Exercises — Kepler's third law — T² ∝ a³ — derivation

(a) Galat. $a$ double karne se $T$ mein $2^{3/2}=2.828$ ka factor aata hai, $2$ nahi. Law squares aur cubes par hai, isliye $T$ par raw factor $3/2$ power hai. (b) Galat. Orbiting mass $m$ gravity $\frac{GMm}{r^2}$ aur centripetal need $\frac{m{v}^2}{r}$ ke beech cancel ho jaati hai. Speed $v=\sqrt{GM/r}$ mein koi $m$ nahi hai. (c) Sach. Kyunki $m$ cancel ho gayi, constant mein sirf $G$ aur central mass $M$ bachte hain.

WHAT: $T^2=a^3$ use karo, toh $T=a^{3/2}$. Compute: $T=5.20^{1.5}=5.20\sqrt{5.20}=5.20\times 2.280=11.86$ years. Real Jupiter year $\approx 11.86$ yr. ✓ WHY trick kaam karti hai: in carefully chosen units mein $\frac{4{\pi }^2}{G{M}_{\odot }}$ exactly $1$ ke barabar hai, isliye ugly constant gaayab ho jaata hai.

WHAT: $T=a^{3/2}=1.524^{1.5}$. $\sqrt{1.524}=1.2345$, toh $T=1.524\times 1.2345=1.881$ years. Days mein: $1.881\times 365.25=687.1$ days. ✓ (Observed Martian year $\approx 687$ days.) WHY: Mars zyada door hai, isliye dono penalties laagoo hoti hain — lambi track aur dheemi speed — jo uska year 1.5 yr se kaafi aage le jaati hain, bhaale hi wo sirf 1.524 AU door hai.

Speed ($v=\sqrt{GM/r}$, kyunki gravity exactly centripetal pull provide karti hai): $v=\sqrt{\frac{3.986\times 10^{14}}{7.0\times 10^6}}=\sqrt{5.694\times 10^7}=7546\text{m/s}.$ Period ($T=2\pi r/v$, uss speed par ek circumference): $T=\frac{2\pi (7.0\times 10^6)}{7546}=\frac{4.398\times 10^7}{7546}=5828\text{s}\approx 97.1\text{min}.$ WHY: ek low-Earth satellite planet ka chakkar lagbhag 90–100 minutes mein lagaati hai — yeh ISS regime hai.

WHAT — ratio form use karo. Kyunki dono usi $M$ ko orbit karte hain, constant cancel ho jaata hai: $\frac{T_B^2}{T_A^2}={\left(\frac{a_B}{a_A}\right)}^3=4^3=64\Rightarrow \frac{T_B}{T_A}=\sqrt{64}=8.$ WHY ratios sabse fast hain: tumhe kabhi $G$, $M$, ya units ki zaroorat nahi — ek hi central body hone ka matlab hai $\frac{4{\pi }^2}{GM}$ cleanly divide ho jaata hai.

WHAT — law ko $r$ ke liye invert karo. $T^2=\frac{4{\pi }^2}{GM}{r}^3$ se: $r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{1/3}.$ Plug in: $GM{T}^2=3.986\times 10^{14}\times (86164{)}^2=3.986\times 10^{14}\times 7.424\times 10^9=2.959\times 10^{24}.$ $4{\pi }^2=39.48$ se divide karo: $7.495\times 10^{22}$. Cube root: $r=(7.495\times 10^{22}{)}^{1/3}=4.216\times 10^7\text{m}=42,160\text{km}$. Altitude $=r-R_{\oplus }=42,160-6,371=35,789\text{km}$. ✓ Dekho Geostationary & Geosynchronous Orbits ki yeh exact height TV dishes ke liye kyun matter karti hai.

Step 1 — period ratio se size ratio nikalo. $T^2\propto a^3\Rightarrow a\propto T^{2/3}$: $\frac{a_X}{a_Y}={\left(\frac{T_X}{T_Y}\right)}^{2/3}={\left(\frac{8}{27}\right)}^{2/3}={\left(\frac{2}{3}\right)}^2=\frac{4}{9}.$ Step 2 — size ko speed mein convert karo. $v=\sqrt{GM/r}\Rightarrow v\propto a^{-1/2}$: $\frac{v_X}{v_Y}={\left(\frac{a_X}{a_Y}\right)}^{-1/2}={\left(\frac{4}{9}\right)}^{-1/2}=\frac{3}{2}=\mathrm{1.5.}$ WHY: X inner planet hai ($a_X<a_Y$), aur inner planets zyada tez chalte hain — uski speed Y ki $1.5\times$ hai. Do proportionalities chain ki gayi, koi constants kabhi nahi chaahiye. (Yeh speed scaling Orbital Energy & Vis-viva Equation ko bhi underlie karti hai.)

Step 1 — shape matter nahi karti, sirf $a$ maayane rakhta hai. Law ko semi-major axis chahiye, do extreme distances ka average: $a=\frac{r_{\min }+r_{\max }}{2}=\frac{0.5+3.5}{2}=2.0\text{AU}.$ Step 2 — Sun-units apply karo. $T=a^{3/2}=2.0^{1.5}=2\sqrt{2}=2.828\text{years}.$ WHY sirf $a$? Orbit ko squash karne se asteroid perihelion ke paas tez hoti hai aur aphelion ke paas dheemi; equal areas se yeh ek chakkar mein exactly trade off karte hain. Eccentricity period mein kabhi enter nahi karti.

WHAT — law ko $M$ isolate karne ke liye rearrange karo. $T^2=\frac{4{\pi }^2}{GM}{a}^3$ se: $M=\frac{4{\pi }^2{a}^3}{G{T}^2}.$ WHY yeh powerful hai: hum Sun ko weigh karte hain sirf ek planet ki distance aur year use karke — koi scale nahi chahiye. Exactly aise hi astronomers door ke stars ka mass nikalte hain. Numerator compute karo: $a^3=(1.496\times 10^{11}{)}^3=3.348\times 10^{33}$. $4{\pi }^2=39.48$ se multiply karo: $1.322\times 10^{35}$. Denominator: $G{T}^2=6.674\times 10^{-11}\times (3.156\times 10^7{)}^2=6.674\times 10^{-11}\times 9.960\times 10^{14}=6.647\times 10^4$. Divide karo: $M=\frac{1.322\times 10^{35}}{6.647\times 10^4}=1.989\times 10^{30}\text{kg}$. ✓ Tabulated Sun mass se teen figures tak match karta hai.

Step 1 — two-body law ko Sun-units mein likhna. $a$ in AU, $T$ in years, aur total mass in $M_{\odot }$ ke saath: $T^2=\frac{a^3}{M_1+M_2}\text{(Sun-units)}\Rightarrow M_1+M_2=\frac{a^3}{T^2}.$ Step 2 — plug in karo. $M_1+M_2=\frac{2.0^3}{1.0^2}=\frac{8.0}{1}=8.0M_{\odot }.$ Step 3 — equally split karo. Kyunki $M_1=M_2=M$: $2M=8.0\Rightarrow M=4.0M_{\odot }$. WHY $M_1+M_2$? Jab masses comparable hoti hain, dono stars apne shared centre of mass ko orbit karte hain; effective central mass total hota hai. Poore reduction ke liye Reduced Mass & Two-Body Problem dekho. Kisi planet ke Sun ke around, $m\ll M$ isliye $M_1+M_2\approx M_{\odot }$ aur hum simple law recover karte hain.

$a$ diya hai, $T$ chahiye (ya ulta) ::: $T^2=\frac{4{\pi }^2}{GM}{a}^3$ mein plug in karo, ek variable at a time Do orbits, ek hi central mass ::: ratio form use karo ${\left(T_1/T_2\right)}^2={\left(a_1/a_2\right)}^3$ $T$ diya hai, $r$ chahiye ::: invert karo: $r={\left(GM{T}^2/4{\pi }^2\right)}^{1/3}$ Central mass $M$ chahiye ::: rearrange karo: $M=4{\pi }^2{a}^3/(G{T}^2)$ Elliptical orbit ::: $a=(r_{\min }+r_{\max })/2$ use karo; eccentricity drop ho jaati hai Comparable masses (binary) ::: $M$ ki jagah $M_1+M_2$ rakhna