Two-body problem — equations of motion, reduction to one-body

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WHAT we are solving

The two key objects:

• Center of mass $\displaystyle \mathbf{R}=\frac{m_1\mathbf{r}_1+m_2\mathbf{r}_2}{m_1+m_2}$
• Relative position $\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1$, with $r=|\mathbf{r}|$.

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HOW: derive the equations of motion from scratch

Step 1 — Newton's law on each body. Gravity on $m_1$ points toward $m_2$, i.e. along $+\mathbf{r}=\mathbf{r}_2-\mathbf{r}_1$:

\qquad m_2\ddot{\mathbf r}_2 = -\frac{G m_1 m_2}{r^2}\,\hat{\mathbf r}$$ *Why the signs?* Each body is pulled toward the other. $m_1$ is pulled in the direction of $m_2$ (that is $+\hat{\mathbf r}$); $m_2$ is pulled back toward $m_1$ ($-\hat{\mathbf r}$). Newton's third law is built in: the forces are equal and opposite. **Step 2 — Center of mass moves freely.** Add the two equations: $$m_1\ddot{\mathbf r}_1 + m_2\ddot{\mathbf r}_2 = 0 \;\Rightarrow\; \frac{d^2}{dt^2}\big(m_1\mathbf r_1+m_2\mathbf r_2\big)=0 \;\Rightarrow\; \ddot{\mathbf R}=0.$$ *Why this matters:* the internal forces cancel, so **no external force** acts on the system → the COM moves with constant velocity. It carries zero useful orbital info, so we set up coordinates *riding along with it* and forget it. This removes 3 of the 6 unknowns. **Step 3 — Relative motion (the heart).** Divide each Newton equation by its mass: $$\ddot{\mathbf r}_1 = +\frac{G m_2}{r^2}\hat{\mathbf r}, \qquad \ddot{\mathbf r}_2 = -\frac{G m_1}{r^2}\hat{\mathbf r}.$$ Subtract (first from second) to get $\ddot{\mathbf r}=\ddot{\mathbf r}_2-\ddot{\mathbf r}_1$: $$\ddot{\mathbf r} = -\frac{G m_1}{r^2}\hat{\mathbf r} - \frac{G m_2}{r^2}\hat{\mathbf r} = -\frac{G(m_1+m_2)}{r^2}\hat{\mathbf r}.$$ > [!formula] The reduced equation of motion > $$\boxed{\;\ddot{\mathbf r} = -\frac{\mu}{r^2}\,\hat{\mathbf r} = -\frac{\mu}{r^3}\,\mathbf r,\qquad \mu \equiv G(m_1+m_2)\;}$$ > This is **identical** to a single particle orbiting a *fixed* mass that produces gravitational parameter $\mu$. The two-body problem is now a one-body problem. Here $\hat{\mathbf r}=\mathbf r/r$. > [!mistake] "$\mu$ should be $Gm_1$ for a satellite around Earth." > *Why it feels right:* the satellite is tiny, Earth seems fixed, so we expect only Earth's mass. **The exact answer is $\mu=G(m_1+m_2)$** — both masses appear, because each accelerates. *Fix/limit:* when $m_2\ll m_1$ (satellite mass $\ll$ Earth), $m_1+m_2\approx m_1$, so $\mu\approx Gm_1$ is an excellent *approximation*, not the exact statement. For binary stars of comparable mass you **must** keep both. --- ## The reduced mass — getting energy/momentum right We can also write the relative dynamics as a *real* one-body problem with an effective mass. > [!definition] Reduced mass > $$\mu_{\text{red}} = \frac{m_1 m_2}{m_1+m_2}.$$ > (Different symbol than $\mu$! One is a mass, the other is $G\cdot M_\text{tot}$. People reuse $\mu$ — watch context.) *Derivation of why it appears.* The force between the bodies has magnitude $F=Gm_1m_2/r^2$. The relative coordinate obeys $\mu_{\text{red}}\,\ddot{\mathbf r}=\mathbf F$: $$\mu_{\text{red}}\ddot{\mathbf r}=\frac{m_1m_2}{m_1+m_2}\Big(-\frac{G(m_1+m_2)}{r^2}\hat{\mathbf r}\Big)=-\frac{Gm_1m_2}{r^2}\hat{\mathbf r}=\mathbf F.\;\checkmark$$ So the relative motion is *exactly* a particle of mass $\mu_{\text{red}}$ in the gravitational force $\mathbf F$. Total kinetic energy splits cleanly: $$T=\tfrac12 m_1\dot{\mathbf r}_1^2+\tfrac12 m_2\dot{\mathbf r}_2^2 = \underbrace{\tfrac12 M\dot{\mathbf R}^2}_{\text{COM}} + \underbrace{\tfrac12\mu_{\text{red}}\dot{\mathbf r}^2}_{\text{internal}},\quad M=m_1+m_2.$$ *Why split this way?* Because $\mathbf R$ and $\mathbf r$ are independent coordinates whose cross-terms vanish — the energy decouples, so the orbit problem (internal part) can be solved alone. --- ## WORKED EXAMPLES > [!example] 1 — Earth's $\mu$ for a low satellite > $m_\oplus=5.97\times10^{24}\,$kg, satellite $m=1000\,$kg. > $\mu=G(m_\oplus+m)=6.674\times10^{-11}(5.97\times10^{24}+10^3)$. > *Why drop the $10^3$?* It's $10^{21}$ times smaller — negligible. So $\mu\approx 3.986\times10^{14}\,\text{m}^3/\text{s}^2$ (the standard $GM_\oplus$). > **Lesson:** satellite mass invisible; the planet–Sun *comparable* case is where both matter. > [!example] 2 — Equal-mass binary stars > $m_1=m_2=m$. Then $\mu=G(2m)$ and $\mu_{\text{red}}=m^2/2m=m/2$. > Where is the COM? Exactly **midway** between them, since masses are equal. Each star orbits the COM on a circle of radius $r/2$ if $\mathbf r$ has length $r$. > *Why radius $r/2$?* $\mathbf r_1-\mathbf R = -\frac{m_2}{M}\mathbf r=-\tfrac12\mathbf r$, so star 1 sits a distance $r/2$ from COM. Each traces a smaller scaled copy of the relative orbit. > [!example] 3 — Recovering individual positions from $\mathbf R,\mathbf r$ > Given $\mathbf R$ and $\mathbf r$, solve the 2×2 linear system: > $$\mathbf r_1=\mathbf R-\frac{m_2}{M}\mathbf r,\qquad \mathbf r_2=\mathbf R+\frac{m_1}{M}\mathbf r.$$ > *Why these coefficients?* Plug into $\mathbf r=\mathbf r_2-\mathbf r_1$: get $\frac{m_1+m_2}{M}\mathbf r=\mathbf r$ ✓. Plug into $m_1\mathbf r_1+m_2\mathbf r_2$: the $\mathbf r$ terms cancel and you get $M\mathbf R$ ✓. The heavier body stays *closer* to the COM (smaller coefficient). --- > [!recall]- Feynman: explain to a 12-year-old > Imagine two kids on ice holding a rope, spinning around. Instead of tracking both kids, you watch the *middle point* between them (it just slides smoothly in a straight line — boring) and the *stretchy rope* between them (this tells you everything about the spin). The two-body problem does the same: middle point = center of mass (drifts freely), rope = relative vector $\mathbf r$. By looking only at the rope, two spinning things become one simple thing going around an invisible center. The heavier kid barely moves; the lighter kid swings wide — that's why a planet looks "fixed" while a satellite zooms. > [!mnemonic] Remember the two $\mu$'s > "**SUM for the force, PRODUCT-over-sum for the mass.**" > Gravitational parameter $\mu=G(m_1+m_2)$ uses the **sum**; reduced mass $\mu_{\text{red}}=\dfrac{m_1m_2}{m_1+m_2}$ uses **product over sum**. --- ## #flashcards/physics Define the relative position vector in the two-body problem ::: $\mathbf r=\mathbf r_2-\mathbf r_1$ (from body 1 to body 2). Why does the center of mass move at constant velocity? ::: Adding the two Newton equations, internal forces cancel ($m_1\ddot{\mathbf r}_1+m_2\ddot{\mathbf r}_2=0$), so $\ddot{\mathbf R}=0$ — no external force. What is the reduced equation of motion? ::: $\ddot{\mathbf r}=-\dfrac{\mu}{r^3}\mathbf r$ with $\mu=G(m_1+m_2)$. Define the gravitational parameter $\mu$ ::: $\mu=G(m_1+m_2)$ — exact two-body value; reduces to $GM$ when one mass dominates. Define the reduced mass ::: $\mu_{\text{red}}=\dfrac{m_1 m_2}{m_1+m_2}$. How does total kinetic energy split? ::: $T=\tfrac12 M\dot{\mathbf R}^2+\tfrac12\mu_{\text{red}}\dot{\mathbf r}^2$ (COM motion + internal motion, decoupled). Recover $\mathbf r_1$ from $\mathbf R,\mathbf r$ ::: $\mathbf r_1=\mathbf R-\dfrac{m_2}{M}\mathbf r$. Why is $\mu\approx GM_\oplus$ valid for satellites? ::: Because $m_\text{sat}\ll m_\oplus$, so $m_1+m_2\approx m_1$; for comparable masses you must keep both. For equal-mass binary, where is the COM and each orbit radius? ::: COM midway; each star orbits at radius $r/2$. --- ## Connections - [[Kepler's laws]] — the reduced equation directly yields conic-section orbits. - [[Conservation of angular momentum (central force)]] — why orbit is planar. - [[Vis-viva equation]] — energy of the reduced one-body system. - [[Center of mass frame]] — the coordinate trick used here. - [[Reduced mass in molecular vibrations]] — same math in quantum/diatomic systems. - [[Three-body problem]] — what breaks when a third mass is added (no clean reduction). ## 🖼️ Concept Map ```mermaid flowchart TD TB[Two-body problem] N1[Newton on m1] N2[Newton on m2] R[Relative vector r = r2 - r1] COM[Center of mass R] ADD[Add equations] SUB[Subtract equations] FREE[COM moves freely, R-ddot = 0] EOM[Reduced EOM r-ddot = -mu/r^3 r] MU[mu = G m1 + m2] KEP[One-body Kepler problem] TB -->|gravity gives| N1 TB -->|gravity gives| N2 N1 -->|equal opposite| N2 N1 --> ADD N2 --> ADD ADD -->|internal forces cancel| FREE FREE -->|removes 3 unknowns| COM N1 --> SUB N2 --> SUB SUB -->|yields| EOM R -->|governs| EOM MU -->|defines| EOM EOM -->|identical to| KEP ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho, two-body problem ka matlab hai do masses ($m_1$ aur $m_2$) jo sirf apni gravity se ek dusre ko kheench rahe hain — jaise Sun aur planet, ya Earth aur satellite. Dono ki position track karna mushkil hai (6 variables). Toh hum ek smart trick lagate hain: motion ko do hisson mein todte hain. Pehla hissa hai **center of mass** $\mathbf R$ — yeh seedhi line mein constant velocity se chalta rehta hai, kyunki internal gravity forces aapas mein cancel ho jaate hain ($\ddot{\mathbf R}=0$). Yeh boring hai, ignore kar do. Dusra hissa hai **relative vector** $\mathbf r=\mathbf r_2-\mathbf r_1$ — isme saari orbit ki information hai. > > Newton ke dono equations ko mass se divide karke subtract karne par milta hai $\ddot{\mathbf r}=-\frac{\mu}{r^3}\mathbf r$, jahan $\mu=G(m_1+m_2)$. Yeh ekdum waisa hi dikhta hai jaise ek single body ek fixed center ke around ghoom rahi ho. Isi liye isko "reduction to one-body" kehte hain — do cheezon ka problem ek cheez ka simple Kepler problem ban gaya. > > Ek important baat: $\mu$ mein **dono** masses ka sum aata hai, sirf Earth ka nahi. Lekin jab satellite bahut halka hai ($m_2\ll m_1$), tab $m_1+m_2\approx m_1$, isliye $\mu\approx GM_\oplus$ chalta hai — yeh approximation hai, exact nahi. Binary stars (comparable mass) mein dono rakhna zaroori hai. Aur energy ke liye ek alag mass aati hai, **reduced mass** $\mu_{red}=\frac{m_1m_2}{m_1+m_2}$ — yaad rakho: force mein **sum**, mass mein **product upon sum**. Bas yahi do formule pakad lo, baaki orbit ki saari physics inhi se nikalti hai. ![[audio/3.2.01-Two-body-problem-—-equations-of-motion,-reduction-to-one-body.mp3]]