Paper 316 (Planetary System Dynamics)

i

Solution

Write

k = μ (1 - β)

. Taking the cross product of the equation of motion with

r

gives

\frac{d}{d t} (r \times \dot{r}) = 0, h = r \times \dot{r} .

Since

r \cdot h = 0

at every time, the orbit lies in the fixed plane perpendicular to the conserved specific angular momentum

h

. Taking the inner product with

\dot{r}

gives the conserved specific orbital energy

C = \frac{1}{2} ∣ \dot{r} ∣^{2} - \frac{μ ( 1 - β )}{r} .

For

β > 1

, the effective inverse-square force is repulsive and the potential term is positive.

ii

Solution

In polar coordinates in the orbital plane,

h = r^{2} \dot{θ}

. Setting

u = 1/ r

, the Binet equation for the effective inverse-square force is

u^{''} + u = \frac{μ ( 1 - β )}{h ^{2}} .

Its general solution is

u = \frac{μ ( 1 - β )}{h ^{2}} [1 + e cos (θ - ϖ)],

and hence

r = \frac{h ^{2} / [ μ ( 1 - β )]}{1 + e cos ( θ - ϖ )} .

This is a branch of a hyperbolic Kepler orbit; because

1 - β < 0

, its physical branch has a negative denominator.

iii

Solution

The parent body's circular Kepler orbit has speed

v_{K} = μ / a_{p}

. Release without a velocity impulse therefore gives

h = a_{p} v_{K} = μ a_{p} .

The release point is the dust orbit's point of closest approach, where

θ - ϖ = π

. Substituting

r = a_{p}

in the orbit equation gives

a_{p} = \frac{a _{p} / ( 1 - β )}{1 - e}, e = \frac{β}{β - 1} .

The longitude of periapsis

ϖ

specifies the symmetry axis of the conic. With this signed repulsive-force convention, it points opposite the release direction, so the closest point is at

θ = ϖ + π

iv

Solution

At release,

C = \frac{μ}{2 a _{p}} - \frac{μ ( 1 - β )}{a _{p}} = \frac{μ}{a _{p}} (β - \frac{1}{2}) .

At large radius the potential vanishes, so

v_{\infty} = 2 C = (2 β - 1) \frac{μ}{a _{p}} ≃ \frac{2 β μ}{a _{p}} .

The asymptote satisfies

1 + e cos f_{\infty} = 0

, where

f = θ - ϖ

. If the angular displacement from the release direction is

Δ = π - f_{\infty}

, then

cos (π - Δ) = - \frac{1}{e} = - 1 + \frac{1}{β} .

For

β ≫ 1

- cos Δ ≃ - 1 + Δ^{2} /2

, and therefore

Δ ≃ \frac{2}{β} .

(4)

v

Solution

Resolving the conserved energy into radial and azimuthal parts gives

C = \frac{1}{2} (\overset{r}{˙}^{2} + \frac{h ^{2}}{r ^{2}}) - \frac{μ ( 1 - β )}{r} .

Using the release values of

C

and

h

, and choosing the outward root,

v_{r} (r) = \overset{r}{˙} = \frac{μ}{a _{p}} [(2 β - 1) - 2 (β - 1) \frac{a _{p}}{r} - (\frac{a _{p}}{r})^{2}] .

Equivalently, with

x = r / a_{p}

v_{r} = \frac{μ}{a _{p}} \frac{( x - 1 ) [( 2 β - 1 ) x + 1 ]}{x} .

vi

Solution

In a steady axisymmetric outflow, mass conservation makes

2 π r Σ (r) v_{r} (r)

independent of

r

. The steady radial dust outflow therefore has

Σ (r) \propto \frac{1}{r v _{r}} \propto \frac{1}{( x - 1 ) [( 2 β - 1 ) x + 1 ]}, x = \frac{r}{a _{p}} .

It has an integrable

(r - a_{p})^{- 1/2}

pile-up just outside the source ring, where the radial speed starts from zero, and approaches

S i g ma \propto r^{- 1}

at large radius.

vii

Solution

The large-distance profile is proportional to

[2 β - 1, x]^{- 1}

. At

x = 2

, the exact profile is proportional to

(4 β - 1)^{- 1/2}

. Thus

\frac{Σ _{actual} ( 2 a _{p} )}{Σ _{r^{- 1} m e x t r a p o l a t i o n} ( 2 a _{p} )} = \frac{2 2 β - 1}{4 β - 1} ⟶ 2

β \to \infty

. The enhancement is the finite-radius remnant of the source-ring pile-up.

3

i

Solution

The disturbing function may be expanded in harmonics of the planets' orbital angles. Its terms fall into three useful classes:

short-period orbital perturbations involve rapidly varying combinations of mean longitudes and produce bounded oscillations associated with conjunctions;
mean-motion resonance terms have a nearly stationary integer combination of mean longitudes and apsidal angles, so repeated conjunctions act coherently and can drive libration, migration, or eccentricity growth;
secular perturbations survive averaging over both mean longitudes and describe slow exchange of eccentricity and apsidal precession.

Well-separated planets far from resonance are governed on long timescales mainly by the secular terms.

ii

Solution

Diagonalize the real Laplace-Lagrange secular matrix

A

. Its eigenvalue s are

g_{1, 2} = \frac{A _{11} + A _{22}}{2} \pm \frac{1}{2} (A_{11} - A_{22})^{2} + 4 A_{12} A_{21},

and choose corresponding real eigenvector s

v_{k}

. If

V = (v_{1} v_{2})

, the initial complex eccentricity vector determines complex mode coefficients

c = V^{- 1} z (0), c_{k} = ∣ c_{k} ∣ e^{i β_{k}} .

The matrix exponential solution is

z (t) = \sum_{k = 1}^{2} c_{k} v_{k} e^{i g_{k} t} .

Thus

z_{j} (t) = k = 1 \sum 2 e_{jk} e^{i (g_{k} t + β_{k})}, e_{jk} = ∣ c_{k} ∣ v_{jk},

(4)

where a negative eigenvector component may equivalently be made positive by adding

π

to its phase. The

g_{k}

are secular precession frequencies, each eigenvector fixes the planets' eccentricity ratio and relative apsidal orientation, and

β_{k}

fixes the phase selected by the initial conditions.

iii

Solution

Insert the planets' two secular eigenmodes and define their forcing strengths

ν_{k} = \sum_{j = 1}^{2} A_{j} e_{jk} .

Solving the first-order linear ordinary differential equation gives

z (t) = e_{free} e^{i (A t + β_{free})} + k = 1 \sum 2 \frac{ν _{k}}{g _{k} - A} e^{i (g_{k} t + β_{k})} .

The first term is the particle's freely precessing eccentricity. The remaining terms are its forced eccentricity, phase-locked to the planets' modes. In the complex plane, their vector sum makes the eccentricity and longitude of periapsis oscillate. A denominator becomes small at a secular resonance

A = g_{k}

; the nonresonant formula then ceases to be uniform.

iv

Solution

The free precession rate

A (a)

tends to zero far inside the inner planet, diverges on approaching

a_{1}

, diverges on both sides of

a_{2}

, and tends to zero far outside the outer planet. Between

a_{1}

and

a_{2}

it diverges at both ends and has at least one minimum. A horizontal line

A = g_{1}

therefore crosses once inside

a_{1}

and once outside

a_{2}

, plus zero, one tangent, or two times between the planets. There are consequently

2, 3, or 4

locations of the corresponding secular resonance. This argument uses the smooth Laplace-Lagrange secular theory away from the immediate neighborhoods of the planets and from mean-motion resonances.

v

Solution

Linear eccentricity damping adds

- z / τ

to the complex equation:

\overset{z}{˙} = (i A - τ^{- 1}) z + i \sum_{k} ν_{k} e^{i (g_{k} t + β_{k})} .

Its solution is

z (t) = C e^{(i A - 1/ τ) t} + k = 1 \sum 2 \frac{ν _{k}}{g _{k} - A - i / τ} e^{i (g_{k} t + β_{k})} .

Unlike the undamped free eccentricity, the homogeneous term decays exponentially. The forced response survives, acquires a phase lag, and has finite amplitude

\frac{∣ ν _{k} ∣}{( g _{k} - A ) ^{2} + τ ^{- 2}} .

vi

Solution

For

t ≫ τ

, the free term has disappeared. Exactly at

A = g_{1}

, the resonant contribution tends to

z_{1} (t) = i τ ν_{1} e^{i (g_{1} t + β_{1})},

with finite eccentricity

∣ ν_{1} ∣ τ

and a quarter-cycle phase shift. Without damping, exact resonance instead gives

z_{1} (t) = e^{i A t} (C + i ν_{1} t e^{i β_{1}}),

whose secular resonance amplitude grows linearly in the ideal linear theory.

vii

Solution

Very near one planet, that planet dominates both the particle's free-precession coefficient and its forcing coefficient, with the leading terms arranged approximately as

\overset{z}{˙} ≃ i A (z - z_{p}) - z / τ

. Since

∣ A ∣ τ ≫ 1

close to the planet, the late-time forced solution approaches

z ≃ z_{p} .

The particle's orbit therefore tends toward the planet's eccentricity and apsidal direction. Extremely close to the planet, close encounters, co-orbital dynamics, and individual mean-motion resonances invalidate the orbit-averaged linear approximation.

4

i

Solution

Treat the just-bound orbit as a parabolic Kepler orbit with

q ≃ a_{p}

. Its specific angular momentum is

h = 2 G M_{*} q

, so its instantaneous angular speed is

h / r^{2}

. Equating this to the planet's mean motion

n_{p} = G M_{*} / a_{p}^{3}

gives

r = (2 q a_{p}^{3})^{1/4} ≃ 2^{1/4} a_{p} .

The parabolic orbit equation

r = 2 q / (1 + cos f)

then gives

f ≃ 4 7^{\circ}

, or

5 0^{\circ}

to the nearest ten degrees.

If the bodies were in conjunction at periapsis, Barker equation gives

n_{p} t = 2 [tan \frac{f}{2} + \frac{1}{3} tan^{3} \frac{f}{2}] ≃ 3 7^{\circ} .

The planetesimal is therefore at its maximum lead, approximately

f - n_{p} t ≃ 1 0^{\circ}

ahead of the planet as seen from the star.

ii

Solution

In the frame rotating with the planet, put the planet at fixed radius

a_{p}

. At periapsis the planetesimal moves prograde faster than the planet and passes just outside it. Its relative longitude initially increases, reaches a turning point near

r = 2^{1/4} a_{p}

at a lead of about

1 0^{\circ}

, and then decreases because its angular speed has fallen below the frame rate. Meanwhile its radius continues to grow along the outgoing parabolic branch. A sketch should therefore show an exterior cusp-like closest passage, a short prograde sweep to maximum lead, and a long outward trail that bends retrograde in the rotating frame.

iii

Solution

Let

v_{K} = G M_{*} / a_{p}

. At periapsis the nearly parabolic planetesimal and planet have parallel speeds

2 v_{K}

and

v_{K}

, so

v_{\infty} ≃ (2 - 1) v_{K} .

For a passage just outside the planet, the impact parameter is

b ≃ δ a_{p}

. In the small-angle limit of the supplied gravitational scattering angle,

\frac{θ}{2} ≃ \frac{G M _{p}}{b v _{\infty}^{2}} = \frac{μ}{δ ( 2 - 1 ) ^{2}} .

Thus

θ ≃ A \frac{μ}{δ}, A = \frac{2}{( 2 - 1 ) ^{2}} = 6 + 42 .

iv

Solution

In the planetocentric frame, the incoming relative velocity

w_{i}

is prograde and the encounter outside the planet bends it inward through

θ

, without changing

w = ∣ w ∣

. In the inertial frame

v = v_{p} + w

, so

Δ (v^{2}) = 2 v_{p} \cdot (w_{f} - w_{i}) = 2 v_{K} v_{\infty} (cos θ - 1) ≃ - v_{K} v_{\infty} θ^{2} .

Substitution of the previous result yields the negative gravitational assist energy kick

Δ (v^{2}) = - B (\frac{μ}{δ})^{2}, B = \frac{4}{( 2 - 1 ) ^{3}} \frac{G M _{*}}{a _{p}} = 4 (7 + 52) \frac{G M _{*}}{a _{p}} .

v

Solution

The specific orbital energy changes by

Δ ε = Δ (v^{2}) /2 < 0

, so a marginally bound orbit becomes more tightly bound. Neglecting its tiny initial binding energy,

a^{'} ≃ \frac{G M _{*}}{B} (\frac{δ}{μ})^{2} = \frac{a _{p}}{4 ( 7 + 5 2 )} (\frac{δ}{μ})^{2} .

To obtain the new periapsis, put

c = 2 - 1

and

s = μ / δ

. Immediately after the encounter,

v_{r}^{'} ≃ - c v_{K} θ, v_{θ}^{'} ≃ v_{K} (2 - \frac{c θ ^{2}}{2}) .

Using the post-encounter energy and angular momentum in the Kepler orbit relations and retaining

O (s^{2})

gives

q^{'} ≃ q [1 - (6 + 42) (\frac{μ}{δ})^{2}] .

The inward radial kick therefore lowers the next periapsis as well as shrinking the semi-major axis.

vi

Solution