Astronomy Answers
Evolution of Lunar Orbit


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1. Angular Momentum ... 1.1. Now ... 2. Energy ... 2.1. Definition and Formulas ... 2.2. Now ... 3. Evolution ... 3.1. Models

\( \newcommand{\dd}[2]{\frac{\text{d}#1}{\text{d}#2}} \)

1. Angular Momentum

The angular momentum \( \vec{L} \) of a point mass, with respect to a certain point of origin, is equal to

\begin{equation} \vec{L} = \vec{r} × m \vec{v} \label{eq:lvec} \end{equation}

where \( \vec{r} \) is the vector from the origin to the location of the point mass, \( m \) is the mass, and \( \vec{v} \) is the velocity of the point mass.

We ignore the influence of the Sun (which is less here) and regard only the Earth and the Moon in their orbits around their common center of gravity. To make things simpler we'll assume that the orbits are circles. Then equation \ref{eq:lvec} simplifies to

\begin{equation} L = m v r = m r^2 Ω \end{equation}

if the distance \( r \) is measured relative to the center of the circular orbits (and hence relative to the barycenter of the system), and \( Ω \) is the angular rotation speed in radians per second. We also assume that the Earth and Moon are spherical and point symmetrical.

For spherical, point-symmetrical objects, the spin angular momentum \( L' \) is equal to

\begin{equation} L' = α m R^2 ω \end{equation}

where \( m \) is the total mass, \( ω \) is the angular rotation speed (radians per second), \( R \) is the radius, and \( α \) is a dimensionless number that indicates how strongly the mass is concentrated toward the center. For a homogeneous sphere (with the same mass density everywhere) \( α = 0.4 \). If the mass is concentrated more toward the center, then \( α < 0.4 \).

We use subscript ₁ for Earth and ₂ for the Moon. The masses are \( m_1 \) and \( m_2 \). The relative masses (compared to the total) are \( μ_1 \) and \( μ_2 \). The radii are \( R_1 = 6378 \text{ km} \) and \( R_2 = 1738 \text{ km} \). The distances to the common center of gravity are \( r_1 \) and \( r_2 \). The sidereal spin periods are \( T_1 \) and \( T_2 \). The sidereal orbital period is \( P \). The orbital speeds are \( v_1 \) and \( v_2 \). The sidereal orbital angular speed is \( Ω \). The sidereal spin angular speeds are \( ω_1 \) and \( ω_2 \). The mass concentration coefficients are \( α_1 = 0.3306 \) and \( α_2 = 0.392 \).

We have:

\begin{align} r & = r_1 + r_2 \\ m & = m_1 + m_2 = 6.0472×10^{24} \text{ kg} \\ μ_1 & = \frac{m_1}{m} = 0.98785 \\ μ_2 & = \frac{m_2}{m} = 0.012151 \\ r_1 & = r μ_2 \\ r_2 & = r μ_1 \\ Ω & = \frac{2π}{P} \\ v_1 & = r_1 Ω = \frac{2π r_1}{P} = \frac{2π μ_2 r}{P} \\ v_2 & = r_2 Ω = \frac{2π r_2}{P} = \frac{2π μ_1 r}{P} \\ v & = v_1 + v_2 \\ v_1 & = v μ_2 \\ v_2 & = v μ_1 \\ ω_1 & = \frac{2π}{T_1} \label{eq:T_1toω_1} \\ ω_2 & = \frac{2π}{T_2} \end{align}

and, from Kepler's Third Law,

\begin{align} P & = \frac{2π}{\sqrt{Gm}} r^{3/2} \\ Ω & = \sqrt{\frac{Gm}{r^3}} \end{align}

where \( G \) is the Universal Gravitational Constant, equal to 6.67259 × 10−11 and \( a \) is the semimajor axis of the orbit; and, because the Moon always shows the same face to us,

\begin{equation} T_2 = P ⇔ ω_2 = Ω \end{equation}

In the Earth-Moon system we can find the following repositories of angular momentum:

orbital angular momentum \( L_1 \) of the Earth

\begin{equation} L_1 = m_1 r_1^2 Ω = \sqrt{G m^3 r} μ_1 μ_2^2 \end{equation}

spin angular momentum \( L_1' \) of the Earth

\begin{equation} L_1' = α_1 m_1 R_1^2 ω_1 \end{equation}

orbital angular momentum \( L_2 \) of the Moon

\begin{equation} L_2 = m_2 r_2^2 Ω = \sqrt{G m^3 r} μ_1^2 μ_2 \end{equation}

spin angular momentum \( L_2' \) of the Moon

\begin{equation} L_2' = α_2 m_2 R_2^2 ω_2 = α_2 m_2 R_2^2 Ω = α_2 μ_2 R_2^2 \sqrt{\frac{G m^3}{r^3}} \end{equation}

total \( L_\text{tot} \)

\begin{equation} L_\text{tot} = L_1 + L_1' + L_2 + L_2' = μ_2 \sqrt{G m^3 r} \left( μ_1 + α_2 \left( \frac{R_2}{r} \right)^2 \right) + α_1 m_1 R_1^2 ω_1 \label{eq:ltot} \end{equation}

1.1. Now

For the current situation we have \( T_1 = 86164.091 \text{ s} \), \( T_2 = P = 2360591.5 \text{ s} \) and \( r = 384748 \text{ km} \). If we fill in all the values then we find \( L_\text{tot} \) = 3.4458 × 1034 kgm²/s. If we then divide by the total of the angular momenta, then we find

\({L_1}\) 0.01008
\({L_1'}\) 0.1700
\({L_2}\) 0.8199
\({L_2'}\) 6.7206 × 10−6

So most of the angular momentum is in the Moon's orbit (82%) and in the Earth's spin (17%).

2. Energy

2.1. Definition and Formulas

The (kinetic) energy contained in motion of a mass \( m \) at speed \( v \) is

\begin{equation} E = \frac{1}{2} m v^2 \end{equation}

The (kinetic) energy contained in spin of a spherical point-symmetrical object with mass \( m \), radius \( R \), spin rate \( ω \), and mass concentration coefficient \( α \) is

\begin{equation} E' = \frac{1}{2} α m R^2 ω^2 \end{equation}

The (potential) energy due to the locations in the gravity field for two masses \( m_1 \) and \( m_2 \) at distance \( r \) from each other is equal to

\begin{equation} U = -\frac{G m_1 m_2}{r} \end{equation}

With the same symbols as in chapter 1 we find for the total energy \( E \) in the Earth-Moon system:

energy due to the orbital speed of the Earth

\begin{equation} E_1 = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} μ_1 μ_2^2 m v^2 \end{equation}

energy due to the spin of the Earth

\begin{equation} E_1' = \frac{1}{2} α_1 m_1 R_1^2 ω_1^2 = \frac{1}{2} α_1 μ_1 m R_1^2 ω_1^2 \end{equation}

energy due to the orbital speed of the Moon

\begin{equation} E_2 = \frac{1}{2} m_2 v_2^2 = \frac{1}{2} μ_1^2 μ_2 m v^2 \end{equation}

energy due to the spin of the Moon

\begin{equation} E_2' = \frac{1}{2} α_2 m_2 R_2^2 ω_2^2 = \frac{1}{2} α_2 m_2 R_2^2 Ω^2 = \frac{1}{2} \frac{α_2 μ_2 R_2^2 G m^2}{r^3} \end{equation}

energy due to the location of the Earth and the Moon in the gravity field

\begin{equation} U = -\frac{G m_1 m_2}{r} = -\frac{μ_1 μ_2 G m^2}{r} \end{equation}

total \( E_\text{tot} \)

\begin{equation} E_\text{tot} = E_1 + E_1' + E_2 + E_2' + U = μ_2 \frac{G m^2}{2r} \left( α_2 \left( \frac{R_2}{r} \right)^2 - μ_1 \right) + \frac{1}{2} α_1 m_1 R_1^2 ω_1^2 \label{eq:etot} \end{equation}

The energy is equal to zero when the Earth and the Moon are infinitely far from each other and no longer orbit around each other or rotate around their own axis.

2.2. Now

If we fill in all of the current values then we find a total of 1.7522 × 1029 J. For comparison: that is about 1700 times more than the whole country of the Netherlands used in 2004. If we divide all values by the total then we find

\({E_1}\) 0.002637
\({E_1'}\) 1.2171
\({E_2}\) 0.2144
\({E_2'}\) 1.7560 × 10−6
\({U}\)−0.4341

so most of the kinetic energy is tied up in the spin of the Earth and (about five times smaller) in the orbital speed of the Moon.

3. Evolution

Formula \ref{eq:ltot} contains only two free parameters: the distance \( r \) between the Earth and the Moon, and the spin rate \( ω_1 \) of the Earth, which is related to the spin period \( T_1 \) of the Earth, i.e., with the length of the day. The other parameters, such as the mass fraction \( μ_1 \) of the Earth or the mass concentration coefficient \( α_2 \) of the Moon belong to those objects and not to their motion, so we expect that they will remain constant. The spin rate (or spin period) of the Moon is not a free parameter, because we have assumed that it is always equal to the orbital angular speed (or orbital period), like it is now. Tidal forces causes those two periods to be equal today, so they can keep them equal also in the future.

Formula \ref{eq:ltot} can be rewritten to

\begin{align} ω_1 & = \frac{L - μ_2 \sqrt{G m^3 r} \left(μ_1 + α_2 \left( \frac{R_2}{r} \right)^2 \right)}{α_1 m_1 R_1^2} = A - Br^{1/2} - Cr^{−3/2} \label{eq:abc} \\ A & = \frac{L}{α_1 m_1 R_1^2} = 0.0004289129 \\ B & = \frac{μ_1 μ_2 \sqrt{G m^3}}{α_1 m_1 R_1^2} = 1.814881×10^{−8} \\ C & = \frac{α_2 μ_2 R_2^2 \sqrt{G m^3}}{α_1 m_1 R_1^2} = 21754.2 \end{align}

with which we can calculate which spin rate \( ω_1 \) of the Earth belongs to a distance \( r \) between the Earth and the Moon. The spin period \( T_1 \) then follows from equation \ref{eq:T_1toω_1}.

3.1. Models

According to the modern theory of tidal evolution, the principal tide of the Moon on the Earth (the so-called M2 tide) generates changes in the semimajor axis \( a \) of the Moon's orbit proportional to

\begin{equation} \dd{r}{t} ∝ \frac{(ω_1 - Ω) τ}{r^5} \end{equation}

where \( τ \) is the lag time of the Earth's tides behind the motion of the Moon.

Today the distance of the Moon from the Earth increases at a rate about 3.7 cm (1.4 in) per year. If we assume that the delay time \( τ \) remains the same, then we can use the \( ω_1 \) and \( Ω \) of today to calculate the constant of proportionality, and then we find the model

\begin{equation} \dd{r}{t} = 1.41×10^{38} \frac{ω_1 - Ω}{r^5} \label{eq:drdt} \end{equation}

A procedure to use this to calculate the past and the future is as follows:

  1. Create an array of distances \( r \), for example with fixed intervals, running between a reasonable lower limit and a reasonable upper limit (for example the 13938 km and 554523 km that we found before).
  2. For each \( r \) from the array, calculate \( \dd{r}{t} \) from formula \ref{eq:drdt}.
  3. If the difference between successive \( r \) is equal to \( ∆r \), then an estimate for the time that it takes the Moon to move from the lower to the higher \( r \) is \( ∆t = \frac{∆r}{\left( \dd{r}{t} \right)} \).
  4. If you now sum all \( ∆t \) for the first until the desired \( r \), then you have an estimate for how much time it takes the Moon to move from the first to the desired \( r \). Subtract the time that goes with the \( r \) of today, and you find the time relative to today. If you use SI units throughout, then you'll find the time measured in seconds.



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Last updated: 2016-02-07