Kepler's second law — equal areas in equal times, from angular momentum conservation: Worked examples

3.2.6 · D3 · Physics › Orbital Mechanics & Astrodynamics › Kepler's second law — equal areas in equal times, from angul

Ye page ek drill hai. parent note ne law ko ek baar prove kiya tha. Yahan hum ise har angle se attack karte hain — har geometric case, har degenerate input, har trap jo exam set kar sakta hai. Har answer se pehle tumhe ek Forecast line milegi: solution cover karo, guess karo, phir check karo.

Parent se do tools hain jinhe hum baar baar use karte hain. Dono bas pictures ke labels hain:

Figure — Kepler's second law — equal areas in equal times, from angular momentum conservation

Figure dekho: coral slice swept triangle hai; uska area sirf $v_{⊥} = v sin ϕ$ use karta hai (woh mint arrow). Lavender component $v cos ϕ$ $r$ ke saath slide karta hai aur kuch paint nahi karta.

Scenario matrix

Is topic ke har problem ka ye cells mein se ek hoga. Neeche ke examples tag karte hain ki konsa cell clear hua.

Cell	Case class	Kya tricky hai	Example
A	Sirf apsides ( $ϕ = 90°$ )	clean $L = m v r$	Ex 1
B	General point ( $ϕ \neq = 90°$ )	$sin ϕ$ use karna padega	Ex 2
C	Do general points compare kiye	$r v sin ϕ$ conserved	Ex 3
D	Areal velocity as a rate (area se time)	$A = \frac{L}{2 m} t$	Ex 4
E	Degenerate: radial "fall" ( $ϕ = 0$ )	$L = 0$ , law trivial	Ex 5
F	Limiting: circular orbit	constant speed, phir bhi equal areas	Ex 6
G	Non-gravity central force (spring)	law phir bhi hold karta hai	Ex 7
H	Real-world word problem	story ko symbols mein translate karo	Ex 8
I	Exam twist: area / time ka fraction	ratio reasoning	Ex 9

Ex 1 — Cell A: apsides, clean case

Worked example Perihelion vs aphelion speed

Ek planet ka perihelion distance $r_{p} = 4.0 \times 1 0^{10} m$ hai aur speed $v_{p} = 60 km/s$ hai. Uska aphelion distance $r_{a} = 6.0 \times 1 0^{11} m$ hai. $v_{a}$ nikalo.

Forecast: aphelion 15× zyada door hai. Kya $v_{a}$ bada hoga ya chota, aur roughly kitne factor se?

Step 1 — Pehchano ki dono points apsides hain. Perihelion aur aphelion par velocity perpendicular hoti hai radius ke saath, isliye $ϕ = 90°$ aur $sin ϕ = 1$ . Ye step kyun? Tabhi $L = m v r$ koi angle factor ke bina hold karta hai — yahi wajah hai ki apsides "easy" points hote hain.

Step 2 — Dono $L$ values ko equal karo. $m v_{p} r_{p} = m v_{a} r_{a} \Rightarrow v_{p} r_{p} = v_{a} r_{a} .$ Ye step kyun? $L$ conserved hai (central force, zero torque). Mass cancel ho jaata hai.

Step 3 — Solve karo. $v_{a} = v_{p} \frac{r _{p}}{r _{a}} = 60 \times \frac{4.0 \times 1 0 ^{10}}{6.0 \times 1 0 ^{11}} = 60 \times \frac{1}{15} = 4.0 km/s .$

Verify: ratio exactly $1/15$ nikla, distance ratio se match karta hai — forecast ke mutabik, slower aur inverse factor se. Units: $(km/s) \times$ dimensionless $=$ km/s. ✓

Ex 2 — Cell B: ek general point, $ϕ \neq = 90°$

Worked example Sirf perpendicular part count karta hai

Orbit par kisi point par ek comet Sun se $r = 2.0 \times 1 0^{11} m$ door hai, speed $v = 25 km/s$ se chal raha hai, aur velocity radius vector se $ϕ = 30°$ ka angle bana rahi hai. $L / m$ (specific angular momentum) aur areal velocity $d A / d t$ nikalo.

Forecast: kyunki $ϕ$ sirf $30°$ hai, kya $L$ apni maximum value $r v$ ke paas hoga ya bahut kam?

Step 1 — Full formula $L = m r v sin ϕ$ use karo. $\frac{L}{m} = r v sin ϕ = (2.0 \times 1 0^{11}) (2.5 \times 1 0^{4}) sin 30°.$ Ye step kyun? Velocity $r$ ke perpendicular nahi hai, isliye $L = m v r$ over-count karega. Sirf $v sin ϕ = v_{⊥}$ hi area sweep karta hai (figure mein mint arrow).

Step 2 — $sin 30° = 0.5$ plug karo. $\frac{L}{m} = (2.0 \times 1 0^{11}) (2.5 \times 1 0^{4}) (0.5) = 2.5 \times 1 0^{15} m^{2} / s .$

Step 3 — Areal velocity ke liye aadha karo. $\frac{d A}{d t} = \frac{L}{2 m} = 1.25 \times 1 0^{15} m^{2} / s .$ Ye step kyun? $d A / d t = \frac{1}{2} ∣ r \times v ∣ = \frac{L}{2 m}$ — swept-area rate hamesha specific angular momentum ka aadha hoti hai.

Verify: $ϕ = 30°$ ke saath $r v = 5.0 \times 1 0^{15}$ maximum ka aadha mila — maximum se kaafi neeche, forecast ke mutabik. Units: $m \cdot (m/s) = m^{2} / s$ . ✓

Ex 3 — Cell C: do general points compare karo

Worked example Do angle factors ke saath conservation

Point 1 par: $r_{1} = 1.0 \times 1 0^{11} m$ , $v_{1} = 40 km/s$ , $ϕ_{1} = 90°$ . Point 2 par: $r_{2} = 2.0 \times 1 0^{11} m$ , $ϕ_{2} = 30°$ . $v_{2}$ nikalo.

Forecast: point 2 zyada door HAI AUR uski velocity tilted hai — kya ye $v_{2}$ ko upar ya neeche push karte hain? (Dhyan se, ye fight karte hain.)

Step 1 — $L / m$ conserved likho dono points ke beech. $r_{1} v_{1} sin ϕ_{1} = r_{2} v_{2} sin ϕ_{2} .$ Ye step kyun? $L$ har jagah same hai; bas uska factored form point to point change karta hai.

Step 2 — $v_{2}$ ke liye solve karo. $v_{2} = \frac{r _{1} v _{1} s i n ϕ _{1}}{r _{2} s i n ϕ _{2}} = \frac{( 1.0 \times 1 0 ^{11} ) ( 40 ) ( 1 )}{( 2.0 \times 1 0 ^{11} ) ( 0.5 )} .$ Ye step kyun? Unknown isolate karo; har factor ko sahi rakho.

Step 3 — Evaluate karo. $v_{2} = \frac{40}{1.0} = 40 km/s .$

Verify: "2× zyada door" $v_{2}$ ko aadha kar deta, lekin tilt $sin 30° = 0.5$ use wapas double kar deta hai — dono exactly cancel ho jaate hain, $v_{2} = v_{1}$ milta hai. Dono effects draw par lad gaye, forecast ke mutabik. ✓

Ex 4 — Cell D: swept area se time nikalo

Ex 5 — Cell E: degenerate radial drop ( $ϕ = 0$ )

Worked example Seedha Sun ki taraf girna

Ek object Sun se bahut door rest se release hoti hai aur seedha Sun ki taraf girti hai, isliye uski velocity hamesha $r$ ke saath align hoti hai ( $ϕ = 0°$ ). Uska angular momentum aur areal velocity kya hai?

Forecast: agar ye kabhi sideways sweep nahi karta, toh uska radius vector kitna area paint karta hai?

Step 1 — $ϕ = 0$ ko $L = m r v sin ϕ$ mein plug karo. $L = m r v sin 0° = 0.$ Ye step kyun? $sin 0° = 0$ : purely $r$ ke saath motion mein koi perpendicular component nahi hai, isliye kuch sweep nahi hota.

Step 2 — Areal velocity follow karta hai. $\frac{d A}{d t} = \frac{L}{2 m} = 0.$ Ye step kyun? Zero angular momentum ⇒ zero swept area. Radius vector bas ek fixed line ke saath shrink karta hai, zero width ka degenerate triangle paint karta hai.

Verify: "orbit" Sun se hoke jaane wali ek straight line hai — $b = 0$ wala degenerate ellipse. Uska area $π ab = 0$ , $d A / d t = 0$ ke consistent hai. Equal-areas law phir bhi hold karta hai: har second same area sweep hota hai, yaani zero. ✓

Ex 6 — Cell F: circular limit

Worked example Equal areas jab speed already constant ho

Ek satellite radius $r = 7.0 \times 1 0^{6} m$ ki circular orbit mein constant speed $v = 7.5 km/s$ se chal rahi hai. $d A / d t$ nikalo, aur confirm karo ki "equal areas" statement vacuous nahi hai.

Forecast: circle mein speed har jagah constant hai — kya equal-areas law tab trivially satisfied ho jaata hai?

Step 1 — Circle mein $v ⊥ r$ har jagah hota hai, isliye $ϕ = 90°$ . $\frac{d A}{d t} = \frac{1}{2} r v sin 90° = \frac{1}{2} r v .$ Ye step kyun? Circular orbit ki velocity hamesha tangent hoti hai, isliye radius ke perpendicular — har point par clean case.

Step 2 — Evaluate karo. $\frac{d A}{d t} = \frac{1}{2} (7.0 \times 1 0^{6}) (7.5 \times 1 0^{3}) = 2.625 \times 1 0^{10} m^{2} / s .$

Step 3 — Full circle ke saath sanity check. Full circle area $π r^{2} = π (7.0 \times 1 0^{6})^{2} = 1.539 \times 1 0^{14} m^{2}$ ; period $T = 2 π r / v = 5.864 \times 1 0^{3} s$ ; isliye $π r^{2} / T = 2.625 \times 1 0^{10} m^{2} / s$ . ✓

Verify: dono routes agree karte hain. Equal areas vacuous nahi hai — ye automatically true hai kyunki circle mein constant $r$ AUR constant $v$ dono hain; law ka interesting content sirf eccentric orbits ke liye dikhta hai (Ex 1, 3). ✓

Ex 7 — Cell G: ek non-gravity central force

Worked example Spring se bandi bead — law phir bhi hold karta hai

Ek bead frictionless plane par move karti hai, ideal spring se origin se attached hai jisme force $F = - k r$ hai (ek central force, lekin $F \propto r$ , $1/ r^{2}$ nahi). Ek instant par $r_{1} = 0.20 m$ , $v_{1} = 3.0 m/s$ , $ϕ_{1} = 90°$ hai. Baad mein $r_{2} = 0.50 m$ hai. Perpendicular speed $v_{2 ⊥}$ nikalo.

Forecast: force ek spring hai, gravity nahi. Kya Kepler's second law phir bhi apply hoga?

Step 1 — Check karo ki kya force central hai. $F = - k r$ , $r$ ke antiparallel hai, isliye $τ = r \times F = 0$ . Dekho Central Forces and Torque. Ye step kyun? Kepler's 2nd law ko sirf zero torque chahiye, inverse-square nahi. Angular Momentum Conservation phir constant $L$ guarantee karta hai.

Step 2 — $L$ conserve karo: $r_{1} v_{1} = r_{2} v_{2 ⊥}$ (dono perpendicular speed use karte hain). $v_{2 ⊥} = \frac{r _{1} v _{1}}{r _{2}} = \frac{( 0.20 ) ( 3.0 )}{0.50} = 1.2 m/s .$ Ye step kyun? $L / m = r v_{⊥}$ force law se independent conserved rahta hai; zyada door matlab choti perpendicular speed.

Verify: $r_{2} v_{2 ⊥} = 0.50 \times 1.2 = 0.60 = r_{1} v_{1} = 0.20 \times 3.0$ . ✓ Yahan orbit ek ellipse centered on origin hai (focus par nahi) — gravity se alag shape, lekin areas phir bhi equal hain. Ye exactly parent ka "any central force" mistake-buster hai. ✓

Ex 8 — Cell H: real-world word problem

Worked example Comet winter mein vs summer mein

Halley-type comet: perihelion par $r_{p} = 8.8 \times 1 0^{10} m$ , speed $v_{p} = 54.6 km/s$ . Aphelion par $r_{a} = 5.3 \times 1 0^{12} m$ . (a) $v_{a}$ nikalo. (b) Specific angular momentum $L / m$ nikalo. (c) Perihelion par ye kitni baar zyada fast hai?

Forecast: aphelion ~60× zyada door hai — expect karo ki comet wahan almost frozen hoga.

Step 1 — (a) Apsides hain, isliye $v_{p} r_{p} = v_{a} r_{a}$ . $v_{a} = v_{p} \frac{r _{p}}{r _{a}} = 54.6 \times \frac{8.8 \times 1 0 ^{10}}{5.3 \times 1 0 ^{12}} = 54.6 \times 0.016604 = 0.9066 km/s .$ Ye step kyun? Dono apsides hain ( $ϕ = 90°$ ), phir wahi clean case.

Step 2 — (b) $L / m = r_{p} v_{p}$ (perihelion par evaluate karo). $\frac{L}{m} = (8.8 \times 1 0^{10}) (5.46 \times 1 0^{4}) = 4.805 \times 1 0^{15} m^{2} / s .$ Ye step kyun? $L$ har jagah same hai; perihelion convenient hai kyunki $ϕ = 90°$ hai.

Step 3 — (c) Speed ratio. $\frac{v _{p}}{v _{a}} = \frac{r _{a}}{r _{p}} = \frac{5.3 \times 1 0 ^{12}}{8.8 \times 1 0 ^{10}} \approx 60.2.$ Ye step kyun? $v_{p} r_{p} = v_{a} r_{a}$ se, speed ratio bas inverse distance ratio hai.

Verify: (b) aphelion par check karo: $r_{a} v_{a} = (5.3 \times 1 0^{12}) (906.6) = 4.805 \times 1 0^{15} m^{2} / s$ — Step 2 se match karta hai. ✓ Comet door ~0.9 km/s par creep karta hai, forecast ke mutabik. ✓

Ex 9 — Cell I: exam twist (period ka fraction area se)

Worked example Ellipse ka quarter sweep karne mein saal ka kitna fraction lagega?

Kyunki $d A / d t$ constant hai, prove karo ki total ellipse area ka koi bhi fraction $f$ sweep karne mein period $T$ ka fraction $f$ lagtaa hai. Phir: Earth ki orbital area $π ab$ hai; us area ka quarter sweep karne mein kitna time ( $f$ aur days mein) lagega? $T = 365.25$ days use karo.

Forecast: kya "area ka quarter" wahin hoga jitna "time ka quarter"? Kya "angle ka quarter" ke barabar hoga? (Inmein se do equal hain, ek nahi.)

Step 1 — Time swept area ke proportional hai. $t = \frac{A _{swept}}{d A / d t}, T = \frac{π ab}{d A / d t} .$ Ye step kyun? Dono same constant $d A / d t$ use karte hain. Divide karne par cancel ho jaata hai.

Step 2 — Ratio lo. $\frac{t}{T} = \frac{A _{swept}}{π ab} = f .$ Ye step kyun? $d A / d t$ cancel ho jaata hai — area fraction equals time fraction, hamesha. Ye exam ka pura point hai.

Step 3 — Area ka quarter, $f = 1/4$ . $t = \frac{1}{4} \times 365.25 = 91.31 days .$

Verify: area-fraction $=$ time-fraction ✓, lekin NOT angle-fraction — perihelion ke paas ka quarter-area slice aphelion ke paas waale se bada angle span karta hai (dekho Kepler's First Law (Ellipse)). Forecast trap: area ↔ time equal; angle alag hota hai. ✓

Connections

Parent: Kepler's Second Law
Angular Momentum Conservation — har example mein kya fixed rehta hai
Central Forces and Torque — kyun spring (Ex 7) phir bhi law obey karta hai
Cross Product and Area — $sin ϕ$ aur swept triangle
Kepler's First Law (Ellipse) — kyun angle-fraction ≠ area-fraction (Ex 9)
Kepler's Third Law (T² ∝ a³) — $π ab = (L /2 m) T$ use karta hai
Vis-viva Equation — kisi bhi $r$ par $v$ nikalo jab $ϕ$ unknown ho