Conservation of energy and angular momentum in gravitational field: Visual walkthrough

3.2.2 · D2 · Physics › Orbital Mechanics & Astrodynamics › Conservation of energy and angular momentum in gravitational

Step 1 — Do arrows aur ek dot: stage set karna

KYA. Ek bhaari body (Sun, mass $M$ ) ko ek fixed point par rakho. Ek choti body (ek planet, mass $m$ ) ko kahin door distance $r$ par rakho. Planet se do arrows draw karo:

$r$ — position arrow, jo Sun se planet ki taraf point karta hai. Iski length $r$ hai.
$v$ — velocity arrow, jo dikhata hai ki planet abhi kis taraf slide kar raha hai, aur kitni tezi se (lamba = tez).

KYU. Hum jo bhi prove karte hain woh sab in do arrows ke beech ka relationship hai. Koi bhi symbol use karne se pehle hume use dekhna hoga. $r$ hai "kitna door hai", $v$ hai "kidhar ja raha hai". Abhi ke liye bas itna hi hamare paas hai.

PICTURE. Figure dekho. $M$ origin par baitha hai. Cyan arrow $r$ planet tak pahunchta hai; amber arrow $v$ planet se ek angle par nikalta hai. Unke beech ka angle matter karta hai — yeh baat yaad rakhna.

Step 2 — Cross product: "twist" measure karna

KYA. Hum ek aisa number invent karte hain jo measure kare ki velocity kitni sideways hai position ke relative, na ki along it. Woh number hai cross product $r \times v$ ka magnitude.

KYU yeh tool? Hume ek aisi quantity chahiye jo zero ho jab do arrows parallel hon aur sabse badi ho jab wo perpendicular hon — kyunki "orbiting" bilkul motion ka sideways part hai. Dot product ulta karta (max jab parallel). Isliye cross product "twist" ke liye sahi instrument hai.

PICTURE. Figure mein $r$ aur $v$ par bana parallelogram dikhaya gaya hai. Iski area equals $∣ r ∣∣ v ∣ sin ϕ$ hai, jahan $ϕ$ arrows ke beech ka angle hai. Chota curved arrow rotation ka sense dikhata hai; right-hand rule se twist-vector seedha page se bahar nikalti hai.

Step 3 — Twist kyun frozen hai: torque zero hai

KYA. Hum dikhate hain ki $L$ kabhi nahi badlta. Jo cheez ise badal sakti hai woh hai torque $τ = r \times F$ — force ki twisting effort.

KYU. Basic mechanics se, angular momentum ke change ki rate torque ke barabar hoti hai: $\frac{d L}{d t} = τ$ . To agar $τ = 0$ hamesha ho, tab $L$ kabhi nahi hilega.

PICTURE. Figure mein, $F$ (amber) bilkul $r$ (cyan) ke saath padi hai — woh ek hi line hain. Ek hi line par do arrows ka cross product ka parallelogram area zero hota hai. Isliye force ki twist kuch bhi nahi hai.

Step 4 — Velocity ko "in-out" aur "around" mein split karna

KYA. Flat plane ke andar, velocity ko do clean pieces mein kaato:

$\overset{r}{˙}$ — radial speed: $r$ kitni tezi se badh ya ghut raha hai (in-out motion).
$r \dot{θ}$ — tangential speed: planet kitni tezi se ghoomta hai, jahan $\dot{θ}$ angle ki turning rate hai.

KYU. Yeh do directions perpendicular hain, isliye Pythagoras se total speed bina kisi cross-term ke split hoti hai. Is tarah split karne se hum "around" part ko seedha $L$ mein feed kar sakte hain.

PICTURE. Figure mein velocity arrow ko ek cyan radial leg (along $r$ ) aur ek amber tangential leg ( $r$ ke perpendicular) mein resolve kiya gaya hai. Unke beech ka right angle mark kiya gaya hai.

Ab tangential part ko $L$ mein feed karo. Kyunki tangential direction $r$ ke perpendicular hai ( $ϕ = 9 0^{\circ}$ , $sin ϕ = 1$ ): $L = m r (r \dot{θ}) = m r^{2} \dot{θ} \Rightarrow \dot{θ} = \frac{L}{m r ^{2}} .$

Step 5 — Potential well banana by infinity se andar aana

KYA. Potential energy $U (r)$ define karo — stored "positional" energy — yeh count karke ki gravity kitna kaam karti hai jab planet bahut door se (jahan hum $U = 0$ kehte hain) drift karta hai.

KYU yeh tool? Kyunki gravity conservative hai (dekho Conservative Forces and Potential Energy): kaam sirf start aur end radius par depend karta hai, kabhi path par nahi. Yahi woh condition hai jo ek single number $U (r)$ exist karne deti hai. Iske bina, koi energy bookkeeping possible nahi.

PICTURE. Figure $U (r) = - GM m / r$ plot karta hai: ek curve jo door par $0$ hai aur Sun ke paas ek deep pit mein ghus jaati hai — ek potential well. Escape karne ke liye, tumhe pit se bahar chadh'na hoga.

Step 6 — Total energy freeze karna

KYA. Moving-energy aur stored-energy add karo. Sum kabhi nahi badlta.

KYU. Work-energy theorem kehta hai ki force kinetic energy ko exactly utni hi rate se feed karta hai jitni rate se woh potential energy drain karta hai — woh ek-ke-badle-ek trade karte hain, isliye unke sum ki change rate zero hai.

PICTURE. Figure mein do "savings accounts" hain, KE (amber bar) aur $U$ (cyan bar), perihelion aur aphelion par. Bars individually height change karte hain, lekin unka combined total line flat rehta hai.

Ab Step 4 se split speed aur $\dot{θ} = L / (m r^{2})$ substitute karo: $E = \frac{1}{2} m \overset{r}{˙}^{2} + U_{eff} (r) \frac{L ^{2}}{2 m r ^{2}} - \frac{GM m}{r} .$

Step 7 — Kepler's 2nd law nikal aata hai (equal areas)

KYA. Frozen $L$ turant Kepler's second law deta hai: Sun→planet line equal areas in equal times sweep karti hai.

KYU. Time $d t$ mein sweep hua ek patla triangular sliver area $d A = \frac{1}{2} r \cdot (r d θ)$ hai (arc ke saath base, height $r$ ). $d t$ se divide karo aur $L = m r^{2} \dot{θ}$ use karo.

PICTURE. Figure mein do shaded slivers hain — perihelion ke paas ek mota chota wala aur aphelion ke paas ek patla lamba wala — equal area ke saath, jo dikhata hai ki planet paas hone par fast aur door hone par slow move karta hai.

Step 8 — Edge aur degenerate cases (kabhi koi gap mat chhodna)

KYA / KYU / PICTURE — ek figure mein sab chaar corner cases:

Formula Energy ka sign shape control karta hai

$E = - \frac{GM m}{2 a}$ (semi-major axis $a$ wali ellipse ki energy) aur $E = \frac{1}{2} m v^{2} - \frac{GM m}{r}$ combine karo padhne ke liye:

Case	$E$	$a$	picture
Circle	$< 0$ (deepest well point)	$a = r$	closed round
Ellipse	$< 0$	finite $> 0$	closed oval
Parabola	$= 0$	$\to \infty$	just escapes, $v \to 0$ at infinity
Hyperbola	$> 0$	$< 0$	flyby, speed left over

$L = 0$ (degenerate). $sin ϕ = 0$ : velocity purely radial hai. Koi orbit nahi — planet seedha Sun mein ghusta hai aur ek line ke saath wapas bahar nikalti hai. Centrifugal barrier gayab ho jaati hai, isliye girne ko rokne wala kuch nahi.
$\overset{r}{˙} = 0$ (turning point). Saari speed tangential hai, $v = L / (m r)$ . Yeh perihelion (sabse paas, sabse tez) aur aphelion (sabse door, sabse slow) hain.
$r \to \infty$ with $E = 0$ . Speed exactly infinity par zero ho jaati hai — trapped aur free ke beech ki boundary: yeh escape define karta hai, $v_{esc} = 2 GM / r$ (dekho Escape Velocity and Hohmann Transfers).

Step 9 — Do boxes vis-viva ban jaate hain

KYA. Dono turning points par $E$ evaluate karo aur $e$ 's aur $L$ 's cancel ho jaate hain, sirf $a$ par depend karne wali energy bachti hai. Phir master speed formula paane ke liye $E$ eliminate karo.

KYU. Perihelion par $r_{p} = a (1 - e)$ aur aphelion par $r_{a} = a (1 + e)$ jahan $r_{p} + r_{a} = 2 a$ hai. Dono par $E$ likhna aur add karna eccentricity $e$ collapse kar deta hai.

PICTURE. Figure ek ellipse par $r_{p}$ , $r_{a}$ , aur $a$ mark karta hai aur dikhata hai ki do energy boxes ek single line $E = - GM m /2 a$ mein kaise merge hote hain.

Ek-picture summary

Yeh single frame har step ko stitch karta hai: flat plane (Step 3), split velocity (Step 4), potential well aur effective barrier (Steps 5–6), equal-area sweep (Step 7), aur $E$ ke sign se set shape dial (Step 8), sab kuch do boxed master results (Step 9) tak le jaate hain.

Recall Feynman retelling — poora walkthrough seedhe alfazon mein

Ek planet ko ek rubber sheet par ek bead socho jisme middle mein ek bhaari ball (Sun) rakkhi hai. Pehle humne notice kiya ki pull hamesha seedha ball ki taraf point karta hai, isliye woh bead ko koi twist nahi de sakta — woh "twist number" (angular momentum) hamesha ke liye lock ho jaata hai, jo poore dance ko ek sheet mein flatten karta hai aur matlab hai ki bead equal fan-shaped areas equal time mein sweep karta hai (paas hone par fast, door hone par slow). Phir humne bead ki motion ko "in-and-out" aur "around" mein split kiya, aur notice kiya ki "around" part actually frozen twist hi hai disguise mein, ek invisible wall ki tarah jo bead ko bilkul andar girne se rokti hai. Aage humne energy count ki: moving-energy plus well-depth-energy, aur paya ki unka sum kabhi nahi badlta — ek account bharti hai jab doosra khali hota hai. Aakhir mein, us energy ko sabse paas aur sabse door ke points par evaluate karne se orbit ki wobbliness cancel ho gayi, ek khoobsurat simple sach chodke: total energy sirf orbit ki width par depend karti hai, uski squish par nahi. Ek dial ghumaao — energy ka sign — aur tum smoothly ek saaf circle se, ek oval se, ek just-barely-free parabola se, ek runaway hyperbola tak jaate ho.

Prove torque vanishes for gravity

F ∥ r

, isliye

r \times F = 0

, hence

d L / d t = 0

Constant

L

geometrically kya force karta hai

Orbit ek flat plane mein hoti hai.

Vis-viva equation

v^{2} = GM (\frac{2}{r} - \frac{1}{a})

Ek bound orbit ki total energy depend karti hai

sirf semi-major axis

a

par, eccentricity

e

par nahi.