Third-body perturbations: Exercises — Physics deep dive

Throughout, we reuse these constants (SI units, so metres, seconds):

The two formulas everything below rests on (from the parent):

Level 1 — Recognition

L1.1 — Spot the perturbing quantity

Problem. A student writes: "The Moon perturbs my satellite with acceleration $μ_{M} / s^{2}$ , where $s$ is the Moon–satellite distance." Is this the perturbing acceleration? Answer yes/no and name the quantity that actually perturbs the orbit.

Recall Solution

No. $μ_{M} / s^{2}$ is the Moon's raw pull on the satellite in an inertial frame. But we track the orbit in an Earth-centred frame, and the Moon also pulls Earth. The quantity that perturbs the orbit is the difference between the Moon's pull on the satellite and its pull on Earth — the tidal (differential) acceleration $a_{3 b} = μ_{3} (\frac{d - r}{∣ d - r ∣ ^{3}} - \frac{d}{d ^{3}})$ . The common part cancels because Earth is free-falling toward the Moon just like the satellite is.

L1.2 — Read the scaling law

Problem. Tidal strength scales as $μ_{3} / d^{n}$ . What is $n$ , and does the perturbation grow or shrink if the third body is moved twice as far away (same mass)?

Recall Solution

$n = 3$ (inverse cube, not inverse square). Doubling $d$ multiplies the tidal strength by $1/ 2^{3} = 1/8$ — it shrinks to one-eighth. The inverse cube is why nearness matters so much: it is the gradient (rate of change) of an inverse-square field, and differentiating $1/ d^{2}$ gives $1/ d^{3}$ .

Level 2 — Application

L2.1 — Which dominates, Sun or Moon?

Problem. Compute the tidal strength factor $μ_{3} / d^{3}$ for the Moon and the Sun, then their ratio. Which wins?

Recall Solution

Moon: $\frac{μ _{M}}{d _{M}^{3}} = \frac{4.90 \times 1 0 ^{12}}{( 3.84 \times 1 0 ^{8} ) ^{3}} = 8.66 \times 1 0^{- 14} s^{- 2}$ . Sun: $\frac{μ _{S}}{d _{S}^{3}} = \frac{1.327 \times 1 0 ^{20}}{( 1.496 \times 1 0 ^{11} ) ^{3}} = 3.96 \times 1 0^{- 14} s^{- 2}$ . Ratio $= 8.66/3.96 \approx 2.19$ . The Moon dominates by about 2×. We only need $μ_{3} / d^{3}$ because in the tidal formula the satellite distance $r$ multiplies both bodies identically and cancels in the ratio.

L2.2 — Peak tidal acceleration at GEO from the Moon

Problem. A GEO satellite sits along the Earth–Moon line ( $\hat{d} \cdot r = r$ ). Find the peak (stretching) tidal acceleration from the Moon.

Recall Solution

Along the line, $3 (\hat{d} \cdot r) \hat{d} - r = 3 r \hat{d} - r \hat{d} = 2 r \hat{d}$ , so the magnitude is the stretch coefficient $\frac{2 μ _{M} r}{d _{M}^{3}}$ . $a = 2 (8.66 \times 1 0^{- 14}) (4.22 \times 1 0^{7}) = 7.3 \times 1 0^{- 6} m/s^{2} .$ Tiny compared with Earth's own pull at GEO ( $μ / r^{2} \approx 0.224$ m/s²), but relentless — over years it precesses the orbit plane and drives the North–South drift that GEO stationkeeping fights.

Level 3 — Analysis

L3.1 — Stretch vs squeeze directions

Problem. Take $\hat{d} = (1, 0, 0)$ (Moon along the $x$ -axis). Write the tidal acceleration for a satellite (a) exactly toward the Moon, $r = (r, 0, 0)$ ; and (b) exactly perpendicular, $r = (0, r, 0)$ . Interpret the signs.

The figure below fixes the geometry we will use. Earth sits at the origin (black dot). The black horizontal arrow is $\hat{d}$ , the direction from Earth toward the Moon (our $+ x$ -axis). Three satellites are drawn: one on the near side along the line ( $x = + r$ ) whose red acceleration arrow points outward, away from Earth (stretch); one on the far side ( $x = - r$ ) whose red arrow also points outward (stretch, opposite direction); and one perpendicular ( $y = + r$ ) whose black arrow points inward, toward Earth (squeeze). Read off from the picture that the two along-line arrows are twice as long as the perpendicular one — that is the 2:1 stretch-to-squeeze ratio you will now confirm algebraically.

Recall Solution

Use $a_{3 b} = \frac{μ _{3}}{d ^{3}} (3 (\hat{d} \cdot r) \hat{d} - r)$ .

(a) $\hat{d} \cdot r = r$ , so $a = \frac{μ _{3}}{d ^{3}} (3 r \hat{d} - r \hat{d}) = \frac{μ _{3}}{d ^{3}} (2 r, 0, 0)$ . Positive $x$ = points away from Earth, toward the Moon side → stretching, coefficient $+ 2 μ_{3} r / d^{3}$ . This is the red near-side arrow in the figure.

(b) $\hat{d} \cdot r = 0$ , so $a = \frac{μ _{3}}{d ^{3}} (0 - r \overset{y}{^}) = \frac{μ _{3}}{d ^{3}} (0, - r, 0)$ . Negative $y$ = points toward Earth → squeezing, coefficient $- μ_{3} r / d^{3}$ . This is the black inward arrow in the figure.

The stretch (along the line) is exactly twice the squeeze magnitude (perpendicular). This 2:1 ratio is the fingerprint of every tidal field.

L3.2 — The near side and far side both stretch

Problem. Does a satellite on the far side of Earth from the Moon ( $r = (- r, 0, 0)$ , still with $\hat{d} = (1, 0, 0)$ ) get pulled toward the Moon or away from Earth? Show it, and explain why "both sides bulge outward."

Recall Solution

$\hat{d} \cdot r = - r$ , so $a = \frac{μ _{3}}{d ^{3}} (3 (- r) \hat{d} - (- r \hat{d})) = \frac{μ _{3}}{d ^{3}} (- 3 r + r, 0, 0) = \frac{μ _{3}}{d ^{3}} (- 2 r, 0, 0)$ . The acceleration points in $- x$ = away from Earth on the far side too. So both the near side (pulled toward Moon, away from Earth) and the far side (Earth pulled toward Moon more than the satellite, leaving the satellite "behind" = outward relative to Earth) get stretched outward. This is the two-bulge tidal pattern — same reason Earth has two ocean tides per day. See Tidal forces.

Level 4 — Synthesis

L4.1 — Exact vs tidal error at GEO

Problem. For a GEO satellite along the Earth–Moon line, compute (a) the ratio $r_{GE O} / d_{M}$ and (b) estimate the fractional error made by using the tidal (first-order) formula instead of the exact one. The leading dropped term is $O ((r / d)^{2})$ , so estimate the error size as roughly $r / d$ .

Recall Solution

(a) $r_{GE O} / d_{M} = \frac{4.22 \times 1 0 ^{7}}{3.84 \times 1 0 ^{8}} = 0.110$ . (b) The tidal formula keeps terms linear in $r / d$ and drops those of order $(r / d)^{2}$ . The relative size of the first dropped term is on the order of $r / d \approx 0.11$ , i.e. roughly an 11% error at GEO. Acceptable for quick estimates; for precision propagation you keep the exact form or add the next (quadrupole) term. This is exactly the small parameter that also governs J2 perturbation and oblateness-style expansions.

L4.2 — Combine Sun and Moon at GEO

Problem. Assume the Moon and Sun are momentarily aligned along the same line as the satellite ( $\hat{d} \cdot r = r$ for both). Find the combined peak stretching acceleration at GEO, and compare to the Moon-only value from L2.2.

Recall Solution

Combined stretch coefficient $= 2 r (\frac{μ _{M}}{d _{M}^{3}} + \frac{μ _{S}}{d _{S}^{3}})$ . $a = 2 (4.22 \times 1 0^{7}) (8.66 \times 1 0^{- 14} + 3.96 \times 1 0^{- 14}) = 2 (4.22 \times 1 0^{7}) (1.262 \times 1 0^{- 13}) = 1.065 \times 1 0^{- 5} m/s^{2} .$ That's $\approx 1.07 \times 1 0^{- 5}$ m/s², about 1.46× the Moon-only $7.3 \times 1 0^{- 6}$ m/s². When aligned they reinforce; when perpendicular they partly cancel — the reason luni-solar stationkeeping demand varies through the year. See GEO stationkeeping.

Level 5 — Mastery

L5.1 — Where does third-body beat $J_{2}$ ?

Problem. Earth's oblateness ( $J_{2}$ ) produces a perturbing acceleration that falls off with altitude roughly as $a_{J 2} \approx \frac{3}{2} J_{2} \frac{μ R _{E}^{2}}{r ^{4}}$ (it scales as $1/ r^{4}$ ), while the third-body tidal term grows as $a_{3 b} \approx \frac{2 μ _{M} r}{d _{M}^{3}}$ (scales as $r$ ). Using $J_{2} = 1.083 \times 1 0^{- 3}$ and $R_{E} = 6.378 \times 1 0^{6}$ m, find the crossover radius $r_{\times}$ where the Moon's tidal acceleration equals $J_{2}$ 's. Interpret.

Recall Solution

Set them equal: $\frac{3}{2} J_{2} \frac{μ R _{E}^{2}}{r ^{4}} = \frac{2 μ _{M} r}{d _{M}^{3}} \Rightarrow r^{5} = \frac{3 J _{2} μ R _{E}^{2} d _{M}^{3}}{4 μ _{M}} .$ Numerator: $3 (1.083 \times 1 0^{- 3}) (3.986 \times 1 0^{14}) (6.378 \times 1 0^{6})^{2} (3.84 \times 1 0^{8})^{3}$ .

$(6.378 \times 1 0^{6})^{2} = 4.068 \times 1 0^{13}$
$(3.84 \times 1 0^{8})^{3} = 5.662 \times 1 0^{25}$
Product $= 3 (1.083 \times 1 0^{- 3}) (3.986 \times 1 0^{14}) (4.068 \times 1 0^{13}) (5.662 \times 1 0^{25}) = 2.983 \times 1 0^{51}$ Divide by $4 μ_{M} = 1.96 \times 1 0^{13}$ : $r^{5} = 1.522 \times 1 0^{38}$ . $r_{\times} = (1.522 \times 1 0^{38})^{1/5} = 4.33 \times 1 0^{7} m \approx 6.8 R_{E} .$ Interpretation: below $\approx 4.3 \times 1 0^{7}$ m, $J_{2}$ oblateness dominates the perturbation budget (LEO cares about $J_{2}$ , not the Moon). Above it — GEO sits right near this crossover, while GTO apogee, Molniya apogee, and lunar transfers are well beyond — third-body luni-solar effects take over. This crossover is why mission designers switch which perturbation they model first depending on orbit altitude. Compare with Gauss and Lagrange planetary equations for how these accelerations feed into slow orbital-element drift, and Restricted three-body problem for the full nonlinear regime when $r$ is no longer $≪ d$ .

L5.2 — Sanity check the tidal limit against the exact form

Problem. For $\hat{d} = (1, 0, 0)$ , $r = (r, 0, 0)$ , show algebraically that the exact $a_{3 b}$ reduces to $+ 2 μ_{3} r / d^{3}$ in the limit $r / d \to 0$ , confirming the tidal coefficient. Give the exact expression and its first-order value at $r / d = 0.11$ .

Recall Solution

Along the line, $∣ d - r ∣ = d - r$ (both along $+ x$ ), so the exact $x$ -component is $a_{x} = μ_{3} (\frac{d - r}{( d - r ) ^{3}} - \frac{d}{d ^{3}}) = μ_{3} (\frac{1}{( d - r ) ^{2}} - \frac{1}{d ^{2}}) .$ Expand $\frac{1}{( d - r ) ^{2}} = \frac{1}{d ^{2}} (1 - r / d)^{- 2} \approx \frac{1}{d ^{2}} (1 + 2 r / d)$ . Then $a_{x} \approx μ_{3} (\frac{1}{d ^{2}} + \frac{2 r}{d ^{3}} - \frac{1}{d ^{2}}) = \frac{2 μ _{3} r}{d ^{3}},$ exactly the tidal stretch coefficient. At $r / d = 0.11$ the exact bracket (in units of $μ_{3} / d^{2}$ ) is $(1 - 0.11)^{- 2} - 1 = 0.263$ vs the linear estimate $2 (0.11) = 0.22$ — a relative discrepancy of $(0.263 - 0.22) /0.263 \approx 16%$ (about $20%$ if measured against the linear estimate). This confirms the first-order approximation is self-consistent and its error grows with $r / d$ , matching the order-of-magnitude estimate in L4.1.

Recall check

Recall Quick self-quiz

Perturbation quantity — raw pull or difference? ::: The difference (tidal/differential acceleration). Scaling of tidal strength with $d$ ? ::: Inverse cube, $μ_{3} / d^{3}$ . Sun-vs-Moon ratio of $μ / d^{3}$ ? ::: About 2.19, Moon wins. Stretch-to-squeeze ratio in a tidal field? ::: 2 to 1. Roughly where does third-body overtake $J_{2}$ for Earth orbits? ::: Near $r \approx 6.8 R_{E}$ ( $\approx 4.3 \times 1 0^{7}$ m, close to GEO).