Third-body perturbations: Worked examples — Physics deep dive

This page is a drill floor. The parent note built the physics; here we hit every case the topic can throw at you — every direction of the satellite relative to the third body, the two limiting geometries (in-line vs sideways), the degenerate cases ( $r \to 0$ , $r$ comparable to $d$ ), a real stationkeeping word problem, and an exam twist that tries to trick you with mass.

Before we start, one reminder of the two workhorses from the parent, so every symbol below is already earned:

The scenario matrix

Every problem in this topic is one (or a mix) of these cells. Each worked example below is tagged with the cell it covers.

Cell	What varies	Question it answers	Example
A. Along-line, near side	$r ∥ \hat{d}$	Max stretch, sign of accel	Ex 1
B. Along-line, far side	$r ∥ - \hat{d}$	Does the far side also stretch?	Ex 2
C. Perpendicular	$r ⊥ \hat{d}$	The squeeze case, sign flip	Ex 3
D. General angle	$\hat{d} \cdot r$ arbitrary	Both components at once	Ex 4
E. Sun vs Moon	mass vs distance	Which body dominates, why cube	Ex 5
F. Degenerate $r \to 0$	satellite at Earth centre	Limiting behaviour, does it vanish?	Ex 6
G. Exact vs tidal	$r$ not $≪ d$	When does the approximation break?	Ex 7
H. Real-world word problem	GEO stationkeeping	Turn accel into a $Δ v$ budget	Ex 8
I. Exam twist	"more massive ⇒ stronger?"	Trap on $μ$ vs $μ / d^{3}$	Ex 9

The geometry cells (A, B, C, D) all live on one picture. Look at it before reading the examples:

The red arrow is $\hat{d}$ (toward the third body). The three coloured satellites sit along $\hat{d}$ , opposite $\hat{d}$ , and perpendicular to it. Watch which way each little tidal arrow points — that is the answer to cells A, B, C.

Worked examples

Throughout we use the standard numbers:

Moon: $μ_{M} = 4.90 \times 1 0^{12} m^{3} / s^{2}$ , $d_{M} = 3.84 \times 1 0^{8}$ m.
Sun: $μ_{S} = 1.327 \times 1 0^{20} m^{3} / s^{2}$ , $d_{S} = 1.496 \times 1 0^{11}$ m.
GEO radius $r = 4.22 \times 1 0^{7}$ m.

For convenience define the tidal coefficient $k \equiv μ_{M} / d_{M}^{3}$ . Let us compute it once and reuse it.

Worked example Ex 1 — Cell A: satellite ON the Earth–Moon line, near side

A GEO satellite sits directly between Earth and Moon, so $r = r \hat{d}$ with $r = 4.22 \times 1 0^{7}$ m. Find its tidal acceleration.

Forecast: guess the direction — toward the Moon, away, or sideways? And is the magnitude $k r$ or $2 k r$ ?

Compute $\hat{d} \cdot r$ . Since $r = r \hat{d}$ and $\hat{d} \cdot \hat{d} = 1$ , we get $\hat{d} \cdot r = r$ . Why this step? The tidal formula needs the projection of $r$ onto $\hat{d}$ ; here it's the whole length.
Plug into the tidal form. $a_{3 b} = k (3 (\hat{d} \cdot r) \hat{d} - r) = k (3 r \hat{d} - r \hat{d}) = 2 k r \hat{d} .$ Why this step? The two terms line up on $\hat{d}$ ; $3 - 1 = 2$ — the stretch coefficient.
Number. $2 k r = 2 (8.65 \times 1 0^{- 14}) (4.22 \times 1 0^{7}) = 7.3 \times 1 0^{- 6} m/s^{2}$ , pointing toward the Moon ( $+ \hat{d}$ ).

Verify: direction $+ \hat{d}$ = toward the Moon, matching the near-side arrow in the figure (stretch pulls the near satellite further out from Earth toward the Moon). Magnitude matches parent Ex 2. ✓

Worked example Ex 2 — Cell B: satellite on the FAR side of Earth

Same GEO satellite but on the opposite side: $r = - r \hat{d}$ . Does it get pushed toward the Moon (falling in) or away?

Forecast: a naive "the Moon pulls everything toward it" says toward the Moon. Is that right?

Projection. $\hat{d} \cdot r = \hat{d} \cdot (- r \hat{d}) = - r$ . Why this step? Now the satellite is on the $- \hat{d}$ side, so the projection is negative — sign matters.
Plug in. $a_{3 b} = k (3 (- r) \hat{d} - (- r \hat{d})) = k (- 3 r + r) \hat{d} = - 2 k r \hat{d} .$ Why this step? Both signs conspire to give $- 2 k r \hat{d}$ — pointing in $- \hat{d}$ , i.e. away from the Moon.
Number. Magnitude again $7.3 \times 1 0^{- 6} m/s^{2}$ , but direction is away from the Moon (outward on the far side).

Verify: this is why tides have two bulges. Near side: pulled toward Moon. Far side: pushed away from Moon. Both bulges point outward from Earth's centre — a net stretch along the line. Look at the two along-line arrows in the figure: they point in opposite absolute directions but both away from Earth. ✓

Worked example Ex 3 — Cell C: satellite PERPENDICULAR to the Moon line

Now $r$ points sideways, perpendicular to $\hat{d}$ , still with $∣ r ∣ = r = 4.22 \times 1 0^{7}$ m. Find the tidal acceleration.

Forecast: stretch or squeeze? Toward Earth or away?

Projection. $\hat{d} \cdot r = 0$ because perpendicular vectors have zero dot product. Why this step? This is the whole point of the perpendicular case — the $3 (\hat{d} \cdot r) \hat{d}$ stretch term switches off.
Plug in. $a_{3 b} = k (3 (0) \hat{d} - r) = - k r .$ Why this step? Only the $- r$ survives — it points back toward Earth's centre, a squeeze.
Number. Magnitude $k r = (8.65 \times 1 0^{- 14}) (4.22 \times 1 0^{7}) = 3.65 \times 1 0^{- 6} m/s^{2}$ , directed inward.

Verify: the squeeze magnitude is exactly half the stretch magnitude ( $k r$ vs $2 k r$ ) and opposite in sense (inward vs outward). That is the classic tidal signature: stretch $+ 2$ , squeeze $- 1$ , sum along three axes $= 2 - 1 - 1 = 0$ — a trace-free (volume-preserving) field, as any pure tidal field must be. ✓

Worked example Ex 4 — Cell D: a general 45° angle (both components at once)

Let $r$ make a $4 5^{\circ}$ angle with $\hat{d}$ , magnitude $r = 4.22 \times 1 0^{7}$ m. Work in a plane with $\hat{d} = (1, 0)$ , so $r = r (cos 4 5^{\circ}, sin 4 5^{\circ})$ . Find $a_{3 b}$ .

Forecast: the answer will have a component along $\hat{d}$ and one perpendicular — not purely stretch or squeeze.

Projection. $\hat{d} \cdot r = r cos 4 5^{\circ} = r / 2$ . Why this step? At an intermediate angle the stretch term is partially on — scaled by $cos 4 5^{\circ}$ .
Assemble the two terms. With $r = (r / 2, r / 2)$ and $\hat{d} = (1, 0)$ : $a_{3 b} = k (3 \frac{r}{2} (1, 0) - (\frac{r}{2}, \frac{r}{2})) = k (\frac{2 r}{2}, - \frac{r}{2}) .$ Why this step? Component-by-component subtraction: along $\hat{d}$ we get $3 \frac{r}{2} - \frac{r}{2} = \frac{2 r}{2}$ (stretch), perpendicular we get $- \frac{r}{2}$ (squeeze).
Magnitude. $∣ a_{3 b} ∣ = k r (2)^{2} + (1/ 2)^{2} = k r 2 + 0.5 = k r 2.5 \approx 1.58 k r .$ Number: $1.58 \times (3.65 \times 1 0^{- 6}) \approx 5.8 \times 1 0^{- 6} m/s^{2}$ .

Verify: the magnitude sits between the pure squeeze ( $1 k r$ ) and pure stretch ( $2 k r$ ) — sanity confirmed. As the angle $\to 0$ it must approach $2 k r$ ; at $4 5^{\circ}$ it's $1.58 k r$ , comfortably in the middle. ✓

Worked example Ex 5 — Cell E: Sun vs Moon, and why the cube wins

Rank the Sun and Moon by tidal strength on any Earth satellite.

Forecast: the Sun is $\sim 27$ million times more massive than the Moon. Surely it wins?

Recognise $r$ cancels. Tidal accel $\propto μ_{3} / d^{3}$ for a fixed satellite, so ratio $= \frac{μ _{S} / d _{S}^{3}}{μ _{M} / d _{M}^{3}}$ . Why this step? We compare bodies, so the common $r$ drops — only the $μ / d^{3}$ factor matters.
Compute each factor. $\frac{μ _{M}}{d _{M}^{3}} = 8.65 \times 1 0^{- 14}, \frac{μ _{S}}{d _{S}^{3}} = \frac{1.327 \times 1 0 ^{20}}{( 1.496 \times 1 0 ^{11} ) ^{3}} = 3.96 \times 1 0^{- 14} .$
Ratio (Moon ÷ Sun). $8.65/3.96 \approx 2.18$ . Why this step? The Moon's tide is about 2.2× the Sun's, despite being tiny in mass — the inverse-cube punishes distance brutally ( $d_{S} / d_{M} \approx 390$ , and $39 0^{3} \approx 5.9 \times 1 0^{7}$ ).

Verify: cross-check with the mass ratio. $μ_{S} / μ_{M} \approx 2.7 \times 1 0^{7}$ ; divide by $39 0^{3} \approx 5.9 \times 1 0^{7}$ → $\approx 0.46$ , i.e. Sun/Moon $\approx 0.46$ , so Moon/Sun $\approx 2.2$ . Two independent routes agree. ✓

Worked example Ex 6 — Cell F: degenerate case

$r \to 0$ What happens to $a_{3 b}$ as the satellite approaches Earth's centre ( $r \to 0$ )? Use the exact form to be safe.

Forecast: a common fear is "the Moon's direct pull $μ_{3} / d^{2}$ blows up or stays finite" — but is the perturbation finite? Does it vanish?

Set $r = 0$ in the exact formula. $a_{3 b} = μ_{3} (\frac{d - 0}{∣ d - 0 ∣ ^{3}} - \frac{d}{d ^{3}}) = μ_{3} (\frac{d}{d ^{3}} - \frac{d}{d ^{3}}) = 0 .$ Why this step? At the centre, satellite and Earth are the same point; the Moon pulls them identically, so the difference is exactly zero.
Confirm with the tidal form. $3 (\hat{d} \cdot 0) \hat{d} - 0 = 0$ . Also zero. Why this step? Both forms must agree in the limit — they do.

Verify: this kills the parent's third "common mistake." The indirect term is precisely what makes the perturbation vanish at $r = 0$ . The naive "direct pull $μ_{3} / d^{2}$ " (using the Earth–Moon distance $d$ , since at the centre the satellite–Moon distance equals $d$ ) would give $μ_{3} / d^{2} \neq = 0$ — clearly wrong for a satellite sitting at Earth's centre, which feels no differential tug at all. ✓

Worked example Ex 7 — Cell G: exact vs tidal, how big is the error?

Take $r = r \hat{d}$ (near-side, along-line) so the exact formula is scalar. Compare the exact and tidal answers for two cases: GEO ( $r / d_{M} = 0.11$ ) and a deep case $r = 0.5 d_{M}$ .

Forecast: the tidal form drops $O ((r / d)^{2})$ . At $r / d = 0.11$ the error should be small; at $r / d = 0.5$ it should be large.

Exact along-line. With $r = r \hat{d}$ , $d - r = (d - r) \hat{d}$ : $a_{exact} = μ_{3} (\frac{1}{( d - r ) ^{2}} - \frac{1}{d ^{2}}) .$ Why this step? Both terms point along $\hat{d}$ , so we can drop vectors and use signed scalars; note the near-side satellite is closer to the Moon, so $(d - r)^{2} < d^{2}$ → positive (toward Moon). ✓
GEO case ( $x \equiv r / d_{M} = 0.110$ ). Exact factor $μ_{M} [(d - r)^{- 2} - d^{- 2}]$ . Numerically the exact along-line accel $\approx 8.7 \times 1 0^{- 6}$ m/s²; tidal gave $2 k r = 7.3 \times 1 0^{- 6}$ m/s². Error $\approx 16%$ — the first neglected term is $O (x) \approx 0.11$ , and the next correction bumps it to roughly one-sixth. Why this step? At GEO the "small" parameter $r / d$ is $0.11$ , so the tidal form is good to only about $15%$ – $16%$ — fine for scaling arguments, not for high-precision propagation.
Deep case $r = 0.5 d_{M}$ . Exact: $μ_{M} [(0.5 d)^{- 2} - d^{- 2}] = μ_{M} / d^{2} \cdot (4 - 1) = 3 μ_{M} / d^{2}$ . Tidal: $2 k r = 2 (μ_{M} / d^{3}) (0.5 d) = μ_{M} / d^{2}$ . Exact is 3× the tidal estimate — the approximation has collapsed. Why this step? When $r$ is no longer $≪ d$ you must use the exact form (this is the regime of lunar-transfer trajectories).

Verify: the tidal form is the leading term of the exact one — it should always be an underestimate on the near side (since $(d - r)^{- 2}$ grows super-linearly). Both cases confirm exact $>$ tidal. ✓

Worked example Ex 8 — Cell H: real-world GEO stationkeeping

$Δ v$ The luni-solar tidal acceleration slowly tilts a GEO orbit's plane, forcing North–South stationkeeping burns. Estimate the yearly $Δ v$ if the effective out-of-plane tidal accel averages $a_{⊥} \approx 3.6 \times 1 0^{- 6} m/s^{2}$ (order of the Moon's squeeze from Ex 3).

Forecast: guess whether it's grams-per-year or tens of m/s per year.

Seconds in a year. $T = 365.25 \times 24 \times 3600 = 3.156 \times 1 0^{7}$ s. Why this step? A continuous acceleration accumulates velocity as $a \cdot t$ ; we need $t$ in SI.
Accumulated velocity change. $Δ v \sim a_{⊥} T = 3.6 \times 1 0^{- 6} \times 3.156 \times 1 0^{7} \approx 114 m/s per year .$ Why this step? Multiply the tiny accel by the huge number of seconds — this is how $1 0^{- 6}$ becomes operationally huge.
Interpret. Real GEO N–S stationkeeping costs $\approx 45$ – $55$ m/s per year. Our crude estimate (114) is the right order of magnitude — the true value is smaller because the accel oscillates and only its slow-secular part must be cancelled.

Verify: units: $(m/s^{2}) (s) = m/s$ . ✓ Order of magnitude (tens of m/s/yr) matches the well-known fact that N–S keeping dominates the GEO fuel budget — confirming the parent note's claim that stationkeeping is "dominated by luni-solar effects." ✓

Worked example Ex 9 — Cell I: the exam trap

Exam statement: "Jupiter is $\sim 318$ Earth masses and vastly more massive than the Moon. Therefore Jupiter's tidal effect on an Earth satellite exceeds the Moon's. True or false?"

Forecast: the word "vastly more massive" is bait. Reach for $μ / d^{3}$ , never $μ$ alone.

Get $μ_{J}$ and $d_{J}$ . $μ_{J} \approx 1.267 \times 1 0^{17} m^{3} / s^{2}$ ; closest approach $d_{J} \approx 5.9 \times 1 0^{11}$ m. Why this step? Tidal strength never uses mass alone — always $μ / d^{3}$ , so we need both numbers.
Compute Jupiter's factor. $\frac{μ _{J}}{d _{J}^{3}} = \frac{1.267 \times 1 0 ^{17}}{( 5.9 \times 1 0 ^{11} ) ^{3}} = \frac{1.267 \times 1 0 ^{17}}{2.05 \times 1 0 ^{35}} \approx 6.2 \times 1 0^{- 19} s^{- 2} .$ Why this step? This is the tidal coefficient for Jupiter, directly comparable to the Moon's $k = 8.65 \times 1 0^{- 14}$ .
Ratio to the Moon. $\frac{6.2 \times 1 0 ^{- 19}}{8.65 \times 1 0 ^{- 14}} \approx 7 \times 1 0^{- 6}$ — Jupiter's tide is about one hundred-thousandth of the Moon's. Why this step? Jupiter's mass advantage ( $\sim 26000 \times$ the Moon) is annihilated by being $\sim 1500 \times$ farther: $150 0^{3} \approx 3 \times 1 0^{9}$ .

Answer: False, and by a huge margin — Jupiter is utterly negligible for Earth satellites.

Verify: the trap is exactly the parent's mistake "Sun dominates because it's massive," escalated. Sanity check the annihilation directly: $(μ_{J} / μ_{M}) / (d_{J} / d_{M})^{3} = (1.267 \times 1 0^{17} /4.90 \times 1 0^{12}) / (5.9 \times 1 0^{11} /3.84 \times 1 0^{8})^{3} \approx 2.6 \times 1 0^{4} /3.6 \times 1 0^{9} \approx 7 \times 1 0^{- 6}$ — matches step 3. Always rank by $μ / d^{3}$ . ✓

Recall

Recall Near side vs far side — which way does tidal accel point?

Near side ::: toward the Moon (outward from Earth); far side ::: away from the Moon (also outward from Earth). Both bulges point outward — two tidal bulges.

Recall Stretch vs squeeze coefficients

Along $\hat{d}$ ::: $+ 2 μ_{3} r / d^{3}$ (stretch); perpendicular ::: $- μ_{3} r / d^{3}$ (squeeze); the trace is zero.

Recall Why is

$a_{3 b} = 0$ at Earth's centre? The direct and indirect terms become identical ::: their difference vanishes; the satellite and Earth feel the same pull.

Recall How do you rank two third bodies by tidal strength?

Compare $μ / d^{3}$ ::: never mass alone — the inverse cube makes nearness dominate.

See also: Tidal forces, J2 perturbation and oblateness, GEO stationkeeping, Gauss and Lagrange planetary equations, Restricted three-body problem, Two-body problem.