Theancient Chinese court astronomers were able to predict lunar and solar eclipses with great reliability. The fact that such eclipses arise only occasionally—while otherwise we have a full moon or a new moon—is caused by the inclination of the orbital plane of the earth–moon system from the

Question

Theancient Chinese court astronomers were able to predict lunar and solar eclipses with great reliability. The fact that such eclipses arise only occasionally—while otherwise we have a full moon or a new moon—is caused by the inclination of the orbital plane of the earth–moon system from the ecliptic, i.e., the orbital plane of the motion of the common center of gravity about the sun. This inclination is about 5.15°. It is not fixed in space but precesses because of the tidal forces exerted by the sun. This leads to the so-called Saros cycle, which is of great importance for the prediction of eclipses.
Consider the earth–moon system as a dumbbell-shaped top which rotates about its center of gravity [latex]S_{p}[/latex]; the center of gravity orbits about the sun on a circular path. The gravitation force between the earth and moon just balances the centrifugal force resulting from the eigenrotation of the system and thus fixes the almost rigid dumbbell length [latex]r_{0}[/latex]. The gravitation of the sun and the centrifugal force due to orbiting about the sun don’t compensate for each body independently but lead to resulting tidal forces. These forces create a torque [latex]M_{0}[/latex] on the top. Calculate [latex]M_{0}[/latex] for the sketched position where it just takes its maximum value. Realize that [latex]M_{0}[/latex] on the (monthly and annual) average has a quarter of this value. Calculate from this the precession period [latex]T_{p}[/latex]. Can you find arguments for why the actual Saros cycle of 18.3 years is notably longer?
Hint: The only data you need for the calculation are the distances [latex]r_{0}[/latex], the length of year, and the length of the sidereal month.

Accepted Answer

[latex]R_{0}[/latex] is defined as the vector pointing from the center of gravity of the sun to the center of gravity of the earth-moon system. The coordinate origin of the system is in the center of gravity of the sun.

Let R be given in cylindrical coordinates:

[latex]R = R_{0} +ΔR[/latex]

[latex]= R_{0}e_{r} + ΔR_{r}e_{r} + ΔR_{\varphi}e_{\varphi} + ΔR_{z}e_{z}[/latex]

with

[latex]|ΔR| \ll |R_{0}|.[/latex]

We then have

[latex]|R| ≈ R_{0} \left(1+ \frac{2ΔR_{r}}{R_{0}} ight)^{1/2}.[/latex]

Hence, we can write for the gravitational force

[latex]F_{Gr}(R)=−\frac{γmM}{|R|^{3}} R[/latex]

[latex]≈−\frac{γmM}{R^{3}_{0}} \left(1− \frac{3ΔR_{r}}{R_{0}} ight) (R_{0}e_{r} +ΔR_{r}e_{r} +ΔR_{\varphi}e_{\varphi} +ΔR_{z}e_{z})[/latex]

[latex]≈ e_{r} \left(−\frac{γmM}{R^{3}_{0}} (R_{0} −2ΔR_{r} ) ight)+ e_{\varphi} \left(-\frac{γmM}{R^{3}_{0}} ΔR_{\varphi} ight)[/latex]

[latex]+e_{z} \left(−\frac{γmM}{R^{3}_{0}} ΔR_{z} ight).[/latex] (13.27)

The two masses m and M don’t yet have a special meaning. We now consider the motion of the earth–moon system as a two-body problem with an external force:

[latex]R_{CM} := R_{0} = \frac{m_{E}R_{E} +m_{M} · R_{M}}{m_{E} +m_{M}}[/latex] (CM means center of mass),

[latex]V_{CM} = \frac{m_{E}V_{E} + m_{M}V_{M}}{m_{E} + m_{M}},[/latex]

[latex]B_{CM} = \frac{m_{E}B_{E} +m_{M}B_{M}}{m_{E} +m_{M}},[/latex]

[latex]= \frac{F_{ES} +F_{EM} +F_{ME} +F_{MS}}{m_{E}+m_{M}}[/latex] (S means sun).

According to (13.27),

[latex]ΔR_{E} = r_{E},[/latex]

[latex]ΔR_{M} = r_{M},[/latex]

[latex]m_{E}r_{E} =−m_{M}r_{M};[/latex]

further from (13.27) we have

[latex]B_{CM} = \frac{1}{m_{E} +m_{M}} \left(−γ \frac{(m_{E} +m_{M})M_{S}}{R^{2}_{0}} · e_{r} ight)[/latex]

[latex]=−\frac{γM_{CM}}{R^{2}_{0}} · e_{r} =−ω^{2}_{CM}R_{0} · e_{r}[/latex] (circular acceleration). (13.28)

The last equality follows from the equilibrium condition for the center of gravity. The magnitude of the gravitational acceleration must be equal to the magnitude of the circular acceleration. From this, it follows that the center of mass CM rotates with the frequency [latex]ω_{CM}[/latex] about the sun at the distance [latex]R_{0}[/latex].

We further know the following values:

[latex]T_{CM} = \frac{2π}{ω_{CM}}= 365 days; R_{0} = 149.6 · 10^{6} km.[/latex] (13.29)

We are mainly interested in the motion of the earth–moon system. To this end we consider the relative distance [latex]r_{rel}[/latex] between the earth and moon.

[latex]r_{rel} = R_{E} −R_{M}, |r_{rel}| = r_{0},[/latex]

[latex]P_{rel} = μ(V_{E} −V_{M}) with μ = \frac{m_{E} · m_{M}}{m_{E} + m_{M}},[/latex]

[latex]\frac{dP_{rel}}{dt}= μ(B_{E} − B_{M}) = μ \left(\frac{F_{ES}}{m_{E}} + \frac{F_{EM}}{m_{E}} − \frac{F_{ME}}{m_{M}}− \frac{F_{MS}}{m_{M}} ight),[/latex]

[latex]F_{EM} =−γ \frac{m_{E}m_{M}}{r^{3}_{0}}r_{rel}=−mω^{2}_{M}r_{rel}.[/latex]

The last equality holds because of the equilibrium condition, as for the center-of-mass acceleration.

The relative distance also performs a circle with the sidereal period of the moon at the distance [latex]r_{0}[/latex]. One has

[latex]T_{M} = \frac{2π}{ω_{M}}= 27 days+ 8 hours,[/latex] (13.30)

[latex]r_{0} = r_{E} +r_{M} = 0.384 · 10^{6} km.[/latex]

We thus obtain the following combined motion:

[latex]\left. \begin{matrix}R_{E} = R_{0} + \frac{m_{M}}{m_{M} +m_{E}}r_{rel} \R_{M} = R_{0} − \frac{m_{E}}{m_{M} +m_{E}}r_{rel} \end{matrix} ight\} \quad epicycle motion.[/latex] (13.31)

The angular momentum with respect to the center of mass CM is given by

[latex]L_{0} = r_{E} ×m_{E}(V_{E} −V_{CM})+ r_{M} ×m_{M}(V_{M} −V_{CM})[/latex]

[latex]= m_{E} \frac{m_{M}}{m_{E} + m_{M}}r_{rel} × \left(\frac{m_{M}}{m_{E} +m_{M}} ight)v_{rel}[/latex]

[latex]+m_{M} \frac{m_{E}}{m_{E} + m_{M}}r_{rel} × \left(\frac{m_{E}}{m_{E} +m_{M}} ight)v_{rel}[/latex]

[latex]= \frac{m_{E}m_{M}}{m_{E} + m_{M}}r_{rel} ×v_{rel}[/latex]

[latex]= μ · r_{rel} ×v_{rel}[/latex]

[latex]≈ \frac{m_{E}m_{M}}{m_{E} + m_{M}} ω_{M} · r^{2}_{0}· l_{EM},[/latex] (13.32)

where [latex]l_{EM}[/latex] represents a normal vector to the orbital plane of the earth–moon system. The relation (13.32) does not hold exactly, since the motion of the earth–moon system is not perfectly circular, but can be well approximated by a circle.

The coordinate system at the center of gravity is oriented just as at the origin. The total angular momentum with respect to the sun is evaluated as

[latex]L_{tot} = m_{E}R_{E} ×V_{E} +m_{M} · R_{M} ×V_{M}[/latex]

[latex]= m_{E} \left(R_{0} + \frac{m_{M}}{m_{M} +m_{E}} r_{E} ight)×\left(V_{CM} + \frac{m_{M}}{m_{M} + m_{E}}v_{rel} ight)[/latex]

[latex]+ m_{M} \left(R_{0} − \frac{m_{E}}{m_{M} + m_{E}}r_{rel} ight)×\left(V_{CM} − \frac{m_{E}}{m_{M} +m_{E}}v_{rel} ight)[/latex]

[latex]= (m_{E} +m_{M})R_{0} ×V_{CM} +μ · r_{rel} ×v_{rel}[/latex]

[latex]= (m_{E} +m_{M})ω_{CM} · R^{2}_{0}· l_{S−CM} + L_{0}[/latex]

[latex]= L_{CM} +L_{0}.[/latex] (13.33)

Similar as in (13.32), here it also holds that [latex]l_{S−CM}[/latex] is a vector normal to the orbital plane defined by the sun and the center of gravity.

[latex]\frac{dL_{tot}}{dt}=(R_{E} ×F_{ES} + R_{M} ×F_{MS}+(R_{E} − R_{M}) ×F_{EM})= 0[/latex];

this means

[latex]\dot{L}_{CM} =−\dot{L}_{0}[/latex]. (13.34)

We now consider the resulting torque [latex]M_{0}[/latex] with respect to the center of mass CM:

[latex]M_{0} = r_{E} ×(F_{ES} + F_{EM} − m_{E}B_{CM})+r_{M} ×(F_{MS} +F_{ME} −m_{M}B_{CM})[/latex]

[latex]= r_{E} ×F_{ES} +r_{M} ×F_{MS}.[/latex]

The second terms in each bracket drop because the force and position vector have the same orientation. The third two terms cancel because

[latex]r_{E}m_{E} =−r_{M}m_{M}.[/latex]

By inserting [latex]r_{rel}[/latex], it follows that

[latex]M_{0} = \frac{m_{M}}{m_{E} +m_{M}}r_{rel} ×F_{ES} − \frac{m_{E}}{m_{E} +m_{M}}r_{rel} ×F_{MS}.[/latex]

Using (13.27) for the two force vectors [latex]F_{ES} and F_{MS}[/latex], one obtains by simplifying [latex]M_{0}[/latex] with respect to cylindrical coordinates:

[latex]M_{0} = \frac{3γm_{E}M}{R^{3}_{0}}(ΔR_{rE}(r_{rel}×e_{r} )).[/latex] (13.35)

Here, [latex]ΔR_{\varphi E} and ΔR_{zE}[/latex] were set to zero, according to the definition of the problem. In order not to complicate the problem unnecessarily, we put the coordinate system at the center of gravity and thereby also that at the origin just so that the angular momentum on the average lies in the x, z-plane. This approach is justified since the precession frequency to be calculated is notably less than [latex]ω_{CM}[/latex]. During one revolution about the sun the angular momentum has changed only insignificantly (by about 20°), so that the ecliptic of the earth–moon system has turned only slightly.

Ansatz:

[latex]r^{′}_{rel}= r_{0} \begin{pmatrix} cos β \ sin β \ 0\end{pmatrix}, where β ∼ ω_{M}t.[/latex]

In the nonprimed system the vector has the components

[latex]r_{rel} =\begin{pmatrix} cos α & 0 & −sin α\ 0 & 1 &0 \sin α & 0& cos α \end{pmatrix}, r^{′}_{rel}= r_{0} \begin{pmatrix}cos α cos β \ sin β \ sin α sin β\end{pmatrix}.[/latex]

Ansatz:

[latex]e_{r} =\begin{pmatrix}cos(γ +\varphi_{0}) \sin(γ +\varphi_{0}) \0\end{pmatrix}, where γ ∼ ω_{CM}t.[/latex]

Using the relation

[latex]ΔR_{rE} = \frac{m_{M}}{m_{E} +m_{M}}(r_{rel} · e_{r} ),[/latex]

one obtains, by using the two approaches, the following new formulation of (13.35):

[latex]M_{0} = \frac{3γm_{E}M}{R^{3}_{0}} r^{2}_{0} \frac{m_{M}}{m_{E} + m_{M}}v[/latex]

with

[latex]v =\begin{pmatrix}−sin α cos α cos^{2} β sin(γ + \varphi_{0}) cos(γ + \varphi_{0})−sin α cos β sin β sin^{2}(γ + \varphi_{0}) \sin α cos α cos^{2} β cos^{2}(γ +\varphi_{0}) +sin α cos β sin β sin(γ +\varphi_{0}) cos(γ +\varphi_{0}) \cos^{2} α cos^{2} β · sin(γ + \varphi_{0}) cos(γ + \varphi_{0})−sin β cos β cos α cos^{2}(γ +\varphi_{0})\ +cos α cos β · sin β sin^{2}(γ + \varphi_{0})−sin^{2} β sin(γ +\varphi_{0}) cos(γ +\varphi_{0}) \end{pmatrix}[/latex].

This clumsy expression can be significantly simplified, assuming again that [latex]ω_{P} \gg ω_{CM}[/latex]. This means that the angular momentum [latex]L_{0}[/latex] changes only slightly during one revolution about the sun.

We first consider β. The revolution period of the moon about the earth is about 28 days. The moment [latex]M_{0}[/latex] changes its orientation with varying β; for the “inert” angular momentum, however, only the average momentum [latex]\left\langle M _{0} ight angle [/latex]β counts; this is obtained by averaging over a full period of β. The moment impact (analogous to the force impact) of [latex]M_{0} ext{ and of } \left\langle M _{0} ight angle[/latex] β has the same value, because of the linearity β = [latex]ω_{M}t[/latex] .

[latex]\left\langle M_{0} ight angleβ = \frac{3γMμ}{R^{3}_{0}}r^{2}_{0}[/latex]

[latex]× \begin{pmatrix} − \sin α \cos α \frac{1}{2} \sin(γ + \varphi_{0}) \cos(γ + \varphi_{0}) \ \sin α \cos α \frac{1}{2} \cos^{2}(γ +\varphi_{0}) \ \cos^{2} α \frac{1}{2} \sin(γ +\varphi_{0}) \cos(γ +\varphi_{0})− \frac{1}{2} \sin(γ + \varphi_{0}) \cos(γ + \varphi_{0}) \end{pmatrix}.[/latex]

The same consideration can be made for the “rotating” angle [latex]γ ∼ ω_{CM}t[/latex] , since we assume that [latex]ω_{p} \gg ω_{CM}[/latex]. We therefore average over [latex]\left\langle M_{0} ight angleβ[/latex]:

[latex]〈〈M_{0}〉β〉γ = \frac{3γMμ}{R^{3}_{0}}r^{2}_{0} \begin{pmatrix} 0 \\frac{1}{4} \sin α \cos α \ 0 \end{pmatrix}[/latex]

[latex]=\frac{3γMμ}{R^{3}_{0}}r^{2}_{0}· \frac{1}{4} \sin α \cos α · e_{y} := 〈M_{0}〉.[/latex] (13.36)

Not very much is left over from the extended expression; the resulting acting moment [latex]〈M_{0}〉[/latex] is exactly perpendicular to [latex]L_{0}[/latex] (see Fig. 13.27) and points along the y-direction. If the angular momentum moved (slowly), we imagine the shifted angular momentum as being embedded in a fixed coordinate system and thus get, according to (13.36), the same result. [latex]〈M_{0}〉[/latex] is constant and always perpendicular to [latex]L_{0}[/latex]. It therefore causes a precession.

But first we will illustrate (13.36) that has been obtained in a rather mathematical way: In this situation, we get a maximum moment. The moment M(β) now points in the [latex]y^{′}[/latex] -direction for all possible β. For β = 90° or β = 270°, the vector M vanishes. On the average, we thus obtain

[latex]〈M(β)〉 = 2〈M_{0}〉[/latex].

The diagram shows that there are also two “maximum” orientations with respect to [latex](γ + \varphi_{0})[/latex]. Between these positions M(β) must vanish. One thus obtains altogether

[latex]〈〈M(β, γ )〉β〉γ = 〈M_{0}〉[/latex].

Thus, the result (13.36) is also clearly understood.

With (13.32) in our defined coordinate system, we have

[latex]L_{0} = μ · ω_{M} · r^{2}_{o} \begin{pmatrix} −\sin α \ 0 \ \cos α \end{pmatrix}[/latex].

The angular momentum [latex]L_{0}[/latex] now precesses:

[latex]ω_{p} = \frac{M_{S}}{L · sin α} =\frac{ 〈M_{0}〉}{L_{0} ·sin α}[/latex]

[latex]= \frac{3}{4} · \frac{γM_{S}μr^{2}_{0} sin α cos α}{R^{3}_{0}μω_{M}r^{2}_{0} sin α}= \frac{3}{4} \frac{γM_{S}}{R^{3}_{0}} cos α · \frac{1}{ω_{M}} .[/latex] (13.37)

It has been shown at the beginning (13.28) that

[latex]\frac{γM_{S}}{R^{2}_{0}} = ω^{2}_{CM}R_{0} ⇒ \frac{γM_{S}}{R^{3}_{0}}= ω^{2}_{CM} =\left(\frac{2π}{T_{CM}} ight)^{2},[/latex]

and with

[latex]ω_{M} = \frac{2π}{T_{M}},[/latex]

one gets

[latex]ω_{p} = \frac{3}{4} cos α\left(\frac{4π^{2}}{T^{2}_{CM}} ight)· \frac{T_{M}}{2π}= \frac{3}{2}cos α \frac{T_{M}}{T^{2}_{CM}},[/latex] (13.38)

and for [latex]T_{p}[/latex]

[latex]T_{p} = \frac{2π}{ω_{p}}= \frac{4}{3}· \frac{1}{cos α} · \frac{T^{2}_{CM}}{T_{M}}.[/latex] (13.39)

With [latex]T_{CM} ≈ 365.25 days, T_{M} ≈ 27.3 days, and α ≈ 5.5°[/latex], one obtains

[latex]T_{p}[/latex] ≈ 17.9 years. (13.40)

The fact that the actual Saros cycle is larger by about 2% is partly due to the approximation when averaging γ (the angular momentum actually moves slightly), but possibly due to the elliptic path of the moon about the earth. In any case, the result is relatively accurate, considering the approximations made.

From (13.34), it further follows that

[latex]\dot{L}_{CM} =−\dot{L}_{0},[/latex]

i.e., the “large” angular momentum vector [latex]L_{CM}[/latex] runs through an opposite precession cone.

Question 13.9: Theancient Chinese court astronomers were able to predict lu...

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