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

3.2.1 · D3 · Physics › Orbital Mechanics & Astrodynamics › Two-body problem — equations of motion, reduction to one-bod

Kuch bhi shuru karne se pehle, do words jo hum baar baar use karenge:

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

Figure dekho: arrow $r$ body 1 se body 2 tak jaata hai, aur amber dot (COM) hamesha unke beech ki line par hota hai, heavier mass ke zyada paas. Neeche sab kuch sirf is picture ka alag alag weights ke saath version hai.

Scenario matrix

Har case jo orbital-reduction mein aa sakta hai, in cells mein se kisi ek mein fit hota hai. Har cell mein kam se kam ek worked example hai.

Cell	Kya vary karta hai	Degenerate / limiting?	Covered by
A — dominant mass ( $m_{2} ≪ m_{1}$ )	bada mass ratio	limit $μ \to G m_{1}$	Ex 1
B — equal masses ( $m_{1} = m_{2}$ )	symmetric	COM exactly midway	Ex 2
C — general unequal masses	finite ratio	none	Ex 3
D — reduced mass vs gravitational $μ$	do alag $μ$ 's	inhe alag rakhna	Ex 4
E — energy split (COM + internal)	velocities	cross-terms vanish karte hain	Ex 5
F — $m_{2} \to 0$ (test particle)	zero mass input	$μ_{red} \to 0$ , COM $\to m_{1}$	Ex 6
G — real-world word problem	Sun–Jupiter wobble	Sun moves karta hai	Ex 7
H — exam twist: recover both orbits	inversion + signs	heavier body closer	Ex 8

Hum jaan-bujhkar do extremes cover karte hain — test-particle limit (Cell F, itna chhota mass ki practically zero hai) aur equal-mass case (Cell B) — kyunki yahi woh jagah hai jahan "planet fixed lagta hai" wali intuition ya bilkul sahi hoti hai ya bilkul fail ho jaati hai.

Examples

Worked example Ex 1 — Cell A: bada mass ratio (low Earth satellite)

$m_{1} = m_{\oplus} = 5.97 \times 1 0^{24} kg$ , satellite $m_{2} = 1000 kg$ , $G = 6.674 \times 1 0^{- 11} m^{3} kg^{- 1} s^{- 2}$ . Exact $μ = G (m_{1} + m_{2})$ nikalo aur dikhao ki satellite ka mass invisible hai.

Forecast: guess karo — kya answer textbook ke $G M_{\oplus}$ se kisi bhi digit mein differ karega jo tum likhoge? (Guess yes/no, phir check karo.)

Masses add karo: $m_{1} + m_{2} = 5.97 \times 1 0^{24} + 1 0^{3} = 5.970 000 \dots \times 1 0^{24} kg$ . Yeh step kyun? $μ$ sum use karta hai (mnemonic: "SUM for the force"), kabhi ek mass akele nahi.
Added $1 0^{3}$ leading digit se $1 0^{21}$ places neeche hai — yeh koi bhi figure change nahi kar sakta jo hum quote karein. Yeh step kyun? Yeh limit $m_{2} ≪ m_{1} \Rightarrow m_{1} + m_{2} \approx m_{1}$ ko concrete form mein dikha raha hai.
$μ = 6.674 \times 1 0^{- 11} \times 5.97 \times 1 0^{24} = 3.984 \times 1 0^{14} m^{3} / s^{2}$ . Yeh step kyun? Yahi "standard gravitational parameter" $G M_{\oplus}$ hai jiske around satellites design kiye jaate hain.

Verify: units $[m^{3} kg^{- 1} s^{- 2}] [kg] = m^{3} s^{- 2}$ ✓. Numeric $μ \approx 3.98 \times 1 0^{14}$ , accepted value $3.986 \times 1 0^{14}$ se match karta hai (chhota offset sirf rounded $G, m_{\oplus}$ ki wajah se).

Worked example Ex 2 — Cell B: equal-mass binary stars

$m_{1} = m_{2} = m = 2.0 \times 1 0^{30} kg$ , separation $r = 1.0 \times 1 0^{11} m$ . $μ$ , $μ_{red}$ , COM location, aur har star ka orbit radius nikalo.

Forecast: COM kahan hai — star 1 ke paas, star 2 ke paas, ya bilkul beech mein? Padhne se pehle guess karo.

$μ = G (m_{1} + m_{2}) = G (2 m) = 6.674 \times 1 0^{- 11} \times 4.0 \times 1 0^{30} = 2.6696 \times 1 0^{20} m^{3} / s^{2}$ . Yeh step kyun? Force parameter hamesha total mass use karta hai.
$μ_{red} = \frac{m \cdot m}{2 m} = \frac{m}{2} = 1.0 \times 1 0^{30} kg$ . Yeh step kyun? "product over sum" equal masses mein half mass ban jaata hai.
COM: $r_{1} = R - \frac{m _{2}}{M} r = R - \frac{1}{2} r$ , toh star 1, COM se $r /2$ par hai. Symmetry se star 2 bhi. COM bilkul beech mein hai. Yeh step kyun? Equal coefficients $m_{2} / M = m_{1} / M = \frac{1}{2}$ balance point ko middle mein force karte hain.
Har orbit radius $= r /2 = 5.0 \times 1 0^{10} m$ . Yeh step kyun? Kisi body ka orbit radius exactly uski COM se distance hoti hai — aur step 3 ne dikhaya ki har body COM se $\frac{m _{2}}{M} r = \frac{1}{2} r$ par hai. Har star usi fixed point ke around circle karta hai, toh circle ka radius wahi same offset hai.

Verify: do orbit radii ka sum $r /2 + r /2 = r = 1 0^{11}$ ✓ (inhe separation mein add hona chahiye). $μ_{red} = 1 0^{30} = m /2$ ✓.

Worked example Ex 3 — Cell C: general unequal masses (Earth–Moon)

$m_{1} = m_{\oplus} = 5.97 \times 1 0^{24} kg$ , $m_{2} = m_{moon} = 7.35 \times 1 0^{22} kg$ , separation $r = 3.84 \times 1 0^{8} m$ . Earth ke center se COM ki distance nikalo.

Forecast: kya COM Earth ke andar hai (radius $6.37 \times 1 0^{6}$ m) ya bahar? Guess karo.

Earth ki COM se distance: $∣ r_{1} - R ∣ = \frac{m _{2}}{M} r$ . Yeh step kyun? Undo formula se, body 1 ka COM se offset $\frac{m _{2}}{M} r$ hai.
$M = 5.97 \times 1 0^{24} + 7.35 \times 1 0^{22} = 6.0435 \times 1 0^{24} kg$ . Yeh step kyun? Offset formula total mass $M$ se divide karta hai, isliye pehle $M = m_{1} + m_{2}$ banana zaroori hai — fraction $m_{2} / M$ literally poochhta hai "pure total ka body 2 ka kitna hissa hai?"
$\frac{m _{2}}{M} = \frac{7.35 \times 1 0 ^{22}}{6.0435 \times 1 0 ^{24}} = 0.012162$ . Yeh step kyun? Hum pehle dimensionless mass fraction isolate karte hain (ek pure number $< 1$ ) taaki agle step mein separation se multiply karne par length mile; yeh ek nazar mein bhi batata hai ki offset kitna chhota hoga ( $\approx 1.2%$ of $r$ ).
Distance $= 0.012162 \times 3.84 \times 1 0^{8} = 4.670 \times 1 0^{6} m$ . Yeh step kyun? Physical offset paane ke liye mass fraction ko separation se multiply karo.

Verify: $4.67 \times 1 0^{6} m < 6.37 \times 1 0^{6} m$ — COM (barycenter) Earth ke andar hai ✓, yeh ek famous fact hai. Units: (dimensionless)×m = m ✓.

Worked example Ex 4 — Cell D: do

$μ$ 's ko confuse nahi karna Ex 3 ke hi Earth–Moon numbers. Dono $μ = G (m_{1} + m_{2})$ (volume-per-time-squared) aur $μ_{red}$ (ek mass) compute karo aur confirm karo ki unke units alag hain.

Forecast: dono mein se kaun mass hai, aur kaun nahi? Zor se bolo.

$μ = G M = 6.674 \times 1 0^{- 11} \times 6.0435 \times 1 0^{24} = 4.033 \times 1 0^{14} m^{3} / s^{2}$ . Yeh step kyun? Yeh orbit ki shape/speed drive karta hai (yeh $\ddot{r} = - μ r / r^{3}$ mein appear hota hai).
$μ_{red} = \frac{m _{1} m _{2}}{M} = \frac{5.97 \times 1 0 ^{24} \times 7.35 \times 1 0 ^{22}}{6.0435 \times 1 0 ^{24}} = 7.261 \times 1 0^{22} kg$ . Yeh step kyun? Yeh effective mass hai jo internal motion ki kinetic energy carry karta hai.
Note karo $μ_{red} < m_{2}$ (lighter body se bhi chhota). Hamesha true hota hai. Yeh step kyun? $μ_{red} = \frac{m _{2}}{1 + m _{2} / m _{1}} < m_{2}$ , ek useful sanity bound.

Verify: $μ \approx 4.03 \times 1 0^{14} m^{3} / s^{2}$ (units of $G \cdot$ mass), $μ_{red} \approx 7.26 \times 1 0^{22} kg$ (ek mass, $m_{2} = 7.35 \times 1 0^{22}$ ke close but neeche) ✓. Diatomics mein same $μ_{red}$ ke liye Reduced mass in molecular vibrations dekho.

Worked example Ex 5 — Cell E: kinetic energy cleanly split hoti hai

Maano == $T$ total kinetic energy== of the two bodies denote karta hai, $T = \frac{1}{2} m_{1} \dot{r}_{1}^{2} + \frac{1}{2} m_{2} \dot{r}_{2}^{2}$ (har term familiar $\frac{1}{2} m v^{2}$ hai). Given $m_{1} = 3 kg$ at velocity $\dot{r}_{1} = (2, 0) m/s$ , $m_{2} = 1 kg$ at $\dot{r}_{2} = (0, 4) m/s$ , verify karo ki $T = \frac{1}{2} M \dot{R}^{2} + \frac{1}{2} μ_{red} \dot{r}^{2}$ .

Forecast: guess karo ki $T$ compute karne ke do tarike same number denge ya nahi (dene chahiye).

Direct: $T = \frac{1}{2} (3) (2^{2}) + \frac{1}{2} (1) (4^{2}) = 6 + 8 = 14 J$ . Yeh step kyun? $T$ ki raw definition, koi trick nahi — hamara ground truth.
$M = 4$ , $\dot{R} = \frac{3 ( 2 , 0 ) + 1 ( 0 , 4 )}{4} = \frac{( 6 , 4 )}{4} = (1.5, 1.0)$ , toh $\dot{R}^{2} = 2.25 + 1.0 = 3.25$ . Yeh step kyun? COM velocity mass-weighted average hoti hai.
COM term: $\frac{1}{2} (4) (3.25) = 6.5 J$ . Yeh step kyun? Pure system ka mass $M$ bodily velocity $\dot{R}$ par drift karta hai; us bulk drift ki kinetic energy $\frac{1}{2} M \dot{R}^{2}$ hai, jo ordinary $\frac{1}{2} m v^{2}$ hai total mass ko ek lump maan ke.
$\dot{r} = \dot{r}_{2} - \dot{r}_{1} = (0, 4) - (2, 0) = (- 2, 4)$ , $\dot{r}^{2} = 4 + 16 = 20$ . $μ_{red} = \frac{3 \cdot 1}{4} = 0.75$ . Yeh step kyun? Internal motion ko single relative-velocity vector $\dot{r}$ describe karta hai (body 2 kitni tezi se body 1 ke nazariye se chalti hai), aur uski energy effective mass $μ_{red}$ ke ek particle ki tarah behave karti hai — isliye $\frac{1}{2} μ_{red} \dot{r}^{2}$ assemble karne se pehle dono pieces chahiye.
Internal term: $\frac{1}{2} (0.75) (20) = 7.5 J$ . Yeh step kyun? Yahi orbit ki energy hai — sirf isi ki vis-viva ko parwah hai.

Verify: $6.5 + 7.5 = 14 J$ = direct answer ✓. Cross-terms exactly cancel karte hain kyunki $R, r$ independent coordinates hain.

Worked example Ex 6 — Cell F: degenerate test-particle limit

$m_{2} \to 0$ Lo $m_{1} = M_{⊙} = 1.99 \times 1 0^{30} kg$ aur $m_{2} \to 0$ jaane do (ek dust grain). $μ$ , $μ_{red}$ , aur COM ka kya hoga?

Forecast: kaun sa quantity zero ho jaata hai — gravitational $μ$ ya reduced mass? Computing se pehle guess karo.

$μ = G (m_{1} + m_{2}) \to G m_{1}$ (finite, nonzero). Yeh step kyun? Orbit ki shape poori tarah bade mass se set hoti hai — grain normally orbit karta hai.
$μ_{red} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} \to \frac{m _{1} \cdot 0}{m _{1}} = 0$ . Yeh step kyun? Zero-mass particle zero internal kinetic energy carry karta hai — energy split degenerate ho jaata hai.
COM: $R = \frac{m _{1} r _{1} + 0}{m _{1}} = r_{1}$ → COM big body par baithta hai. Yeh step kyun? Doosri taraf koi mass nahi, toh balance point heavy body khud hai; "planet fixed at the focus" wali picture exact ban jaati hai.

Verify: numerically, $m_{2} = 1 kg$ par: $μ_{red} = \frac{1.99 \times 1 0 ^{30}}{1.99 \times 1 0 ^{30} + 1} \approx 1.0000 kg \to$ $m_{2}$ ko track karta hai, $m_{2} \to 0$ ke saath $0$ ki taraf jaata hai ✓. $μ \to G m_{1} = 1.328 \times 1 0^{20}$ , unchanged ✓.

Worked example Ex 7 — Cell G: real-world word problem (kya Sun hilta hai?)

Jupiter $m_{2} = 1.898 \times 1 0^{27} kg$ , Sun $m_{1} = 1.989 \times 1 0^{30} kg$ , mean separation $r = 7.78 \times 1 0^{11} m$ . Sun–Jupiter barycenter se Sun kitna door hai, aur kya woh Sun ke andar hai ya bahar (radius $6.96 \times 1 0^{8}$ m)?

Forecast: guess karo — barycenter Sun ke andar hai, ya uski surface ke upar empty space mein?

Sun ka COM se offset $= \frac{m _{2}}{M} r$ , $M = 1.989 \times 1 0^{30} + 1.898 \times 1 0^{27} = 1.9909 \times 1 0^{30} kg$ . Yeh step kyun? Same undo-formula offset jaisa Ex 3 mein, ab giant planet ke saath.
$\frac{m _{2}}{M} = \frac{1.898 \times 1 0 ^{27}}{1.9909 \times 1 0 ^{30}} = 9.533 \times 1 0^{- 4}$ . Yeh step kyun? Sun ka wobble Jupiter ke total mass mein share se drive hota hai, isliye hum planet-to-total ratio $m_{2} / M$ pehle banate hain — ek pure number — phir use $r$ se scale karte hain; alag banana yeh bhi reveal karta hai ki wobble orbit size ka sirf $\approx 0.095%$ hai.
Offset $= 9.533 \times 1 0^{- 4} \times 7.78 \times 1 0^{11} = 7.417 \times 1 0^{8} m$ . Yeh step kyun? Yeh Sun ka physical wobble radius hai.

Verify: $7.42 \times 1 0^{8} m > 6.96 \times 1 0^{8} m$ — barycenter Sun ki surface ke bilkul bahar hai ✓. Toh Sun genuinely ek space mein point ke around orbit karta hai — woh wobble jo astronomers exoplanets detect karne ke liye use karte hain. Yahi jagah hai jahan Cell A ki "planet fixed" intuition ek baal se fail hoti hai.

Worked example Ex 8 — Cell H: exam twist — ratio se dono orbit radii recover karo

Ek binary mein mass ratio $m_{1} / m_{2} = 4$ aur separation $r = 1.0 \times 1 0^{10} m$ hai. Individual masses jaane bina, har body ki COM se distance nikalo aur confirm karo ki heavier body closer hai.

Forecast: kaun sa body chhota circle orbit karta hai — heavy wala ya light wala? Guess karo.

Maano $m_{2} = 1$ unit $\Rightarrow m_{1} = 4$ , $M = 5$ . Yeh step kyun? Offsets mein sirf ratio enter karta hai, isliye hum convenient units choose kar sakte hain.
Body 1 (heavy) distance $= \frac{m _{2}}{M} r = \frac{1}{5} \times 1 0^{10} = 2.0 \times 1 0^{9} m$ . Yeh step kyun? Body 1 ka offset doosra mass fraction $m_{2} / M$ carry karta hai.
Body 2 (light) distance $= \frac{m _{1}}{M} r = \frac{4}{5} \times 1 0^{10} = 8.0 \times 1 0^{9} m$ . Yeh step kyun? Har body ka offset doosri body ke mass fraction se weighted hota hai (undo formulas se: $r_{1}$ uses $m_{2} / M$ , $r_{2}$ uses $m_{1} / M$ ), isliye body 1 se body 2 par jaate waqt hum fractions swap karte hain — exactly isi wajah se heavier partner COM ko apne paas pin karta hai aur lighter wala wider swing karta hai.
Radii ka ratio $= \frac{8 \times 1 0 ^{9}}{2 \times 1 0 ^{9}} = 4 = m_{1} / m_{2}$ : light body 4× wider swing karta hai. Yeh step kyun? COM se distances masses ke inversely jaati hain — heavier = closer.

Verify: radii sum $2 \times 1 0^{9} + 8 \times 1 0^{9} = 1.0 \times 1 0^{10} = r$ ✓. Heavy-body radius ( $2 \times 1 0^{9}$ ) < light-body radius ( $8 \times 1 0^{9}$ ) ✓, confirm karta hai "heavier COM ke paas rehta hai."

Recall Kaun sa cell kaun sa tha?

Scenario ko uske lesson se match karo. Big mass ratio ::: Cell A — $μ \approx G m_{1}$ , satellite mass invisible (Ex 1). Equal masses ::: Cell B — COM midway, har body $r /2$ par orbit karti hai (Ex 2). One mass $\to 0$ ::: Cell F — $μ_{red} \to 0$ , COM big body par baithta hai (Ex 6). Sun wobbles ::: Cell G — barycenter Sun ki surface ke bilkul bahar (Ex 7). Distances from COM ::: masses ke inversely jaati hain — heavier closer hota hai (Ex 8).

Connections

Parent topic — woh derivation jo yeh examples exercise karti hai.
Center of mass frame — woh coordinate trick jo yahan har offset formula ke peeche hai.
Vis-viva equation — Ex 5 mein isolated internal energy term use karta hai.
Kepler's laws — orbit shapes jo reduced equation produce karta hai.
Conservation of angular momentum (central force) — kyun har orbit planar rehti hai.
Reduced mass in molecular vibrations — Ex 4 ka same $μ_{red}$ diatomics mein.
Three-body problem — jahan Cell-style clean reduction kaam karna band kar deta hai.