Visual walkthrough — Third-body perturbations

Before we begin, here is the entire cast of characters, drawn once so you never wonder what a letter means.

One more piece of grammar we will lean on constantly:

Now we derive.

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Step 1 — Write every real pull on the satellite

WHAT. In the true inertial frame, the satellite is pulled by two masses at once: Earth and the third body. Add both arrows.

${\ddot{\vec{r}}}_s=\underset{\text{Earth’s pull on satellite}}{\underbrace{-\mu \frac{{\vec{r}}_s-{\vec{r}}_E}{|{\vec{r}}_s-{\vec{r}}_E{|}^3}}}+\underset{\text{third body’s pull on satellite}}{\underbrace{{\mu }_3\frac{{\vec{r}}_3-{\vec{r}}_s}{|{\vec{r}}_3-{\vec{r}}_s{|}^3}}}$

• ${\ddot{\vec{r}}}_s$ = the acceleration of the satellite (the double-dot means "rate of change of the rate of change of position" — how the velocity arrow is bending). This is what we ultimately want.
• ${\vec{r}}_s-{\vec{r}}_E$ points from Earth to the satellite, so the minus sign flips it to point from satellite back to Earth — Earth pulls the satellite inward. Good.
• ${\vec{r}}_3-{\vec{r}}_s$ points from satellite to the third body — no minus needed, the pull already aims that way.

WHY. This is just "list all forces, divide by mass." Nothing subtle yet — we are honest about every tug the satellite feels.

PICTURE. Two arrows leaving the satellite: a long one toward Earth, a faint one toward the Moon.

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Step 2 — Earth is not nailed down — it falls too

WHAT. The very frame we watch from (Earth's centre) is itself being pulled by the third body. Write Earth's own acceleration.

${\ddot{\vec{r}}}_E=\underset{\text{third body’s pull on Earth}}{\underbrace{{\mu }_3\frac{{\vec{r}}_3-{\vec{r}}_E}{|{\vec{r}}_3-{\vec{r}}_E{|}^3}}}$

• ${\ddot{\vec{r}}}_E$ = how Earth's centre accelerates. The third body yanks the whole planet toward it.
• ${\vec{r}}_3-{\vec{r}}_E$ = arrow from Earth to the third body — Earth accelerates that way.

WHY. We measure the satellite relative to Earth. If our ruler's zero-point is accelerating, we must account for that motion or our numbers are wrong. The Moon does not just pull the satellite — it pulls the platform under our feet.

PICTURE. The whole Earth drifting toward the Moon; the satellite drawn faded to remind us it drifts too.

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Step 3 — Subtract: watch the common pull vanish

WHAT. We want the satellite's motion as seen from Earth: the relative acceleration $\ddot{\vec{r}}$, where $\vec{r}={\vec{r}}_s-{\vec{r}}_E$. Subtract Step 2 from Step 1.

$\ddot{\vec{r}}={\ddot{\vec{r}}}_s-{\ddot{\vec{r}}}_E=-\mu \frac{\vec{r}}{r^3}+\underset{\text{third-body perturbation }{\vec{a}}_{3b}}{\underbrace{{\mu }_3\left(\frac{{\vec{r}}_3-{\vec{r}}_s}{|{\vec{r}}_3-{\vec{r}}_s{|}^3}-\frac{{\vec{r}}_3-{\vec{r}}_E}{|{\vec{r}}_3-{\vec{r}}_E{|}^3}\right)}}$

• $\vec{r}={\vec{r}}_s-{\vec{r}}_E$ = the satellite's position relative to Earth (this is the orbit we actually track). $r=|\vec{r}|$ is its length.
• $-\mu \vec{r}/r^3$ = the plain Keplerian pull — the clean ellipse the satellite would follow if the Moon didn't exist.
• The bracket = (pull on satellite) minus (pull on Earth). This difference is the entire perturbation.

WHY. This subtraction is the heart of everything. Because the satellite and Earth both fall toward the Moon, the shared part of that fall cancels — exactly like astronauts in orbit feeling weightless while plummeting alongside their capsule. What survives is only how the Moon's pull differs across the Earth-to-satellite gap.

PICTURE. Two nearly-parallel Moon arrows (one on the satellite, one on Earth); their difference is a tiny residual arrow — the real culprit.

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Step 4 — Rename for clarity (the exact result)

WHAT. Let $\vec{d}={\vec{r}}_3-{\vec{r}}_E$ = "arrow from Earth to the third body," with length $d$. Then ${\vec{r}}_3-{\vec{r}}_s=\vec{d}-\vec{r}$. Rewrite the bracket:

WHY. Both terms genuinely matter. Keep only the direct term and the formula misbehaves as $r\to 0$ (a satellite at Earth's centre should feel zero perturbation — and only the difference gives that). The two terms together make a gradient, not a raw force.

PICTURE. $\vec{d}$ pointing to the Moon, $\vec{r}$ pointing to the satellite, and $\vec{d}-\vec{r}$ closing the triangle from satellite to Moon.

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Step 5 — Use $r\ll d$: expand the awkward length

WHAT. The satellite–Earth gap $r$ is tiny next to the Earth–Moon gap $d$ (for GEO, $r/d\approx 0.11$; for the Sun, far smaller). We approximate $|\vec{d}-\vec{r}{|}^{-3}$ keeping only the first correction.

First the length-squared, using the dot product $\vec{d}\cdot \vec{r}$ (which measures how much $\vec{r}$ points along $\vec{d}$): $|\vec{d}-\vec{r}{|}^2=d^2-2\vec{d}\cdot \vec{r}+r^2.$

Factor out $d^2$ and write $\hat{d}=\vec{d}/d$ (the unit arrow toward the Moon): $|\vec{d}-\vec{r}{|}^{-3}=d^{-3}(1-2\frac{\vec{d}\cdot \vec{r}}{d^2}+\frac{r^2}{d^2}{)}^{-3/2}.$

WHY these tools.

• Why the dot product? We need to know how much of $\vec{r}$ lies along the Moon's direction — that is precisely what stretches versus squeezes. The dot product is the one operation that extracts "shadow length along a direction," so it is the correct tool, not the cross product (which would give area) or plain multiplication.
• Why the binomial approximation $(1+x{)}^{-3/2}\approx 1-\frac{3}{2}x$? Because $x=-2\frac{\vec{d}\cdot \vec{r}}{d^2}+\frac{r^2}{d^2}$ is small ($\sim r/d$), so its square is negligible. This is the standard "zoom in on a curve and it looks like its tangent line" idea — we keep the straight-line part and throw away the curvature. That is the whole meaning of "first order in $r/d$."

Applying it and dropping the $r^2/d^2$ piece (it is second-order): $|\vec{d}-\vec{r}{|}^{-3}\approx d^{-3}(1+3\frac{\vec{d}\cdot \vec{r}}{d^2}).$

• The $+3$ appears because $-\frac{3}{2}\times (-2)=+3$: the $-\frac{3}{2}$ from the binomial times the $-2$ in front of the dot product.

PICTURE. A curve $(1+x{)}^{-3/2}$ and its straight tangent line hugging it near $x=0$ — the approximation we are making.

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Step 6 — Substitute and collect: the tidal pattern appears

WHAT. Put the expansion into the exact formula and keep only first-order terms: ${\vec{a}}_{3b}\approx {\mu }_3\left[\frac{\vec{d}-\vec{r}}{d^3}(1+3\frac{\vec{d}\cdot \vec{r}}{d^2})-\frac{\vec{d}}{d^3}\right].$

Multiply out. The leading $\vec{d}/d^3$ cancels the indirect term. Among the rest, keep terms linear in $\vec{r}$ (throw away the $\vec{r}\times$ dot-product cross-term, which is second-order): $\boxed{{\vec{a}}_{3b}\approx \frac{{\mu }_3}{d^3}(3(\hat{d}\cdot \vec{r})\hat{d}-\vec{r})}$

• $\hat{d}\cdot \vec{r}$ = the length of $\vec{r}$'s shadow along the Moon-direction (a single number).
• $3(\hat{d}\cdot \vec{r})\hat{d}$ = an arrow pointing along $\hat{d}$, three times that shadow long — the stretch.
• $-\vec{r}$ = pulls straight back toward Earth's centre — the squeeze.
• ${\mu }_3/d^3$ = the overall strength. Note the cube of $d$, not the square: this is the fingerprint of a gradient.

WHY. This is the pattern we chased since Step 3: along the Earth–Moon line the field pulls outward (stretch), sideways it pulls inward (squeeze). Two special directions, opposite effects.

PICTURE. The stretch-squeeze field: arrows pointing out along the Moon line, in along the perpendicular.

Let us confirm the two extreme cases explicitly. First define one more unit arrow: $\hat{r}=\vec{r}/r$ is the unit arrow along $\vec{r}$ itself (the satellite's own direction from Earth) — same idea as $\hat{d}=\vec{d}/d$, just pointing along $\vec{r}$.

The stretch is exactly twice the squeeze — the signature 2-to-1 ratio of every tidal field, the same one behind ocean tides.

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Step 7 — The degenerate and edge cases (never leave a gap)

WHAT & WHY & PICTURE, three quick limits that must behave sensibly:

1. $r\to 0$ (satellite at Earth's centre). Both exact terms in Step 4 become ${\mu }_3\vec{d}/d^3-{\mu }_3\vec{d}/d^3=0$. The perturbation vanishes — correct: something sitting exactly where Earth's centre is falls with Earth, feeling no differential pull.

2. Sun vs Moon (strength is ${\mu }_3/d^3$, not ${\mu }_3$). Using ${\mu }_M=G{m}_{\text{Moon}}$ and ${\mu }_S=G{m}_{\text{Sun}}$ (the earlier ${\mu }_3$, subscripted by body), the Sun's mass gives ${\mu }_S$ about $2.7\times 10^7$ times the Moon's — but the Sun is ~390× farther, and $390^3\approx 5.9\times 10^7$. The cube of distance more than eats the mass advantage. Compute the ratio:

   • Moon: ${\mu }_M/d_M^3=4.90\times 10^{12}/(3.84\times 10^8{)}^3=8.65\times 10^{-14}$.
   • Sun: ${\mu }_S/d_S^3=1.327\times 10^{20}/(1.496\times 10^{11}{)}^3=3.96\times 10^{-14}$.
   • Ratio $\approx 2.2$ — the Moon dominates by about twofold.

3. LEO vs GEO (why altitude decides). Tidal accel $\propto r$. For GEO ($r=4.22\times 10^7$ m) the Moon's peak stretch is $2(8.65\times 10^{-14})(4.22\times 10^7)\approx 7.3\times 10^{-6}{\text{m/s}}^2$. For LEO ($r=7\times 10^6$ m) it is only $\approx 1.2\times 10^{-6}{\text{m/s}}^2$, dwarfed by Earth's ~8.9 m/s² and $J_2$'s ~$10^{-2}$ m/s². Third-body effects matter for high, eccentric orbits (GEO, GTO, Molniya) — the domain of luni-solar North–South stationkeeping.

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The one-picture summary

Everything above in a single frame: raw pulls (Step 1–2) → subtract the common fall (Step 3–4) → zoom in with $r\ll d$ (Step 5–6) → the stretch-and-squeeze tidal field (Step 6–7).

Recall Feynman retelling — say it in plain words

Your satellite is not orbiting a lonely Earth; the Moon pulls it too. But here is the trick: the Moon also pulls the whole Earth — the very platform your ruler sits on. Both of you fall toward the Moon nearly together, so that big shared pull cancels out and you feel almost nothing. Almost. The satellite and Earth's centre are a little apart, so the Moon pulls them by slightly different amounts and in slightly different directions. That leftover — the difference of two nearly-equal arrows — is the whole perturbation. Do the honest bookkeeping: write both pulls on the satellite, write the pull on Earth, subtract. Then use that the satellite is close to Earth compared with how far the Moon is, so you keep only the first tiny correction. What pops out is a beautifully simple field: it stretches things along the Earth–Moon line and squeezes them sideways, exactly twice as hard stretching as squeezing. Its strength falls off as one-over-distance-cubed, which is why the near Moon beats the mighty-but-distant Sun by about two to one, and why only tall, stretched-out orbits ever really feel it.

Recall Quick self-check

Why is the perturbation a difference of two pulls, not just the Moon's pull? ::: Because we measure the satellite relative to Earth, and Earth is itself falling toward the Moon; the common pull cancels, leaving only the tidal gradient. Why does strength go as ${\mu }_3/d^3$ and not ${\mu }_3/d^2$? ::: Because the perturbation is the variation (gradient) of the Moon's inverse-square field across the small gap $r$; differentiating $1/d^2$ gives $1/d^3$. Stretch coefficient vs squeeze coefficient? ::: $+2$ along $\hat{d}$, $-1$ perpendicular — the 2-to-1 tidal signature. Which limit forces the perturbation to zero, and why? ::: $r\to 0$: a body at Earth's centre falls identically with Earth, so no differential force.

Related: Third-body perturbations · Restricted three-body problem · Gauss and Lagrange planetary equations · 3.2.36 Third-body perturbations (Hinglish)