
(ii) the distance between the moon and the earth is assumed to be constant,
(iii) the earth is assumed to be completely covered by an ocean,
(iv) the dynamic effects of the rotation of the earth around its axis are neglected, and
(v) the gravitational attraction of the earth can be determined as if all mass were concentrated at the centre of the earth.

The following data are given:

Mass of the earth: \(M = 5.98 \cdot 10^{24}\ \mathrm{kg}\)
Mass of the moon: \(M_m = 7.3 \cdot 10^{22}\ \mathrm{kg}\)
Radius of the earth: \(R = 6.37 \cdot 10^6\ \mathrm{m}\)
Distance between centre of the earth and centre of the moon:
\(L = 3.84 \cdot 10^8\ \mathrm{m}\)
The gravitational constant: \(G = 6.67 \cdot 10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\).

a) The moon and the earth rotate with angular velocity \(\omega\) about their common centre of mass, \(C\). How far is \(C\) from the centre of the earth? (Denote this distance by \(l\).)

Determine the numerical value of \(\omega\). (2 points)

We now use a frame of reference that is co-rotating with the moon and the center of the earth around \(C\). In this frame of reference the shape of the liquid surface of the earth is static.

Figure 1

In the plane \(P\) through \(C\) and orthogonal to the axis of rotation the position of a point mass on the liquid surface of the earth can be described by polar coordinates \(r,\varphi\) as shown in the figure. Here \(r\) is the distance from the centre of the earth.

We will study the shape

\[ r(\varphi)=R+h(\varphi) \]

of the liquid surface of the earth in the plane \(P\).

b) Consider a mass point (mass \(m\)) on the liquid surface of the earth (in the plane \(P\)). In our frame of reference it is acted upon by a centrifugal force and by gravitational forces from the moon and the earth. Write down an expression for the potential energy corresponding to these three forces.

Note: Any force \(F(r)\), radially directed with respect to some origin, is the negative derivative of a spherically symmetric potential energy \(V(r)\):
\(F(r)=-V'(r)\). (3 points)

c) Find, in terms of the given quantities \(M,M_m\), etc, the approximate form \(h(\varphi)\) of the tidal bulge. What is the difference in meters between high tide and low tide in this model?

You may use the approximate expression

\[ \frac{1}{\sqrt{1+a^2-2a\cos\theta}} \approx 1+a\cos\theta+\frac12a^2(3\cos^2\theta-1), \]

valid for \(a\) much less than unity.

In this analysis make simplifying approximations whenever they are reasonable. (5 points)

Official English text of the problem.

Answers

  • a) The centre of mass lies toward the Moon, at \[ \boxed{l=\frac{M_mL}{M+M_m}\approx4.6\times10^6\ \mathrm{m}} \] from the Earth’s centre. The angular velocity is \[ \boxed{\omega=\sqrt{\frac{G(M+M_m)}{L^3}} \approx2.67\times10^{-6}\ \mathrm{rad\,s^{-1}}.} \]

  • b) Up to an additive constant, \[ \boxed{ V(r,\varphi)= -\frac{GMm}{r} -\frac{GM_m m}{\sqrt{L^2+r^2-2Lr\cos\varphi}} -\frac12m\omega^2\left(r^2+l^2-2lr\cos\varphi\right). } \]

  • c) The leading tidal shape is \[ \boxed{ h(\varphi)=h_{\min} +\frac{3M_mR^4}{2ML^3}\cos^2\varphi, } \] where the constant \(h_{\min}\) fixes the reference sea level. Equivalently, relative to the mean height \(\bar h_P\) around the section, \[ \boxed{ h(\varphi)-\bar h_P =\frac{3M_mR^4}{4ML^3}\cos2\varphi \approx(0.27\ \mathrm{m})\cos2\varphi. } \] High tides occur at \(\varphi=0,\pi\), and low tides at \(\varphi=\pi/2,3\pi/2\). Their height difference is \[ \boxed{\Delta h=\frac{3M_mR^4}{2ML^3}\approx0.53\ \mathrm{m}.} \]

Solution

a)

The centre of mass lies on the line joining the Earth and Moon. Taking distances from it gives

\[ Ml=M_m(L-l), \]

and therefore

\[ l=\frac{M_m}{M+M_m}L =\frac{7.3\times10^{22}}{5.98\times10^{24}+7.3\times10^{22}} (3.84\times10^8) =4.631\ldots\times10^6\ \mathrm{m}. \]

Thus \(C\) lies inside the Earth, since \(l<R\).

The Earth moves in a circle of radius \(l\) about \(C\). Its gravitational attraction to the Moon supplies the centripetal force:

\[ M\omega^2l=\frac{GMM_m}{L^2}. \]

Substituting the expression for \(l\),

\[ \boxed{\omega^2=\frac{G(M+M_m)}{L^3}.} \]

Numerically,

\[ \omega =\sqrt{ \frac{(6.67\times10^{-11})(5.98\times10^{24}+7.3\times10^{22})} {(3.84\times10^8)^3} } =\boxed{2.67\times10^{-6}\ \mathrm{rad\,s^{-1}}}. \]

We will also use the exact relation

\[ \boxed{\omega^2l=\frac{GM_m}{L^2}.} \]

b)

Place the origin at the Earth’s centre and take \(\varphi=0\) toward the Moon. The Moon and \(C\) then lie at distances \(L\) and \(l\) along that direction.

By the cosine rule, the distance from the mass point to the Moon is

\[ d=\sqrt{L^2+r^2-2Lr\cos\varphi}. \]

Because the point lies in the plane perpendicular to the rotation axis through \(C\), its distance from that axis is

\[ s=\sqrt{r^2+l^2-2lr\cos\varphi}. \]

The gravitational potential energies are

\[ V_E=-\frac{GMm}{r}, \qquad V_m=-\frac{GM_m m}{d}. \]

The centrifugal force points away from the rotation axis and has magnitude \(m\omega^2s\). Hence

\[ -\frac{dV_{\mathrm{cf}}}{ds}=m\omega^2s \quad\Longrightarrow\quad V_{\mathrm{cf}}=-\frac12m\omega^2s^2. \]

Adding the three contributions gives

\[ \boxed{ V(r,\varphi)= -\frac{GMm}{r} -\frac{GM_m m}{\sqrt{L^2+r^2-2Lr\cos\varphi}} -\frac12m\omega^2(r^2+l^2-2lr\cos\varphi) +\text{constant}. } \]

c)

For a fluid at rest, hydrostatic equilibrium requires

\[ \nabla p=-\rho\nabla\Phi, \qquad \Phi=\frac Vm. \]

Pressure is constant along the free surface, so that surface must be an equipotential of \(\Phi\).

Since

\[ \frac RL=0.0166\ll1, \]

we expand the lunar potential through second order in \(r/L\):

\[ -\frac{GM_m}{d} \approx -\frac{GM_m}{L} -\frac{GM_mr}{L^2}\cos\varphi -\frac{GM_mr^2}{2L^3}(3\cos^2\varphi-1). \]

The centrifugal potential per unit mass is

\[ -\frac12\omega^2(r^2+l^2-2lr\cos\varphi) = -\frac12\omega^2r^2-\frac12\omega^2l^2 +\omega^2lr\cos\varphi. \]

Using the relation found in part a),

\[ -\frac{GM_mr}{L^2}\cos\varphi +\omega^2lr\cos\varphi=0. \]

Thus the uniform part of the Moon’s gravitational field is cancelled by the corresponding centrifugal term. After absorbing terms independent of position into a constant,

\[ \Phi(r,\varphi) \approx -\frac{GM}{r} -\frac12\omega^2r^2 -\frac{GM_mr^2}{2L^3}(3\cos^2\varphi-1) +\text{constant}. \]

Now put \(r=R+h(\varphi)\), with \(|h|\ll R\). The Earth’s potential becomes

\[ -\frac{GM}{R+h} \approx -\frac{GM}{R}+\frac{GM}{R^2}h = -\frac{GM}{R}+gh, \]

where

\[ g=\frac{GM}{R^2}=9.83\ \mathrm{m\,s^{-2}}. \]

In the smaller centrifugal and lunar terms, we may set \(r=R\). For example, the centrifugal correction to the coefficient of \(h\), relative to \(g\), is only

\[ \frac{\omega^2R}{g}=4.62\times10^{-6}. \]

The centrifugal potential evaluated at \(R\) is independent of \(\varphi\) within \(P\), so it contributes only to the constant surface level. The equipotential condition therefore reduces to

\[ gh(\varphi) -\frac{GM_mR^2}{2L^3}(3\cos^2\varphi-1) =\text{constant}. \]

Consequently,

\[ h(\varphi) = H+\frac{M_mR^4}{2ML^3}(3\cos^2\varphi-1), \]

where \(H\) is independent of \(\varphi\). This constant is fixed by the reference sea level or a volume normalization; the angular tidal variation does not depend on it.

Writing the result relative to its minimum gives

\[ \boxed{ h(\varphi)=h_{\min} +\frac{3M_mR^4}{2ML^3}\cos^2\varphi. } \]

Alternatively, using \(\cos^2\varphi=(1+\cos2\varphi)/2\),

\[ \boxed{ h(\varphi)-\bar h_P =\frac{3M_mR^4}{4ML^3}\cos2\varphi. } \]

The surface therefore has two equal maxima along the Earth–Moon line: one facing the Moon and one on the opposite side. Its two minima lie perpendicular to that line in \(P\). Equality of the two bulges holds to the retained order in \(R/L\).

The difference between maximum and minimum is

\[ \begin{aligned} \Delta h &=\frac{3M_mR^4}{2ML^3}\\ &=\frac32 \frac{7.3\times10^{22}}{5.98\times10^{24}} \frac{(6.37\times10^6)^4}{(3.84\times10^8)^3} \ \mathrm{m}\\ &=0.532447\ldots\ \mathrm{m}. \end{aligned} \]

Hence

\[ \boxed{\Delta h\approx0.53\ \mathrm{m}.} \]

The amplitude relative to the mean in \(P\) is half this value, approximately \(0.27\ \mathrm{m}\). This small deformation confirms the validity of the linear expansion in \(h/R\).

The solution was checked against the official solution.