Assignment 02 — Derivations in the N-body and Two-body Problems

Notation follows the lecture slides: \(\mathbf{R}_i\) are inertial positions, \(\mathbf{r}_{ij} = \mathbf{R}_j - \mathbf{R}_i\), \(\hat{\mathbf{r}}_{ij} = \mathbf{r}_{ij}/r_{ij}\), \(\mu = G(m_1 + m_2)\), \(\mathbf{h} = \mathbf{r} \times \dot{\mathbf{r}}\), \(\varepsilon = v^2/2 - \mu/r\), \(p = h^2/\mu = a(1-e^2)\), and \(f\), \(E\), \(M\) are the true, eccentric and mean anomalies. Every answer must be a derivation: state what is assumed, and justify each step.

Part A — The N-body Problem

Problem 1 — The tenth integral

The slides list ten classical integrals of the N-body problem but derive only nine of them (the six for the centre of mass, three for angular momentum). Supply the tenth.

Starting from

\[ m_i \ddot{\mathbf{R}}_i = G \sum_{j=1,\, j\neq i}^{n} \frac{m_i m_j}{r_{ij}^{2}}\, \hat{\mathbf{r}}_{ij}, \qquad i = 1,\ldots,n, \]

take the dot product with \(\dot{\mathbf{R}}_i\) and sum over \(i\).

Show that the double sum on the right collapses, when reorganised over unordered pairs, to a total time derivative. You will need Newton’s third law \(\mathbf{F}_{ji} = -\mathbf{F}_{ij}\) and the identity \(\mathbf{r}\cdot\dot{\mathbf{r}} = r\dot r\).
Hence prove that \[ E \;=\; \underbrace{\sum_{i=1}^{n} \tfrac12 m_i V_i^{2}}_{T} \;+\; \underbrace{\sum_{i<j} \Phi_{ij}}_{U}, \qquad \Phi_{ij} = -\frac{G m_i m_j}{r_{ij}}, \] is a constant of the motion.
Explain why the sum in \(U\) runs over \(i<j\) and not over all ordered pairs, and what would go wrong numerically if you got this wrong.

Problem 2 — The Lagrange–Jacobi identity and Jacobi’s escape criterion

Let \(\boldsymbol{\rho}_i = \mathbf{R}_i - \mathbf{R}_{cm}\) and define the polar moment of inertia of the system about its barycentre,

\[ I \;=\; \sum_{i=1}^{n} m_i \, \left|\boldsymbol{\rho}_i\right|^{2}. \]

Prove the Lagrange–Jacobi identity \[ \ddot I \;=\; 4T + 2U \;=\; 4E - 2U , \] where \(T\), \(U\), \(E\) are as in Problem 1, evaluated in the barycentric frame. (The step \(\sum_i \boldsymbol{\rho}_i \cdot \mathbf{F}_i = U\) is Euler’s theorem for a potential homogeneous of degree \(-1\); derive it directly rather than quoting it.)
Jacobi’s criterion. Suppose the system remained bounded for all time, so that \(r_{ij} \le D\) for every pair and all \(t\). Show that \(\ddot I\) is then bounded below by a positive constant whenever \(E \ge 0\), and hence that \(I \to \infty\). Conclude: no N-body system with \(E \ge 0\) can remain bounded.
Virial theorem. For a bounded system (so that \(I\) and \(\dot I\) stay bounded), show that the long-time averages satisfy \[ \langle 2T \rangle = -\langle U \rangle, \qquad E = -\langle T \rangle = \tfrac12 \langle U \rangle . \]
A star cluster of total mass \(M\) and radius \(R\) has measured root-mean-square speed \(v_{rms}\). Use (c) to estimate \(M\) in terms of \(v_{rms}\) and \(R\), and comment on what this measurement told Zwicky in 1933.

Problem 3 — Where the invariable plane really lives

The slides prove that \(\mathbf{H} = \sum_i m_i \mathbf{R}_i \times \dot{\mathbf{R}}_i\) is constant, and then declare the invariable plane to be the plane through the barycentre perpendicular to \(\mathbf{H}\). Justify that sentence.

With \(\boldsymbol{\rho}_i = \mathbf{R}_i - \mathbf{R}_{cm}\), prove the König decomposition \[ \mathbf{H} \;=\; \underbrace{M\, \mathbf{R}_{cm} \times \dot{\mathbf{R}}_{cm}}_{\text{orbital}} \;+\; \underbrace{\sum_{i} m_i\, \boldsymbol{\rho}_i \times \dot{\boldsymbol{\rho}}_i}_{\mathbf{H}_{cm}} \] and show that both terms are separately constant.
Show that under a shift of origin \(\mathbf{R}_i \mapsto \mathbf{R}_i - \mathbf{b}\) one gets \(\mathbf{H} \mapsto \mathbf{H} - \mathbf{b}\times\mathbf{C}_1\), and under a Galilean boost \(\dot{\mathbf{R}}_i \mapsto \dot{\mathbf{R}}_i - \mathbf{u}\) that \(\mathbf{H}\) also changes — but that \(\mathbf{H}_{cm}\) is invariant under both.
Hence explain why the invariable plane is a property of the solar system and not of the observer, whereas “the plane perpendicular to \(\mathbf{H}\) through the origin” is not.

Problem 4 — Lagrange’s equilateral solution for unequal masses

In Assignment 01 you found \(\omega_0\) for three equal masses at the vertices of an equilateral triangle. Now let \(m_1, m_2, m_3\) be arbitrary, placed at the vertices of an equilateral triangle of side \(d\), and let the whole configuration rotate rigidly with angular velocity \(\omega\) about the axis through the barycentre, perpendicular to the plane of the triangle.

Taking the origin at the barycentre, show that the gravitational acceleration of body \(i\) is exactly \[ \ddot{\mathbf{R}}_i \;=\; -\frac{GM}{d^{3}}\,\mathbf{R}_i, \qquad M = m_1+m_2+m_3 , \] i.e. every body is pulled straight towards the barycentre with a force proportional to its distance from it. Identify precisely where the equilateral assumption is used.
Deduce that circular motion is consistent iff \[ \omega^{2} = \frac{GM}{d^{3}}, \] independent of the individual masses and of the (unequal) orbital radii \(\rho_i = |\mathbf{R}_i|\). Check that this reduces to your Assignment 01 answer.
The result of (a) is a linear (Hooke) force law, which is also solved by ellipses. What does this suggest about non-circular equilateral solutions of the three-body problem?

Part B — The Two-body Problem

Problem 5 — The orbit equation without the eccentricity vector

The slides obtain \(r(f)\) from the Laplace–Runge–Lenz vector. Obtain it a second, independent way.

Using \(h = r^2 \dot f\) and the substitution \(u = 1/r\), show that \(\dot r = -h\,\dfrac{du}{df}\) and \(\ddot r = -h^{2}u^{2}\dfrac{d^{2}u}{df^{2}}\).
Substitute into the radial equation \(\ddot r - r\dot f^{2} = -\mu/r^{2}\) to obtain Binet’s equation \[ \frac{d^{2}u}{df^{2}} + u = \frac{\mu}{h^{2}} . \]
Solve it and show the general solution is a conic with \(p = h^2/\mu\). Identify the two constants of integration, and show that they are precisely the magnitude and direction of the eccentricity vector \(\mathbf{e}\) of the slides.
Where, in this route, does the sixth constant (\(t_p\)) enter? Relate your answer to the “\(5\) (geometry) \(+\;1\) (timing)” count on the Counting the integrals slide.

Problem 6 — Why the inverse square is special: apsidal precession

The slides remark that any departure from \(1/r^2\) makes \(\dot{\mathbf{e}} \ne \mathbf{0}\) and the apse line rotate. Make that quantitative.

Consider the perturbed central acceleration \[ \ddot{\mathbf{r}} = -\left(\frac{\mu}{r^{2}} + \frac{\delta}{r^{3}}\right)\hat{\mathbf{r}}, \qquad |\delta| \ll h^{2}, \] (this is the field of a slightly oblate primary, to leading order). Show first that \(\mathbf{h}\) is still conserved, and then that Binet’s equation becomes \[ \frac{d^{2}u}{df^{2}} + k^{2} u = \frac{\mu}{h^{2}}, \qquad k^{2} = 1 - \frac{\delta}{h^{2}} . \]
Solve exactly and show the orbit is \(r = p'/(1 + e\cos kf)\). Deduce that successive periapsis passages are separated by \(\Delta f = 2\pi/k\), so that the apse line advances per revolution by \[ \Delta\varpi = 2\pi\left(\frac{1}{k} - 1\right) \;\approx\; \frac{\pi\delta}{h^{2}} . \] For which sign of \(\delta\) does the orbit precess forwards?
Mercury. General relativity instead adds a \(u^{2}\) term, \[ \frac{d^{2}u}{df^{2}} + u = \frac{\mu}{h^{2}} + \frac{3\mu}{c^{2}}u^{2} . \] Treating the last term as a small perturbation about the Kepler solution \(u_0 = (\mu/h^2)(1 + e\cos f)\), identify the resonant term on the right and show that the secular part of the correction gives \[ \Delta\varpi = \frac{6\pi\mu}{c^{2}p}\;\;\text{per revolution}. \] Evaluate it for Mercury (\(a = 5.79\times10^{10}\,\mathrm{m}\), \(e = 0.2056\), \(\mu_{\odot} = 1.327\times10^{20}\,\mathrm{m^3 s^{-2}}\), \(T = 87.97\) d) in arcseconds per century.

Problem 7 — The hyperbolic Kepler equation

Repeat, for \(e > 1\), the auxiliary-circle construction the slides carried out for the ellipse. Write \(\bar a = |a| = -a > 0\), so that \(p = \bar a\,(e^{2}-1)\), and define the hyperbolic anomaly \(H\) through the perifocal coordinates

\[ x_P = \bar a\left(e - \cosh H\right), \qquad y_P = \bar a\sqrt{e^{2}-1}\,\sinh H . \]

Verify that this parametrises the correct branch of the hyperbola and that \[ r = \bar a\left(e\cosh H - 1\right), \] the exact analogue of \(r = a(1 - e\cos E)\).
Using \(r^{2}df = x_P\,dy_P - y_P\,dx_P\) and \(h = \sqrt{\mu p}\), show that the awkward factor again becomes linear, and integrate to obtain the hyperbolic Kepler equation \[ M_h \;\equiv\; \sqrt{\frac{\mu}{\bar a^{3}}}\,\left(t - t_p\right) \;=\; e\sinh H - H . \]
Derive the half-angle relation \[ \tanh\frac{H}{2} = \sqrt{\frac{e-1}{e+1}}\,\tan\frac{f}{2} . \]
From the orbit equation obtain the asymptote direction \(\cos f_\infty = -1/e\), the hyperbolic excess speed \(v_\infty = \sqrt{\mu/\bar a}\), and the turn angle \(\delta\) satisfying \(\sin(\delta/2) = 1/e\).

Problem 8 — Barker’s equation: the one conic that yields

The slides argue that \(f(t)\) has no closed form. The parabola is the exception.

Put \(e = 1\) in \(t - t_p = \dfrac{p^{2}}{h}\displaystyle\int_0^f \frac{df'}{(1+e\cos f')^{2}}\) and evaluate the integral in closed form using \(D = \tan(f/2)\), obtaining Barker’s equation \[ t - t_p = \frac{1}{2}\sqrt{\frac{p^{3}}{\mu}}\left(D + \frac{D^{3}}{3}\right). \]
Show that the resulting cubic in \(D\) has exactly one real root for every \(t\), and write that root down explicitly (Cardano, or the \(\sinh\) form).
Explain, in one paragraph, what is structurally different about \(e = 1\) that makes the inversion elementary here but transcendental for the ellipse.

Problem 9 — Rectilinear fall: the case \(h = 0\)

On the angular-momentum slide we asked whether \(h = 0\) is possible. It is: release two bodies, of masses \(m_1\) and \(m_2\), from rest at separation \(r_0\).

Show that \(\mathbf{h} = \mathbf{0}\), that the motion is therefore confined to the line joining them, and that \(e = 1\) while \(a\) stays finite. What conic is this?
From the energy integral, obtain \(\dot r\) and hence show by direct quadrature (substitute \(r = r_0\sin^{2}\theta\)) that the two bodies collide after \[ t_{\text{fall}} = \frac{\pi}{2}\sqrt{\frac{r_0^{3}}{2\mu}} . \]
Obtain the same answer in one line from Kepler’s third law, by regarding the fall as half a revolution of a degenerate ellipse. What is its semi-major axis?
Evaluate \(t_{\text{fall}}\) for the Earth released from rest at \(1\,\mathrm{AU}\) from the Sun, in days.

Problem 10 — Three different averages of \(r\), and why they differ

For a bound orbit of semi-major axis \(a\) and eccentricity \(e\), compute the mean distance — averaged three different ways.

Over true anomaly: show \(\displaystyle \langle r \rangle_f = \frac{1}{2\pi}\int_0^{2\pi} r\,df = a\sqrt{1-e^{2}} = b\).
Over eccentric anomaly: show \(\langle r \rangle_E = a\).
Over time: using \(dt = (1-e\cos E)\,dE/n\), show \[ \langle r \rangle_t = \frac{1}{T}\int_0^{T} r \, dt = a\left(1 + \frac{e^{2}}{2}\right). \]
Order the three and explain physically why the time average is the largest and the true-anomaly average the smallest.
Show also that \(\langle 1/r\rangle_t = 1/a\), and use it together with vis-viva to prove that \(\langle v^{2}\rangle_t = \mu/a\). Verify that this is exactly the virial theorem of Problem 2(c) applied to the two-body problem.