IPhO 2011 · Theory problem 1

A Three-body Problem and LISA

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Figure 1

FIGURE 1 Coplanar orbits of three bodies.

1.1 Two gravitating masses \(M\) and \(m\) are moving in circular orbits of radii \(R\) and \(r\), respectively, about their common centre of mass. Find the angular velocity \(\omega_0\) of the line joining \(M\) and \(m\) in terms of \(R,r,M,m\) and the universal gravitational constant \(G\).

[1.5 points]

1.2 A third body of infinitesimal mass \(\mu\) is placed in a coplanar circular orbit about the same centre of mass so that \(\mu\) remains stationary relative to both \(M\) and \(m\) as shown in Figure 1. Assume that the infinitesimal mass is not collinear with \(M\) and \(m\). Find the values of the following parameters in terms of \(R\) and \(r\):

[3.5 points]

1.2.1 distance from \(\mu\) to \(M\).

1.2.2 distance from \(\mu\) to \(m\).

1.2.3 distance from \(\mu\) to the centre of mass.

1.3 Consider the case \(M=m\). If \(\mu\) is now given a small radial perturbation (along \(O\mu\)), what is the angular frequency of oscillation of \(\mu\) about the unperturbed position in terms of \(\omega_0\)? Assume that the angular momentum of \(\mu\) is conserved.

[3.2 points]

The Laser Interferometry Space Antenna (LISA) is a group of three identical spacecrafts for detecting low frequency gravitational waves. Each of the spacecrafts is placed at the corners of an equilateral triangle as shown in Figure 2 and Figure 3. The sides (or ‘arms’) are about 5.0 million kilometres long. The LISA constellation is in an earth-like orbit around the Sun trailing the Earth by \(20^\circ\). Each of them moves on a slightly inclined individual orbit around the Sun. Effectively, the three spacecrafts appear to roll about their common centre one revolution per year.

They are continuously transmitting and receiving laser signals between each other. Overall, they detect the gravitational waves by measuring tiny changes in the arm lengths using interferometric means. A collision of massive objects, such as blackholes, in nearby galaxies is an example of the sources of gravitational waves.

Figure 2

FIGURE 2 Illustration of the LISA orbit. The three spacecrafts roll about their centre of mass with a period of 1 year. Initially, they trail the Earth by \(20^\circ\) (Picture from D.A. Shaddock, “An Overview of the Laser Interferometer Space Antenna”, Publications of the Astronomical Society of Australia, 2009, 26, pp.128-132.).

Figure 3

FIGURE 3 Enlarged view of the three spacecrafts trailing the Earth. A, B and C are the three spacecrafts at the corners of the equilateral triangle.

1.4 In the plane containing the three spacecrafts, what is the relative speed of one spacecraft with respect to another?

[1.8 point]

Official English text of the problem.

Answers

  • 1.1 \[ \boxed{\omega_0=\sqrt{\frac{G(M+m)}{(R+r)^3}}} \]

  • 1.2.1 \[ \boxed{d_{\mu M}=R+r} \]

  • 1.2.2 \[ \boxed{d_{\mu m}=R+r} \]

  • 1.2.3 \[ \boxed{d_{\mu O}=\sqrt{R^2+Rr+r^2}} \]

  • 1.3, using the stipulated conservation of angular momentum: \[ \boxed{\omega_{\mathrm{osc}}=\frac{\sqrt7}{2}\,\omega_0} \]

  • 1.4, with arm length \(L=5.0\times10^9\,\mathrm{m}\) and rotation period \(T=1\,\mathrm{year}\): \[ \boxed{v_{\mathrm{rel}}=\frac{2\pi L}{T} \simeq1.0\times10^3\,\mathrm{m\,s^{-1}} =1.0\,\mathrm{km\,s^{-1}}} \]

Solution

1.1

The distance between the masses is \(d=R+r\). Since \(O\) is their centre of mass, \[ MR=mr. \]

Gravity supplies the centripetal acceleration of each mass: \[ M\omega_0^2R=\frac{GMm}{d^2}, \qquad m\omega_0^2r=\frac{GMm}{d^2}. \] Dividing by the respective masses and adding gives \[ \omega_0^2(R+r)=\frac{G(M+m)}{d^2}. \] Therefore, \[ \boxed{\omega_0=\sqrt{\frac{G(M+m)}{(R+r)^3}}}. \]

1.2.1 and 1.2.2

Because \(\mu\) is infinitesimal, it does not affect the motion of \(M\) and \(m\). To remain stationary relative to them, it must revolve about \(O\) with angular velocity \(\omega_0\).

At a chosen instant, take \[ O=(0,0),\qquad M=(-R,0),\qquad m=(r,0),\qquad \mu=(x,y), \] where \(y\ne0\). Write its distances from the two masses as \[ a=\sqrt{(x+R)^2+y^2}, \qquad b=\sqrt{(x-r)^2+y^2}. \]

Its acceleration is \(-\omega_0^2(x,y)\). Newton’s law, after cancelling the test mass \(\mu\), therefore gives \[ \omega_0^2x=\frac{GM(x+R)}{a^3}+\frac{Gm(x-r)}{b^3}, \] \[ \omega_0^2y=\frac{GMy}{a^3}+\frac{Gmy}{b^3}. \] Since \(y\ne0\), the second equation implies \[ \omega_0^2=\frac{GM}{a^3}+\frac{Gm}{b^3}. \] Subtracting \(x\) times this equation from the first yields \[ 0=\frac{GMR}{a^3}-\frac{Gmr}{b^3}. \] Using \(MR=mr\), we obtain \(a=b\). Consequently, \[ \omega_0^2=\frac{G(M+m)}{a^3}. \] Comparison with part 1.1 gives \(a=R+r\), and hence \[ \boxed{\text{1.2.1: }d_{\mu M}=R+r}, \qquad \boxed{\text{1.2.2: }d_{\mu m}=R+r}. \]

Thus the three bodies occupy the vertices of an equilateral triangle.

1.2.3

The midpoint of the segment joining the two finite masses has coordinate \[ x_{\mathrm{mid}}=\frac{r-R}{2}. \] The third vertex of an equilateral triangle of side \(R+r\) lies a perpendicular distance \(\sqrt3(R+r)/2\) from this midpoint. Therefore the two possible positions of \(\mu\) are \[ x=\frac{r-R}{2}, \qquad y=\pm\frac{\sqrt3}{2}(R+r). \] Its distance from \(O\) satisfies \[ d_{\mu O}^2=x^2+y^2 =\frac{(r-R)^2+3(R+r)^2}{4} =R^2+Rr+r^2. \] Thus, \[ \boxed{d_{\mu O}=\sqrt{R^2+Rr+r^2}}. \]

1.3

When \(M=m\), the centre-of-mass condition gives \(R=r\). The unperturbed orbital radius of \(\mu\) is therefore \[ \rho_0=\sqrt3R, \] and part 1.1 gives \[ \omega_0^2=\frac{GM}{4R^3}. \]

On the perpendicular bisector of the line joining the equal masses, a point at distance \(\rho\) from \(O\) is at distance \(\sqrt{R^2+\rho^2}\) from each mass. Their transverse gravitational components cancel, so the inward radial acceleration is \[ g(\rho)=\frac{2GM\rho}{(R^2+\rho^2)^{3/2}}. \] This determines the linear radial force: by reflection symmetry, a small angular departure from the perpendicular bisector changes the radial component only at second order.

Use the angular-momentum conservation prescribed in the question. The angular momentum per unit mass is \[ h=\rho^2\dot\theta=\rho_0^2\omega_0. \] The radial equation of motion, \[ \ddot\rho-\rho\dot\theta^2=-g(\rho), \] becomes \[ \ddot\rho=\frac{h^2}{\rho^3} -\frac{2GM\rho}{(R^2+\rho^2)^{3/2}}. \]

Put \[ \rho=\rho_0+\xi,\qquad |\xi|\ll\rho_0, \] and keep only terms linear in \(\xi\). The first term expands as \[ \frac{h^2}{\rho^3} =\rho_0\omega_0^2-3\omega_0^2\xi+O(\xi^2). \] For the inward gravitational acceleration, \[ g'(\rho) =\frac{2GM}{(R^2+\rho^2)^{3/2}} \left(1-\frac{3\rho^2}{R^2+\rho^2}\right). \] Since \(\rho_0^2=3R^2\), \[ g'(\rho_0) =\omega_0^2\left(1-\frac94\right) =-\frac54\omega_0^2. \] Thus \[ g(\rho)=\rho_0\omega_0^2-\frac54\omega_0^2\xi+O(\xi^2). \]

The equilibrium terms cancel, leaving \[ \ddot\xi =\left(-3+\frac54\right)\omega_0^2\xi =-\frac74\omega_0^2\xi. \] This is the simple harmonic oscillator equation, so \[ \boxed{\omega_{\mathrm{osc}}=\frac{\sqrt7}{2}\,\omega_0}. \]

This result uses the imposed angular-momentum-conserving radial approximation. Angular momentum about \(O\) is not generally conserved for unrestricted motion in the rotating two-mass gravitational field.

1.4

Treat the constellation as an equilateral triangle of constant side length \(L\), rotating in its plane with angular velocity \[ \Omega=\frac{2\pi}{T}. \] Let \(\mathbf r_A\) and \(\mathbf r_B\) be the positions of two spacecraft relative to the triangle’s centre. Their velocities are \[ \mathbf v_A=\mathbf V+\boldsymbol\Omega\times\mathbf r_A, \qquad \mathbf v_B=\mathbf V+\boldsymbol\Omega\times\mathbf r_B, \] where \(\mathbf V\) is the common translational velocity of the centre.

Subtracting eliminates this common motion: \[ \mathbf v_B-\mathbf v_A =\boldsymbol\Omega\times(\mathbf r_B-\mathbf r_A). \] The separation vector lies in the triangle’s plane and has magnitude \(L\), so it is perpendicular to \(\boldsymbol\Omega\). Hence \[ v_{\mathrm{rel}}=\Omega L=\frac{2\pi L}{T}. \]

With \[ L=5.0\times10^6\,\mathrm{km}=5.0\times10^9\,\mathrm m, \qquad T\simeq365\times24\times3600\,\mathrm s, \] we obtain \[ v_{\mathrm{rel}}\simeq996\,\mathrm{m\,s^{-1}}. \] To two significant figures, \[ \boxed{v_{\mathrm{rel}}\simeq1.0\times10^3\,\mathrm{m\,s^{-1}} =1.0\,\mathrm{km\,s^{-1}}}. \]

The relative velocity is perpendicular to the arm joining the spacecraft, consistent with the arm’s constant length in this idealized motion.

The solution was checked against the official solution.