Kepler's third law — T² ∝ a³ — derivation: Worked examples

3.2.7 · D3 · Physics › Orbital Mechanics & Astrodynamics › Kepler's third law — T² ∝ a³ — derivation

Yeh ek companion page hai the main derivation ke liye. Yahaan hum sirf ek kaam karte hain: hum $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ law ke har tarah ke problem ko grind karke khatam karte hain, taaki koi bhi exam question aur koi bhi real-world scenario unfamiliar na lage.

Shuru karne se pehle, yaad karo is law ke teen roop (har ek alag unknown solve karta hai):

Yahaan $T$ = period (ek poora chakkar lagane ka time), $a$ = semi-major axis (orbit size; circle ke liye bas radius), $G$ = gravitational constant $6.674 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}$ , aur $M$ = mass of the central body being orbited. Hum $v$ = orbital speed se bhi milenge, yaani ek body apne path par per second kitni distance cover karti hai (metres per second mein).

The scenario matrix

Is law ke har problem ka koi na koi cell mein jagah hoti hai. Neeche ke examples us cell ke label ke saath hain jo woh cover karta hai. (Agar neeche ki table render na ho, toh har Cell — distinguishing feature → Example line ko ek row samjho.)

Cell A — Absolute period: real $a$ aur $M$ SI units mein given hai, $T$ find karo. → Ex 1
Cell B — Absolute size (inverse): $T$ aur $M$ given hai, $a$ find karo. → Ex 2 (GEO)
Cell C — Ratio, same $M$ : do orbits compare karo, constant cancel ho jaata hai. → Ex 3
Cell D — Kepler's units trick: $a$ AU mein, $T$ years mein, Sun ⇒ $T^{2} = a^{3}$ . → Ex 4
Cell E — Ellipse, is it $a$ ?: perihelion & aphelion given hai, $a$ build karo. → Ex 5
Cell F — Solve for central mass: $M$ ke liye rearrange karo ("Sun/planet ko weigh karo"). → Ex 6
Cell G — Degenerate / limiting: $e \to 0$ (circle), $e \to 1$ (radial fall), $a \to \infty$ . → Ex 7
Cell H — Exam twist: comparable masses ⇒ $M_{1} + M_{2}$ use karo; unit trap. → Ex 8

Cell	Isse alag kya banata hai	Example
A. Absolute period	Real $a$ aur $M$ (SI units) given, $T$ find karo	Ex 1
B. Absolute size (inverse)	$T$ aur $M$ given, $a$ find karo	Ex 2 (GEO)
C. Ratio, same $M$	Do orbits compare karo, constant cancel	Ex 3
D. Kepler's units trick	$a$ AU mein, $T$ years mein, Sun → $T^{2} = a^{3}$	Ex 4
E. Ellipse — is it $a$ ?	Perihelion & aphelion given, $a$ build karo	Ex 5
F. Solve for central mass	$M$ ke liye rearrange ("Sun/planet ko weigh karo")	Ex 6
G. Degenerate / limiting	$e \to 0$ (circle), $e \to 1$ (radial fall), $a \to \infty$	Ex 7
H. Exam twist	Comparable masses → $M_{1} + M_{2}$ use karo; unit trap	Ex 8

Cell A — absolute period in SI units

Worked example Ex 1 — Low-Earth-orbit satellite ka period

Ek satellite Earth ke around $r = 7.00 \times 1 0^{6}$ m radius par ghoom rahi hai (approximately 620 km upar). Earth ka $G M_{\oplus} = 3.986 \times 1 0^{14} m^{3} / s^{2}$ (standard gravitational parameter). Period $T$ find karo.

Forecast: International Space Station (ISS), jo ek crewed satellite hai low orbit mein, lagbhag 90 minutes mein chakkar lagaata hai. Compute karne se pehle guess karo — yeh minutes mein hoga ya hours mein?

Step 1 — Law ka sahi roop chuno. Hume $a = r$ aur $M$ pata hai; hum $T$ chahte hain. Toh $T = 2 π a^{3} / GM$ use karo. Yeh step kyun? $T$ isolated unknown hai, isliye hum woh form choose karte hain jisme $T$ already left side par akela ho.

Step 2 — Radius ko cube karo. $r^{3} = (7.00 \times 1 0^{6})^{3} = 3.43 \times 1 0^{20} m^{3}$ . Yeh step kyun? Law $a^{3}$ par depend karta hai, na ki $a$ par — woh cube hi Kepler's law ki poori personality hai.

Step 3 — $GM$ se divide karo aur root lo. $\frac{r ^{3}}{GM} = \frac{3.43 \times 1 0 ^{20}}{3.986 \times 1 0 ^{14}} = 8.605 \times 1 0^{5} s^{2}, = 927.6 s .$ Yeh step kyun? Root ke andar ki units $m^{3} \div (m^{3} / s^{2}) = s^{2}$ hain, toh square root honestly seconds mein hai — yeh ek achha sign hai ki hum ne algebra mein galti nahi ki.

Step 4 — $2 π$ se multiply karo. $T = 2 π \times 927.6 = 5828 s \approx 97 min .$ Yeh step kyun? Root ne humein $a^{3} / GM$ diya, lekin formula mein leading factor $2 π$ hota hai. Woh $2 π$ is fact se aata hai ki orbital speed $v$ (path par per second metres travel) period se $v = 2 π r / T$ se relate hoti hai: body ek period $T$ mein poori circumference $2 π r$ cover karti hai. $2 π$ se multiply karna raw timescale $a^{3} / GM$ ko ek complete chakkar ke time mein convert karta hai.

Verify: $\approx 97$ minutes ISS-class result se match karta hai jo humne forecast kiya tha — real low-Earth-orbit periods 88–100 min hote hain. Units seconds mein aayi. ✓

Cell B — inverse problem, orbit size ke liye solve karo

Worked example Ex 2 — Geostationary radius (real engineering)

Ek communications satellite ka period $T = 86400$ s (ek din) hona chahiye taaki woh Earth's equator ke upar ek jagah hover kare. Uska orbital radius find karo. ( $G M_{\oplus} = 3.986 \times 1 0^{14}$ .)

Forecast: Ex 1 mein low Earth orbit 97 min ke liye 7000 km tha. Ek poora din lagbhag 15× longer hai. Kyunki $T \propto a^{3/2}$ , kya $a$ 15× badhega ya kaafi kam?

Step 1 — Law ko invert karo. $T^{2} = \frac{4 π ^{2}}{GM} a^{3}$ ko $a$ ke liye solve karo: $a = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3} .$ Yeh step kyun? Ab $a$ unknown hai, toh hum algebraically ise nikaalte hain — across multiply karo, phir cube root lo kyunki $a$ cubed tha.

Step 2 — Numerator build karo. $GM T^{2} = 3.986 \times 1 0^{14} \times (86400)^{2} = 3.986 \times 1 0^{14} \times 7.465 \times 1 0^{9} = 2.976 \times 1 0^{24} .$ Yeh step kyun? Hum pehle woh sab assemble karte hain jo law fraction bar ke upar rakhta hai, kyunki divide karne se pehle numerator group karna powers-of-ten bookkeeping ko clean aur error-free rakhta hai.

Step 3 — $4 π^{2} = 39.478$ se divide karo. $\frac{2.976 \times 1 0 ^{24}}{39.478} = 7.538 \times 1 0^{22} m^{3} .$ Yeh step kyun? Yeh $a^{3}$ hai; units $m^{3} / s^{2} \times s^{2} = m^{3}$ hain — ek volume-jaisi quantity, exactly wahi jo ek length ke cube ko hona chahiye.

Step 4 — Cube root. $a = (7.538 \times 1 0^{22})^{1/3} = 4.224 \times 1 0^{7} m = 42, 240 km .$ Yeh step kyun? Step 3 ne $a^{3}$ deliver kiya, lekin hum $a$ khud chahte hain, toh hum cube root se cube undo karte hain — inverse operation. Tabhi number ek genuine length banta hai jise hum Earth's radius se compare kar sakein.

Verify: Earth's radius $6371$ km ghatao → altitude $\approx 35, 870$ km, textbook GEO belt. Aur note karo: period ~15× badha lekin radius sirf ~6× badha (7000 se 42200 km tak) — kyunki $1/3$ power ise tame karta hai, exactly as forecast. ✓ Dekho Geostationary & Geosynchronous Orbits.

Cell C — ratio form, same central mass

Worked example Ex 3 — Jupiter ke do moons

Moon A Jupiter ke around $1.77$ days ki period ke saath $a_{A} = 4.22 \times 1 0^{5}$ km par orbit karta hai. Moon B $a_{B} = 1.070 \times 1 0^{6}$ km par hai. $T_{B}$ find karo bina $G$ ya $M_{Jupiter}$ use kiye.

Forecast: B lagbhag 2.5× zyada door hai. Period 3/2 power ke saath badhta hai, toh roughly $2. 5^{1.5} \approx 4$ × longer expect karo.

Step 1 — Ratio form use karo. Same central body ⇒ constant $4 π^{2} / GM$ dono ke liye identical hai, toh divide out ho jaata hai: $\frac{T _{B}^{2}}{T _{A}^{2}} = (\frac{a _{B}}{a _{A}})^{3} .$ Yeh step kyun? Humein koi $G$ aur koi $M$ nahi diya gaya — lekin hume unki zaroorat nahi, kyunki woh dono moons ke liye same hain aur ratio mein cancel ho jaate hain. Yeh hamesha fastest route hai jab ek central mass do orbits host kare.

Step 2 — Size ratio compute karo. $\frac{a _{B}}{a _{A}} = \frac{1.070 \times 1 0 ^{6}}{4.22 \times 1 0 ^{5}} = 2.536.$ Yeh step kyun? Law sirf do orbits ke relative sizes ki parwah karta hai, toh hum dono distances ko ek single dimensionless number mein collapse karte hain — B ki orbit kitni baar badi hai. Ek baar yeh ratio mil jaaye, actual kilometres irrelevant hain.

Step 3 — 3/2 power par raise karo. $T_{B} = T_{A} \times (2.536)^{3/2} = 1.77 \times 4.038 = 7.15 days .$ Yeh step kyun? $(\cdot)^{3}$ lena dono effects — longer track aur slower speed — ko ek single exponent $3/2$ mein fold karta hai.

Verify: Yeh Jupiter ke real moons Io ( $T_{A} = 1.77$ d) aur Ganymede ( $T_{B} = 7.15$ d observed) hain. ✓ Prediction rounding ke andar land karti hai.

Cell D — Kepler's natural-units trick

Worked example Ex 4 — Mars ka saal, apne dimaag mein karo

Mars Sun ke around $a = 1.524$ AU par orbit karta hai. Uska saal Earth-years mein find karo.

Forecast: Mars Earth se thoda zyada door hai, toh uska saal 1 se thoda zyada hona chahiye — 2 years se kam.

Step 1 — Natural units mein switch karo. $a$ ko AU mein aur $T$ ko Sun ke around years mein measure karo. Tab $\frac{4 π ^{2}}{G M _{⊙}} = 1$ by construction (kyunki Earth ka $a = 1$ , $T = 1$ in units define karta hai). Law collapse hokar $T^{2} = a^{3}$ ban jaata hai. Yeh step kyun? Ugly constant exactly woh number hai jo Earth ko fit karta hai; aise units choose karna jahan Earth "1" ho, ise hamesha ke liye chhupa deta hai. Yeh sirf Sun-orbiting bodies ke liye in units ke saath kaam karta hai.

Step 2 — Apply karo. $T = a^{3/2} = 1.52 4^{1.5} = 1.524 \times 1.524 = 1.524 \times 1.234 = 1.881 years .$ Yeh step kyun? Constant chale jaane ke baad, $T^{2} = a^{3}$ dono sides ka square root leke $T = a^{3/2}$ ban jaata hai. Hum $a^{3/2} = a \cdot a^{1/2}$ (ek full factor of $a$ times uska square root) split karte hain taaki arithmetic haath se ki ja sake — yahi trick ka poora point hai.

Verify: $1.881$ yr $\times 365.25 = 687$ days — observed Martian year. ✓

Cell E — ellipse: perihelion aur aphelion se $a$ build karo

Worked example Ex 5 — Halley's Comet

Halley's Comet ka perihelion (sabse paas) $r_{m i n} = 0.586$ AU aur aphelion (sabse door) $r_{m a x} = 35.08$ AU Sun se hai. Uska period find karo.

Forecast: Yeh Neptune se bhi aage jaata hai; decades expect karo, na ki years.

Step 1 — Semi-major axis do extremes ka AVERAGE hota hai. $a = \frac{r _{m i n} + r _{m a x}}{2} = \frac{0.586 + 35.08}{2} = 17.83 AU .$ Yeh step kyun? Ek ellipse ke liye, "Sun se distance" ek number nahi hai — yeh $r_{m i n}$ se $r_{m a x}$ tak range karti hai. Law $a$ use karta hai, jo exactly us range ka midpoint hai.

Figure — ellipse se $a$ padhna. Neeche ki picture mein Sun elongated orbit ke ek focus par baitha hai (centre par nahi). Green dot perihelion ( $r_{m i n}$ ) mark karta hai, red dot aphelion ( $r_{m a x}$ ), aur yellow bar semi-major axis $a$ hai jo geometric centre se far vertex tak jaata hai. Dhyaan do ki yellow bar literally poore long axis ka aadha hai, aur $a$ green aur red distances ka average hai — yeh woh quantity hai jo Kepler's law maangta hai, na ki koi bhi akela extreme.

Step 2 — Kepler AU/years mein apply karo (Cell D trick). $T = a^{3/2} = 17.8 3^{1.5} = 75.3 years .$ Yeh step kyun? Eccentricity appear hi nahi karta — Sun ke paas squashing (faster) aur door stretching (slower) ek lap mein exactly trade off karte hain, isliye sirf $a$ matter karta hai.

Verify: Halley's real period ~75–76 years hai (last seen 1986, next 2061). ✓ Eccentricity $e = \frac{r _{m a x} - r _{m i n}}{r _{m a x} + r _{m i n}} = \frac{34.49}{35.67} = 0.967$ kabhi hamari arithmetic mein nahi aaya — proof ki law shape ignore karta hai.

Cell F — "central body ko weigh karo"

Worked example Ex 6 — Ek moon se Jupiter ki mass

Io Jupiter ke around $a = 4.22 \times 1 0^{8}$ m aur $T = 1.77$ days $= 1.529 \times 1 0^{5}$ s ke saath orbit karta hai. Jupiter ki mass find karo. ( $G = 6.674 \times 1 0^{- 11}$ .)

Forecast: Jupiter Solar System ka giant hai, hundreds of Earth-masses. ~ $1 0^{27}$ kg expect karo.

Step 1 — Law ko $M$ ke liye solve karo. $T^{2} = \frac{4 π ^{2}}{GM} a^{3} \Rightarrow M = \frac{4 π ^{2} a ^{3}}{G T ^{2}} .$ Yeh step kyun? $M$ ab unknown hai; har observable (orbit ka $a$ aur $T$ ) right side par hai. Is tarah hum actually planets aur stars weigh karte hain jinhein hum kabhi touch nahi kar sakte — Kepler's law ek cosmic scale hai.

Step 2 — Numerator. $4 π^{2} a^{3} = 39.478 \times (4.22 \times 1 0^{8})^{3} = 39.478 \times 7.515 \times 1 0^{25} = 2.967 \times 1 0^{27} .$ Yeh step kyun? Hum pehle fraction bar ke upar ki sab cheezein gather karte hain — $4 π^{2}$ aur cubed orbit size — taaki Step 4 mein division ek single clean operation ho na ki teen tangled ones. Cube woh jagah hai jahan orbit ki size enter hoti hai, exactly har Kepler problem ki tarah.

Step 3 — Denominator. $G T^{2} = 6.674 \times 1 0^{- 11} \times (1.529 \times 1 0^{5})^{2} = 6.674 \times 1 0^{- 11} \times 2.338 \times 1 0^{10} = 1.560.$ Yeh step kyun? Ab hum bar ke neeche ki sab cheezein assemble karte hain — constant $G$ aur squared period. $T$ ko square karna wahi hai jo law demand karta hai ( $T^{2}$ ), aur ise alag karna tiny $G$ aur huge $T^{2}$ ko ek line mein muddled hone se bachata hai.

Step 4 — Divide karo. $M = \frac{2.967 \times 1 0 ^{27}}{1.560} = 1.90 \times 1 0^{27} kg .$ Yeh step kyun? Numerator aur denominator dono fully build hone ke baad, last act woh division hai jo formula prescribe karta hai — aur abhi units kilograms mein resolve hoti hain, jo hume bataati hai ki humne sach mein ek mass solve kiya hai.

Verify: Jupiter ki accepted mass $1.898 \times 1 0^{27}$ kg hai. ✓ Teen figures tak match — aur note karo $m_{Io}$ kabhi appear nahi hua, kyunki force balance mein cancel ho gaya (yeh cell H ki warning pehle se).

Cell G — degenerate aur limiting cases

Worked example Ex 7 — Law apni edges par kya karta hai

Teen limits examine karo: (a) circle limit $e \to 0$ ; (b) radial-fall limit $e \to 1$ fixed $a$ par; (c) escape limit $a \to \infty$ .

Forecast: Guess karo kaun sa limit infinite period deta hai.

(a) $e \to 0$ (perfect circle). Tab $r_{m i n} = r_{m a x} = a$ , toh $a = \frac{r + r}{2} = r$ . General law $T = 2 π a^{3} / GM$ circular wala $T = 2 π r^{3} / GM$ ban jaata hai bina kisi badlaav ke. Yeh step kyun? Yeh confirm karta hai ki ellipse formula circle ko contain karta hai as a special case — ek sanity anchor, naya physics nahi.

(b) Fixed $a$ par $e \to 1$ (degenerate "orbit" = seedhi plunge aur wapassi). Ellipse squash hokar line segment ban jaati hai; body seedha centre tak girती hai aur wapas aati hai. Period phir bhi $T = 2 π a^{3} / GM$ hai — unchanged, kyunki $e$ kabhi appear hi nahi karta. Yeh step kyun? Yeh astonishing part hai: semi-major axis $a$ ki razor-thin ellipse par ek near-radial comet utna hi time per lap leta hai jitna radius $a$ ka ek circle. Shape irrelevant hai; sirf $a$ count karta hai. (Note karo yeh abhi bhi ek bound $e < 1$ ellipse hai; jis moment $e$ exactly $1$ reach karta hai, orbit band hona band ho jaata hai — upar ki definition dekho.)

(c) $a \to \infty$ . $T = 2 π a^{3} / GM \to \infty$ . Jaise-jaise $a$ badhta hai bina bound ke, ellipse ka far end infinity tak chala jaata hai aur orbit band hona band kar deta hai — yeh $e = 1$ parabola (escape) aur usse aage ban jaata hai. Infinite period law ka honest tarika hai yeh kehne ka "no return."

Figure — teen orbits, ek $a$ . Picture mein ek green circle, ek blue moderate ellipse, aur ek red near-radial sliver dikhti hai, sab ek hi semi-major axis $a$ aur ek hi central mass (yellow dot ek focus par) share karte hue. Wildly alag dikhne ke bawajood, teeno ek identical time mein ek lap complete karte hain — visual proof ki period sirf $a$ par depend karta hai, kabhi eccentricity par nahi.

(b) vs (a) ke liye numeric anchor: $GM = 1$ , $a = 1$ lo. Circle: $T = 2 π 1 = 6.283$ . Squashed ellipse, same $a$ : $T = 2 π 1 = 6.283$ . Identical.

Verify: Dono $T = 6.283$ in units mein dete hain — equal, $e$ -independence confirm karta hai. ✓ Aur $a \to \infty$ se $T \to \infty$ bound aur escape orbits ke beech physical boundary se match karta hai — Orbital Energy & Vis-viva Equation dekho.

Cell H — the exam twist (comparable masses)

Worked example Ex 8 — Binary stars: jab

$m$ cleanly cancel nahi hota Do stars apne common centre ke around $a = 3.00 \times 1 0^{11}$ m separation aur $T = 2.00 \times 1 0^{7}$ s period ke saath orbit karte hain. Total mass $M_{1} + M_{2}$ find karo. ( $G = 6.674 \times 1 0^{- 11}$ .)

Forecast: Planets ke liye hum chhoti mass drop karte hain. Yahaan dono bodies heavy hain — toh law mein kaun si mass jaati hai?

Step 1 — Law ka two-body form use karo. Jab dono masses matter karte hain, simple $M$ ko sum se replace kiya jaata hai: $T^{2} = \frac{4 π ^{2}}{G ( M _{1} + M _{2} )} a^{3} \Rightarrow M_{1} + M_{2} = \frac{4 π ^{2} a ^{3}}{G T ^{2}} .$ Yeh step kyun? Planet case mein humne assume kiya ki Sun still baitha hai. Jab masses comparable hain, dono ek shared centre ke around orbit karte hain; honest derivation (dekho Reduced Mass & Two-Body Problem) $M$ ko $M_{1} + M_{2}$ se replace karta hai, jahan $a$ unke beech ka separation hai. Yeh bhool jaana classic exam trap hai.

Step 2 — Numerator. $4 π^{2} a^{3} = 39.478 \times (3.00 \times 1 0^{11})^{3} = 39.478 \times 2.70 \times 1 0^{34} = 1.066 \times 1 0^{36} .$ Yeh step kyun? Ex 6 ki tarah hum pehle bar ke upar ki sab cheezein build karte hain — $4 π^{2}$ times cubed separation. Huge $3 \times 1 0^{11}$ m separation ko cube karna woh jagah hai jahan orbit ki size answer mein dominate karti hai, isliye yeh carefully aur akele kiya jaata hai.

Step 3 — Denominator. $G T^{2} = 6.674 \times 1 0^{- 11} \times (2.00 \times 1 0^{7})^{2} = 6.674 \times 1 0^{- 11} \times 4.00 \times 1 0^{14} = 2.670 \times 1 0^{4} .$ Yeh step kyun? Ab hum bar ke neeche ki sab cheezein build karte hain — constant $G$ aur squared period. Minuscule $G$ aur large $T^{2}$ ko alag line mein rakhna exponents ko jab hum finally divide karein tab collide hone se rokta hai.

Step 4 — Divide karo. $M_{1} + M_{2} = \frac{1.066 \times 1 0 ^{36}}{2.670 \times 1 0 ^{4}} = 3.99 \times 1 0^{31} kg .$ Yeh step kyun? Formula ek badi fraction hai; top aur bottom dono assembled hone ke baad, closing division combined mass deta hai — aur units kilograms mein land karti hain, confirm karta hai ki humne ek mass solve kiya kuch aur nahi.

Verify: Yeh lagbhag $20 M_{⊙}$ hai ( $M_{⊙} = 1.99 \times 1 0^{30}$ kg) do stars ke beech split — ek plausible massive binary. Units: $m^{3} / (N\cdotm^{2} / kg^{2} \cdot s^{2}) = kg$ . ✓ Trap avoid kiya: agar tum single planet ka $M$ -formula use karte toh waise bhi sirf combined system measure karte — law akele $M_{1}$ ko $M_{2}$ se kabhi separate nahi kar sakta; uske liye har star ke individual orbit radius chahiye.

Recall

Recall Inme se har ek kaun sa cell hai? (answers chhupao)

$a$ (AU mein) given aur Sun ke around orbit, years mein $T$ chahiye ::: Cell D — $T = a^{3/2}$ use karo Perihelion aur aphelion given, period chahiye ::: Cell E — pehle $a = (r_{m i n} + r_{m a x}) /2$ form karo Ek moon ka $a$ aur $T$ given, planet ki mass chahiye ::: Cell F — $M = 4 π^{2} a^{3} / (G T^{2})$ Saturn ke do moons compare karo, $G$ nahi diya ::: Cell C — ratio $(T_{2} / T_{1})^{2} = (a_{2} / a_{1})^{3}$ Similar mass ke do stars ::: Cell H — $M_{1} + M_{2}$ use karo, $a$ = separation Kya orbit ko zyada elliptical banane se (fixed $a$ ) $T$ change hota hai? ::: Nahi — Cell G, $e$ drop out ho jaata hai

Connections

3.2.07 Kepler's third law — T² ∝ a³ — derivation (Hinglish) — parent derivation
Newton's Law of Universal Gravitation — $GM m / r^{2}$ supply karta hai jo har derivation start karta hai
Centripetal Force & Circular Motion — $m v^{2} / r$ jise hum gravity ke equal set karte hain
Kepler's Second Law (equal areas) — kyun eccentricity Cells E aur G mein cancel hoti hai
Orbital Energy & Vis-viva Equation — Cell G mein $a \to \infty$ escape boundary
Geostationary & Geosynchronous Orbits — Cell B ka destination
Reduced Mass & Two-Body Problem — Cell H ka $M_{1} + M_{2}$ justify karta hai