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

3.2.1 · D5 · Physics › Orbital Mechanics & Astrodynamics › Two-body problem — equations of motion, reduction to one-bod

Shuru karne se pehle, ek reminder taaki is page ka koi bhi symbol unearned na lage. Hamare paas do point masses hain: pehle mass ko $m_{1}$ kehte hain (maano heavier body, jaise Earth ya koi star) aur doosre ko $m_{2}$ (lighter wala, jaise satellite ya companion star). Dono sirf kuch kilograms mein numbers hain.

$r_{1}, r_{2}$ = do masses $m_{1}$ aur $m_{2}$ ke position vectors.
$r = r_{2} - r_{1}$ = relative position (body 1 se body 2 ki taraf arrow), $r = ∣ r ∣$ .
Center of mass (COM) = mass-weighted average point $R = \frac{m _{1} r _{1} + m _{2} r _{2}}{M}$ , jahan total mass $M = m_{1} + m_{2}$ hai. Aage se "center of mass" ko COM likhenge.
$μ = G (m_{1} + m_{2})$ = gravitational parameter (ek sum, units $m^{3} / s^{2}$ ).
$μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ = reduced mass (ek product-over-sum, units kg). Gravitational parameter $μ$ se clash avoid karne ke liye, is page par reduced mass ko hamesha $μ_{red}$ likhenge — kabhi bare $μ$ nahi.
Dot ka matlab hai "time mein rate of change": $\dot{r}$ = velocity, $\ddot{r}$ = acceleration.

Neeche di gayi figure in objects ki geometry fix karti hai — items solve karte waqt ise dekhte raho.

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

Aur yeh doosri figure do individual orbits aur COM ka free drift dikhati hai, taaki baad mein "scaled copies" wali baat ka ek picture ho.

True or false — justify

Ek isolated two-body system ka center of mass (COM) hamesha constant speed par straight line mein move karta hai.

True. Do Newton equations add karne par, internal gravity cancel ho jaati hai (

m_{1} \ddot{r}_{1} + m_{2} \ddot{r}_{2} = 0

), isliye

\ddot{R} = 0

— zero acceleration ka matlab hai straight, uniform drift.

Earth ke around ek satellite ke liye gravitational parameter exactly

G M_{\oplus}

hai.

False. Yeh exactly

μ = G (m_{\oplus} + m_{sat})

hai;

G M_{\oplus}

sirf ek excellent approximation hai kyunki

m_{sat} ≪ m_{\oplus}

. Comparable masses ke liye yeh galat hoga.

Reduced equation

\ddot{r} = - μ r / r^{3}

ek aisi body describe karti hai jo actual center of mass par baithi hai.

False. Yeh ek fictitious particle describe karti hai jo

r

ki tip par hai aur ek fixed center ke around orbit karti hai jo parameter

μ

produce karta hai. Koi bhi real body wahan nahi baithi; COM ek alag point hai.

Agar

m_{2} \to 0

, toh relative motion parameter

μ

ban jaata hai

G m_{1}

True.

μ = G (m_{1} + m_{2}) \to G m_{1}

, "fixed mass ke around test particle" wala limit recover ho jaata hai — light body ka koi back-reaction nahi hota.

Do equal masses ke liye, reduced mass ek mass ki aadhi hoti hai.

True.

μ_{red} = \frac{m \cdot m}{m + m} = \frac{m}{2}

. Product-over-sum effective mass ko smaller body ki taraf shrink kar deta hai.

Heavier body, lighter body ke mukable center of mass se zyada door orbit karti hai.

False. Heavier body paas baithi hoti hai:

r_{1} - R = - \frac{m _{2}}{M} r

, isliye uski distance doosre mass ke saath scale hoti hai. Bada mass, choti swing.

μ

aur

μ_{red}

donon symbols ke units same hain.

False.

μ = G (m_{1} + m_{2})

ke units

m^{3} / s^{2}

hain;

μ_{red}

kg mein ek mass hai. Same dikhne wale letters, bilkul alag cheezein — yehi wajah hai ki hum subscript rakhte hain.

System ki kinetic energy

\frac{1}{2} μ_{red} \dot{r}^{2}

hai jab drifting COM ko ignore kar dete hain.

True.

T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}

; COM term drop karne par

μ_{red}

se carry hone wali internal (orbital) energy bachti hai.

Reduced-mass one-body reduction kisi bhi central force ke liye kaam karta hai donon bodies ke beech, sirf inverse-square attraction ke liye nahi.

True. Derivation mein sirf "force directed along

\hat{r}

, equal and opposite" (Newton's third law) use hua. Koi bhi central force

F (r) \hat{r}

same tarah reduce ho jaata hai — inverse-square sirf gravitational special case hai; Reduced mass in molecular vibrations mein spring-like force ke liye bhi identical algebra same reduced mass deta hai.

Reduce karne ke baad, ek three-body system bhi ek clean one-body equation mein collapse ho jaata hai.

False. Yeh trick rely karti hai internal forces ke pairwise cancel hone par sirf ek relative vector ke saath. Teen mutual pulls ke saath coordinates decouple nahi hote — dekho Three-body problem.

Spot the error

"Kyunki Earth barely move karti hai, sirf satellite accelerate karta hai, isliye

μ = G m_{sat}

Do errors hain: Earth zaroor accelerate karti hai (tiny but nonzero), aur

μ

sum

G (m_{\oplus} + m_{sat}) \approx G M_{\oplus}

use karta hai. Sirf satellite ki mass use karna ulta hai.

"Center of mass gravity se inward pull hota hai, isliye yeh orbit ki taraf accelerate karta hai."

Internal forces COM ko move nahi kar sakti — woh sum mein cancel ho jaati hain. Sirf ek external force

R

ko accelerate kar sakti hai; yahan koi nahi hai, isliye

\ddot{R} = 0

r = r_{1} - r_{2}

body 2 se body 1 ki taraf point karta hai, lekin physics nahi badlti."

Definition har downstream formula ka sign fix karti hai.

r = r_{2} - r_{1}

ke saath reduced particle par force attractive hai (

- \hat{r}

); ise flip karo aur sare reconstruction coefficients bhi flip karne padenge, warna bodies galat sides par place ho jaayengi.

μ_{red} \ddot{r} = F

mein force

- \frac{G ( m _{1} + m _{2} )}{r ^{2}} \hat{r}

hai."

Nahi — physical force hai

F = - \frac{G m _{1} m _{2}}{r ^{2}} \hat{r}

. Extra

(m_{1} + m_{2})

\ddot{r} = - μ r / r^{3}

mein belong karta hai;

μ_{red}

se multiply karne par yeh wapas

G m_{1} m_{2}

par cancel ho jaata hai.

"Kyunki

μ

mein donon masses appear hain, donon masses ko double karne par orbit period unchanged rehti hai."

Galat.

μ = G (m_{1} + m_{2})

double ho jaata hai, isliye relative acceleration double hoti hai aur orbit period choti ho jaati hai (

T \propto μ^{- 1/2}

fixed size par, via Kepler's laws).

"Relative orbit aur har body ki individual orbit ka size same hota hai."

Yeh scaled copies hain. Body 1 ki orbit ka radius relative orbit ka

\frac{m _{2}}{M}

times hai, body 2 ki

\frac{m _{1}}{M}

times — dono

r

ke trace se choti hain jab tak doosri mass zero na ho.

Why questions

Hum do Newton equations directly solve karne ki bajaye subtract kyun karte hain?

Subtract karne par do coupled acceleration equations ek equation mein

r

ke liye collapse ho jaati hain, ek 6-variable mess ko ek solvable one-body problem mein badal deta hai.

Equations add karne par COM motion kyun milti hai, jabki subtract karne par relative (orbital) motion milti hai?

Add karein → equal-and-opposite forces cancel → force-free center-of-mass motion (uniform drift). Subtract karein → forces

(m_{1} + m_{2})

factor ke saath reinforce → relative motion (actual orbit). Do combinations system ko center-of-mass motion plus relative motion mein decouple kar dete hain.

Reduced mass hamesha kisi bhi individual mass se choti kyun honi chahiye?

μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} < min (m_{1}, m_{2})

hamesha, kyunki product ko (bade) sum se divide karne par yeh chota ho jaata hai. Physically, acceleration ko do movable bodies ke beech share karne se pair "respond" karta hai jaise kuch dono se lighter ho.

Energy ko center-of-mass part + relative part mein split karna legitimate kyun hai?

Kyunki

R

aur

r

independent coordinates hain jinke cross-terms vanish ho jaate hain, kinetic energy decouple ho jaati hai:

\frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}

. Isse hum orbit solve kar sakte hain jabki center-of-mass drift ko bilkul ignore karte hain — yeh Center of mass frame ke peeche key idea hai.

Reduced problem angular momentum kyun conserve karta hai, orbit ko ek plane mein rakhte hue?

Force

\hat{r}

ke along hai (central), isliye yeh center ke baare mein koi torque exert nahi karta — dekho Conservation of angular momentum (central force). Constant angular momentum vector motion ko ek single plane mein pin kar deta hai.

Vis-viva relation mein

μ = G (m_{1} + m_{2})

use hota hai, sirf

GM

nahi — kyun?

Vis-viva equation reduced equation of motion se derive hota hai, jiska parameter full sum

μ

hai. Tabhi yeh familiar

GM

form mein simplify hota hai jab ek mass dominate kare.

Edge cases

Jab

m_{2} = 0

exactly ho toh reduction ka kya hoga?

μ_{red} = 0

aur

μ = G m_{1}

: reduced equation phir bhi hold karta hai, lekin body 2 ek massless test particle hai jo exactly relative orbit trace karta hai jabki body 1 (ab coincident) COM par fixed rehti hai.

Agar do masses equal hoon — har body ki orbit center kahan hogi?

COM bilkul beech mein hoga, aur har star ek circle par radius

r /2

ke saath orbit karta hai — relative orbit of size

r

ki mirror-image scaled copies.

Agar initial relative velocity zero ho (bodies momentarily at rest)?

Relative angular momentum zero hai, isliye

r

mein koi sideways motion nahi — bodies seedha ek doosre ki taraf unhe join karne wali line ke along girte hain (ek degenerate radial "orbit"), jabki COM apna drift maintain karta hai.

Agar

r \to 0

ho (bodies collide)?

Force

\propto 1/ r^{2}

blow up ho jaati hai — point-mass model break down ho jaata hai. Real bodies finite size ki hoti hain, isliye

r = 0

se pehle woh touch karti hain; idealized equation tab tak hi hold hoti hai jab tak separated rahen.

Jab ek mass without bound grow kare (

m_{1} \to \infty

) toh

μ_{red}

ka kya hoga?

μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} \to m_{2}

. Reduced mass lighter body ki mass ke paas pahunch jaata hai, us intuition se match karta hai ki bahut heavy partner ek fixed anchor jaisa kaam karta hai.

Kya yeh two-body reduction ek three-body system ke andar apply ki ja sakti hai ek close pair ko ek object maanke?

Sirf approximately, aur sirf tab jab third body bahut door ho. Tight pair ko ek single mass as uske COM par model kiya ja sakta hai, lekin true reduction tab fail ho jaati hai jab third pull comparable ho — Three-body problem ka hallmark.

Connections

Parent: full derivation
Kepler's laws · Vis-viva equation · Conservation of angular momentum (central force)
Center of mass frame · Reduced mass in molecular vibrations · Three-body problem