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IPhO 2011, Теория — Задача 1. A Three-body Problem and LISA

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Theoretical Competition: 12 July 2011 — Question 1 · 12 юли 2011 г.

Условие

Three concentric circular orbits about the point O. Mass M lies on the smallest circle and mass m on the middle circle, on opposite sides of O along a horizontal line; braces mark the distances R (from M to O) and r (from O to m). The small mass μ sits on the largest circle, with lines drawn from μ to M, to O and to m. Arrows on each circle show counter-clockwise motion. — 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 т.]

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 $r$ :

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. [3,5 т.]

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 т.]

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.

Perspective drawing of the Earth's orbit around the Sun (yellow disc labelled Sun) with the Earth marked on it. Along the orbit, a sequence of dashed circles each contains a red triangle joining three spacecraft (blue and black dots), showing the triangle rolling as it moves around the Sun. A line of length 1 AU joins the Sun to the LISA constellation, making an angle of 20° with the line to the Earth. — FIGURE 2 Illustration of the LISA orbit. The three spacecraft 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.).

Boxed diagram: on the left a grey disc labelled Earth with an arrow pointing left; on the right an arrow pointing left and three grey spacecraft labelled A (bottom), B (upper right) and C (upper left) on a dotted circle with arrows showing rotation A → B → C. — 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 т.]

Отговори

Решение

I. Solution

Circular orbit of μ about O with arrow showing counter-clockwise motion. M and m lie on a horizontal line through O at distances R and r from O. Lines from μ to M (length r1), to O (length ρ) and to m (length r2); angles θ1 and θ2 at μ, ψ1 at M, φ at O and ψ2 at m are marked.

1.1 Let O be their centre of mass. Hence

MR - mr = 0 \qquad (1)

m\omega_0^2 r = \frac{GMm}{(R+r)^2}, \qquad M\omega_0^2 R = \frac{GMm}{(R+r)^2} \qquad (2)

From Eq. (2), or using reduced mass, $\omega_0^2 = \dfrac{G(M+m)}{(R+r)^3}$

Hence,

\omega_0^2 = \frac{G(M+m)}{(R+r)^3} = \frac{GM}{r(R+r)^2} = \frac{Gm}{R(R+r)^2}. \qquad (3)

1.2 Since $\mu$ is infinitesimal, it has no gravitational influences on the motion of neither $M$ nor $m$ . For $\mu$ to remain stationary relative to both $M$ and $m$ we must have:

\frac{GM\mu}{r_1^2}\cos\theta_1 + \frac{Gm\mu}{r_2^2}\cos\theta_2 = \mu\omega_0^2\rho = \frac{G(M+m)\mu}{(R+r)^3}\rho \qquad (4)

\frac{GM\mu}{r_1^2}\sin\theta_1 = \frac{Gm\mu}{r_2^2}\sin\theta_2 \qquad (5)

Substituting $\dfrac{GM}{r_1^2}$ from Eq. (5) into Eq. (4), and using the identity $\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2 = \sin(\theta_1+\theta_2)$ , we get

m\frac{\sin(\theta_1+\theta_2)}{r_2^2} = \frac{(M+m)}{(R+r)^3}\rho\sin\theta_1 \qquad (6)

The distances $r_2$ and $\rho$ , the angles $\theta_1$ and $\theta_2$ are related by two Sine Rule equations

\frac{\sin\psi_1}{\rho} = \frac{\sin\theta_1}{R}, \qquad \frac{\sin\psi_1}{r_2} = \frac{\sin(\theta_1+\theta_2)}{R+r} \qquad (7)

Substitute (7) into (6)

\frac{1}{r_2^3} = \frac{R}{(R+r)^4}\frac{(M+m)}{m} \qquad (10)

Since $\dfrac{m}{M+m} = \dfrac{R}{R+r}$ ,Eq. (10) gives

r_2 = R + r \qquad (11)

By substituting $\dfrac{Gm}{r_2^2}$ from Eq. (5) into Eq. (4), and repeat a similar procedure, we get

r_1 = R + r \qquad (12)

Alternatively,

\frac{r_1}{\sin(180^{\circ}-\phi)} = \frac{R}{\sin\theta_1} \quad \text{and} \quad \frac{r_2}{\sin\phi} = \frac{r}{\sin\theta_2}

\frac{\sin\theta_1}{\sin\theta_2} = \frac{R}{r}\times\frac{r_2}{r_1} = \frac{m}{M}\times\frac{r_2}{r_1}

Combining with Eq. (5) gives $r_1 = r_2$

Hence, it is an equilateral triangle with

\psi_1 = 60^{\circ}, \qquad \psi_2 = 60^{\circ} \qquad (13)

The distance $\rho$ is calculated from the Cosine Rule.

\rho^2 = r^2 + (R+r)^2 - 2r(R+r)\cos 60^{\circ}, \qquad \rho = \sqrt{r^2 + rR + R^2} \qquad (14)

Alternative Solution to 1.2

Since $\mu$ is infinitesimal, it has no gravitational influences on the motion of neither $M$ nor $m$ .For $\mu$ to remain stationary relative to both $M$ and $m$ we must have:

\frac{GM\mu}{r_1^2}\cos\theta_1 + \frac{Gm\mu}{r_2^2}\cos\theta_2 = \mu\omega^2\rho = \frac{G(M+m)\mu}{(R+r)^3}\rho \qquad (4)

\frac{GM\mu}{r_1^2}\sin\theta_1 = \frac{Gm\mu}{r_2^2}\sin\theta_2 \qquad (5)

Note that

\frac{r_1}{\sin(180^{\circ}-\phi)} = \frac{R}{\sin\theta_1}

\frac{r_2}{\sin\phi} = \frac{r}{\sin\theta_2} \qquad \text{(see figure)}

\frac{\sin\theta_1}{\sin\theta_2} = \frac{R}{r}\times\frac{r_2}{r_1} = \frac{m}{M}\times\frac{r_2}{r_1} \qquad (6)

Equations (5) and (6):

r_1 = r_2 \qquad (7)

\frac{\sin\theta_1}{\sin\theta_2} = \frac{m}{M} \qquad (8)

\psi_1 = \psi_2 \qquad (9)

The equation (4) then becomes:

M\cos\theta_1 + m\cos\theta_2 = \frac{(M+m)}{(R+r)^3}r_1^2\rho \qquad (10)

Equations (8) and (10):

\sin(\theta_1+\theta_2) = \frac{M+m}{M}\frac{r_1^2\rho}{(R+r)^3}\sin\theta_2 \qquad (11)

Note that from figure,

\frac{\rho}{\sin\psi_2} = \frac{r}{\sin\theta_2} \qquad (12)

Equations (11) and (12):

\sin(\theta_1+\theta_2) = \frac{M+m}{M}\frac{r_1^2 r}{(R+r)^3}\sin\psi_2 \qquad (13)

Also from figure,

(R+r)^2 = r_2^2 - 2r_1r_2\cos(\theta_1+\theta_2) + r_1^2 = 2r_1^2\left[1-\cos(\theta_1+\theta_2)\right] \qquad (14)

Equations (13) and (14):

\sin(\theta_1+\theta_2) = \frac{\sin\psi_2}{2\left[1-\cos(\theta_1+\theta_2)\right]} \qquad (15)

\theta_1+\theta_2 = 180^{\circ} - \psi_1 - \psi_2 = 180^{\circ} - 2\psi_2 \qquad \text{(see figure)}

\therefore \quad \cos\psi_2 = \frac{1}{2}, \quad \psi_2 = 60^{\circ}, \quad \psi_1 = 60^{\circ}

Hence $M$ and $m$ from an equilateral triangle of sides $(R+r)$

Distance $\mu$ to $M$ is $R+r$

Distance $\mu$ to $m$ is $R+r$

Distance $\mu$ to O is

\rho = \sqrt{\left(\frac{R+r}{2} - R\right)^2 + \left\{(R+r)\frac{\sqrt{3}}{2}\right\}^2} = \sqrt{R^2 + Rr + r^2}

1.3 The energy of the mass $\mu$ is given by

E = -\frac{GM\mu}{r_1} - \frac{Gm\mu}{r_2} + \tfrac{1}{2}\mu\left(\left(\frac{d\rho}{dt}\right)^2 + \rho^2\omega^2\right) \qquad (15)

Since the perturbation is in the radial direction, angular momentum is conserved ( $r_1 = r_2 = \Re$ and $m = M$ ),

E = -\frac{2GM\mu}{\Re} + \tfrac{1}{2}\mu\left(\left(\frac{d\rho}{dt}\right)^2 + \frac{\rho_0^4\omega_0^2}{\rho^2}\right) \qquad (16)

Equilateral triangle with two masses at the base vertices, the base split by O into two halves each labelled R, a vertical line of length ρ from O to the apex, a side labelled ℜ, and the base angle marked 60°.

Since the energy is conserved,

\frac{dE}{dt} = 0

\frac{dE}{dt} = \frac{2GM\mu}{\Re^2}\frac{d\Re}{dt} + \mu\frac{d\rho}{dt}\frac{d^2\rho}{dt^2} - \mu\frac{\rho_0^4\omega_0^2}{\rho^3}\frac{d\rho}{dt} = 0 \qquad (17)

\frac{d\Re}{dt} = \frac{d\Re}{d\rho}\frac{d\rho}{dt} = \frac{d\rho}{dt}\frac{\rho}{\Re} \qquad (18)

\frac{dE}{dt} = \frac{2GM\mu}{\Re^3}\rho\frac{d\rho}{dt} + \mu\frac{d\rho}{dt}\frac{d^2\rho}{dt^2} - \mu\frac{\rho_0^4\omega_0^2}{\rho^3}\frac{d\rho}{dt} = 0 \qquad (19)

Since $\dfrac{d\rho}{dt} \neq 0$ , we have

\frac{2GM}{\Re^3}\rho + \frac{d^2\rho}{dt^2} - \frac{\rho_0^4\omega_0^2}{\rho^3} = 0 \quad \text{or}

\frac{d^2\rho}{dt^2} = -\frac{2GM}{\Re^3}\rho + \frac{\rho_0^4\omega_0^2}{\rho^3}. \qquad (20)

The perturbation from $\Re_0$ and $\rho_0$ gives $\Re = \Re_0\left(1 + \dfrac{\Delta\Re}{\Re_0}\right)$ and $\rho = \rho_0\left(1 + \dfrac{\Delta\rho}{\rho_0}\right)$ .

Then

\frac{d^2\rho}{dt^2} = \frac{d^2}{dt^2}(\rho_0 + \Delta\rho) = -\frac{2GM}{\Re_0^3\left(1+\dfrac{\Delta\Re}{\Re_0}\right)^3}\rho_0\left(1+\frac{\Delta\rho}{\rho_0}\right) + \frac{\rho_0^4\omega_0^2}{\rho_0^3\left(1+\dfrac{\Delta\rho}{\rho_0}\right)^3} \qquad (21)

Using binomial expansion $(1+\varepsilon)^n \approx 1 + n\varepsilon$ ,

\frac{d^2\Delta\rho}{dt^2} = -\frac{2GM}{\Re_0^3}\rho_0\left(1+\frac{\Delta\rho}{\rho_0}\right)\left(1-\frac{3\Delta\Re}{\Re_0}\right) + \rho_0\omega_0^2\left(1-\frac{3\Delta\rho}{\rho_0}\right). \qquad (22)

Using $\Delta\rho = \dfrac{\Re}{\rho}\Delta\Re$ ,

\frac{d^2\Delta\rho}{dt^2} = -\frac{2GM}{\Re_0^3}\rho_0\left(1+\frac{\Delta\rho}{\rho_0} - \frac{3\rho_0\Delta\rho}{\Re_0^2}\right) + \rho_0\omega_0^2\left(1-\frac{3\Delta\rho}{\rho_0}\right). \qquad (23)

Since $\omega_0^2 = \dfrac{2GM}{\Re_0^3}$ ,

\frac{d^2\Delta\rho}{dt^2} = -\omega_0^2\rho_0\left(1+\frac{\Delta\rho}{\rho_0} - \frac{3\rho_0\Delta\rho}{\Re_0^2}\right) + \omega_0^2\rho_0\left(1-\frac{3\Delta\rho}{\rho_0}\right) \qquad (24)

\frac{d^2\Delta\rho}{dt^2} = -\omega_0^2\rho_0\left(\frac{4\Delta\rho}{\rho_0} - \frac{3\rho_0\Delta\rho}{\Re_0^2}\right) \qquad (25)

\frac{d^2\Delta\rho}{dt^2} = -\omega_0^2\Delta\rho\left(4 - \frac{3\rho_0^2}{\Re_0^2}\right) \qquad (26)

From the figure, $\rho_0 = \Re_0\cos 30^{\circ}$ or $\dfrac{\rho_0^2}{\Re_0^2} = \dfrac{3}{4}$ ,

\frac{d^2\Delta\rho}{dt^2} = -\omega_0^2\Delta\rho\left(4 - \frac{9}{4}\right) = -\frac{7}{4}\omega_0^2\Delta\rho. \qquad (27)

Angular frequency of oscillation is $\dfrac{\sqrt{7}}{2}\omega_0$ .

Alternative solution:

$M = m$ gives $R = r$ and $\omega_0^2 = \dfrac{G(M+M)}{(R+R)^3} = \dfrac{GM}{4R^3}$ . The unperturbed radial distance of $\mu$ is $\sqrt{3}R$ , so the perturbed radial distance can be represented by $\sqrt{3}R + \zeta$ where $\zeta \ll \sqrt{3}R$ as shown in the following figure.

Using Newton’s 2 $^{\mathrm{nd}}$ law,

-\frac{2GM\mu}{\{R^2 + (\sqrt{3}R+\zeta)^2\}^{3/2}}(\sqrt{3}R+\zeta) = \mu\frac{d^2}{dt^2}(\sqrt{3}R+\zeta) - \mu\omega^2(\sqrt{3}R+\zeta).

(1)

The conservation of angular momentum gives $\mu\omega_0(\sqrt{3}R)^2 = \mu\omega(\sqrt{3}R+\zeta)^2$ .

(2)

Manipulate (1) and (2) algebraically, applying $\zeta^2 \approx 0$ and binomial approximation.

-\frac{2GM}{\{R^2 + (\sqrt{3}R+\zeta)^2\}^{3/2}}(\sqrt{3}R+\zeta) = \frac{d^2\zeta}{dt^2} - \frac{\omega_0^2\sqrt{3}R}{(1+\zeta/\sqrt{3}R)^3}

-\frac{2GM}{\{4R^2 + 2\sqrt{3}\zeta R\}^{3/2}}(\sqrt{3}R+\zeta) \approx \frac{d^2\zeta}{dt^2} - \frac{\omega_0^2\sqrt{3}R}{(1+\zeta/\sqrt{3}R)^3}

-\frac{GM}{4R^3}\sqrt{3}R\frac{(1+\zeta/\sqrt{3}R)}{(1+\sqrt{3}\zeta/2R)^{3/2}} = \frac{d^2\zeta}{dt^2} - \frac{\omega_0^2\sqrt{3}R}{(1+\zeta/\sqrt{3}R)^3}

-\omega_0^2\sqrt{3}R\left(1-\frac{3\sqrt{3}\zeta}{4R}\right)\left(1+\frac{\zeta}{\sqrt{3}R}\right) \approx \frac{d^2\zeta}{dt^2} - \omega_0^2\sqrt{3}R\left(1-\frac{3\zeta}{\sqrt{3}R}\right)

\frac{d^2}{dt^2}\zeta = -\left(\frac{7}{4}\omega_0^2\right)\zeta

1.4 Relative velocity

Let $v$ = speed of each spacecraft as it moves in circle around the centre O.
The relative velocities are denoted by the subscripts A, B and C.
For example, $v_{\mathrm{BA}}$ is the velocity of B as observed by A.

The period of circular motion is 1 year $T = 365\times 24\times 60\times 60\ \mathrm{s}$ . $\qquad (28)$

The angular frequency $\omega = \dfrac{2\pi}{T}$

The speed $v = \omega\dfrac{L}{2\cos 30^{\circ}} = 575\ \mathrm{m/s}$ $\qquad (29)$

The speed is much less than the speed light → Galilean transformation.

In Cartesian coordinates, the velocities of B and C (as observed by O) are

Boxed diagram of the three spacecraft A (bottom), B (upper right) and C (upper left) at the corners of an equilateral triangle with dashed sides labelled L, centre O, and unit vectors î and ĵ. At each spacecraft a bold arrow shows its velocity v and thin arrows show the relative velocities: vCB and vCA at C, vBA and vBC at B, vAB and vAC at A.

For B, $\vec{v}_B = v\cos 60^{\circ}\,\hat{\mathbf{i}} - v\sin 60^{\circ}\,\hat{\mathbf{j}}$

For C, $\vec{v}_C = v\cos 60^{\circ}\,\hat{\mathbf{i}} + v\sin 60^{\circ}\,\hat{\mathbf{j}}$

Hence $\vec{v}_{\mathrm{BC}} = -2v\sin 60^{\circ}\,\hat{\mathbf{j}} = -\sqrt{3}v\,\hat{\mathbf{j}}$

The speed of B as observed by C is $\sqrt{3}v \approx 996\ \mathrm{m/s}$ $\qquad (30)$

Notice that the relative velocities for each pair are anti-parallel.

Alternative solution for 1.4

One can obtain $v_{\mathrm{BC}}$ by considering the rotation about the axis at one of the spacecrafts.

v_{\mathrm{BC}} = \omega L = \frac{2\pi}{365\times 24\times 60\times 60\ \mathrm{s}}(5\times 10^{6}\ \mathrm{km}) \approx 996\ \mathrm{m/s}

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf