Worked examples — Conservation of energy and angular momentum in gravitational field

3.2.2 · D3 · Physics › Orbital Mechanics & Astrodynamics › Conservation of energy and angular momentum in gravitational

Kuch bhi shuru karne se pehle, main un chaar tools ko plain words mein re-state karta hoon jinpe hum rely karte hain, taaki koi bhi symbol bina kamaaye use na ho.

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The scenario matrix

Is topic ka har problem in cells mein se ek hai. Neeche har worked example us cell ke saath tagged hai jise woh fill karta hai.

#	Cell class	Kya special hai	Example
A	Sign: $E<0$ (bound)	ellipse, finite $a$	Ex 1
B	Sign: $E=0$ (parabolic, $e=1$)	escape "just barely", $a\to \infty$	Ex 4
C	Sign: $E>0$ (unbound, $e>1$)	hyperbola, $a<0$	Ex 5
D	Degenerate: circle ($e=0$)	$r=a$ har jagah	Ex 2
E	Degenerate: radial drop ($L=0$)	bilkul bhi sideways motion nahi	Ex 6
F	Turning points (perihelion/aphelion)	$\dot{r}=0$, poori speed sideways hai	Ex 3
G	Limiting value ($r\to \infty$)	escaper ki infinity pe speed	Ex 5 (part 2)
H	Real-world word problem	English → symbols mein translate karo	Ex 7
I	Exam twist (same $a$, different $e$)	trap: energy unchanged	Ex 8

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Example 1 — Bound orbit: $a$ nikalo, phir $r_p$ aur $v_p$ (cells A, F)

Forecast: $E<0$ to orbit bound hai — ek closed ellipse. Guess karo: kya $r_p$, $a$ se bada hoga ya chhota? (Chhota hona chahiye — perihelion close point hai.)

1. Energy se $a$ nikalo. $E=-\frac{GMm}{2a}\Rightarrow a=-\frac{GMm}{2E}$. Yeh step kyun? Ek bound orbit ki $E$ sirf $a$ par depend karti hai, isliye energy akele orbit size unlock kar deti hai. $a=-\frac{(3.99\times 10^{14})(1000)}{2(-2.0\times 10^9)}=9.975\times 10^7\text{m}.$

2. Turning-point equation setup karo. Perihelion par $\dot{r}=0$, to poori speed sideways hai: $v_p=\frac{L}{m{r}_p}$. Yeh step kyun? Turning point woh jagah hai jahan messy radial speed vanish ho jaati hai, to $L$ humein $v$ free mein de deta hai.

3. Use karo energy mein. $v_p=\frac{L}{m{r}_p}$ ko $E=\frac{1}{2}m{v}_p^2-\frac{GMm}{r_p}$ mein substitute karo: $E=\frac{L^2}{2m{r}_p^2}-\frac{GMm}{r_p}.$ Yeh step kyun? Ab $r_p$ ek hi unknown hai — yeh $U_{\text{eff}}(r_p)=E$ hai turning point par evaluate hua.

4. Isse $x=1/r_p$ mein standard quadratic banao. Multiply karo aur powers of $x$ collect karo: $\frac{L^2}{2m}{x}^2-(GMm)x-E=0.$ Yeh step kyun? Physically, $r$ energy mein sirf $1/r$ aur $1/r^2$ ke roop mein aata hai; $x=1/r$ substitute karne se "energy = const at a turning point" ek ordinary quadratic ban jaata hai jiske do roots exactly do turning radii hain $1/r_p$ aur $1/r_a$ — us hi ellipse ke close aur far ends. Signs note karo: $L^2/2m>0$ (woh $+x^2$ centrifugal term), $-GMm<0$ (attractive gravity), aur $-E>0$ yahan kyunki $E<0$. Numbers daalo ten ke careful powers ke saath: $\frac{L^2}{2m}=\frac{(4.0\times 10^{13}{)}^2}{2(1000)}=\frac{1.6\times 10^{27}}{2\times 10^3}=8.0\times 10^{23},$ $GMm=3.99\times 10^{17},-E=2.0\times 10^9.$

5. Quadratic formula apply karo. $A=8.0\times 10^{23}$, $B=-3.99\times 10^{17}$, $C=-E=2.0\times 10^9$ ke saath: $x=\frac{GMm\pm \sqrt{(GMm{)}^2-4\cdot \frac{L^2}{2m}\cdot (-E)}}{2\cdot \frac{L^2}{2m}}=\frac{3.99\times 10^{17}\pm \sqrt{(3.99\times 10^{17}{)}^2-4(8.0\times 10^{23})(2.0\times 10^9)}}{1.6\times 10^{24}}.$ Perihelion kaun sa root hai? Perihelion chhota radius = bada $x$ hai, to $+$ root lo; $-$ root $1/r_a$ (aphelion) deta hai. Evaluate karne par: $x_p=\frac{1}{r_p}\approx 3.06\times 10^{-7}{\text{m}}^{-1}\Rightarrow r_p\approx 3.27\times 10^6\text{m},$ $x_a=\frac{1}{r_a}\approx 1.94\times 10^{-7}\times 10^0\cdots \Rightarrow r_a\approx 1.966\times 10^8\text{m}.$

6. Perihelion par speed. $v_p=\frac{L}{m{r}_p}=\frac{4.0\times 10^{13}}{1000\cdot 3.27\times 10^6}\approx 1.22\times 10^4\text{m/s}.$

Verify karo: Check karo $r_p+r_a=2a$: $3.27\times 10^6+1.966\times 10^8\approx 1.999\times 10^8\approx 2a=1.995\times 10^8$ ✔ (rounding). $v_p$ ke units: $\frac{{\text{kg⋅m}}^2/\text{s}}{\text{kg⋅m}}=\text{m/s}$ ✔.

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Example 2 — Circle (degenerate, $e=0$) (cell D)

Forecast: Circle mein loop ke around kuch bhi nahi badalta — ek constant speed. Kya yeh famous "$\sim 7.5$ km/s low-Earth-orbit" number ke paas honi chahiye? Guess: haan.

1. Circle ke liye $r=a$ use karo. Circle ek aisi ellipse hai jisme longest aur shortest width equal hain, to $a=r$ (aur $e=0$). Yeh step kyun? Yeh vis-viva ko instantly collapse kar deta hai.

2. Vis-viva mein plug karo. $v^2=GM\left(\frac{2}{r}-\frac{1}{a}\right)=GM\left(\frac{2}{r}-\frac{1}{r}\right)=\frac{GM}{r}.$ Yeh step kyun? $\frac{2}{r}-\frac{1}{r}=\frac{1}{r}$ — general speed law ka circular special case.

3. Compute karo. $v=\sqrt{\frac{3.99\times 10^{14}}{7.0\times 10^6}}\approx 7550\text{m/s}=7.55\text{km/s}.$

Verify karo: Circle ke liye, gravity centripetal need ke equal honi chahiye: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}\Rightarrow v^2=\frac{GM}{r}$ — same formula, alag route se ✔. LEO value se match karta hai ✔.

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Example 3 — $L$ se perihelion vs aphelion speeds (cell F)

Neeche figure (s01): ek ellipse jisme Earth right-hand focus par hai (yellow dot). Ek chhota pink radius arm near point (perihelion) tak pahunchta hai jahan velocity arrow lamba aur fast hai; ek lamba blue radius arm far point (aphelion) tak pahunchta hai jahan velocity arrow chhota aur slow hai. Picture dikhata hai kyun chhote arm ko bada speed carry karna padta hai taaki $L=mr{v}_{\perp }$ constant rahe.

Forecast: Figure dekho — perihelion par (close, pink) radius arm chhota hai, aphelion par (far, blue) lamba hai. Kyunki $L=mr{v}_{\perp }$ dono par same hai, chhote arm mein badi speed honi chahiye. To $v_p/v_a>1$.

1. Dono ends turning points hain. Har jagah, velocity purely sideways hai ($\dot{r}=0$), to $L=m{r}_p{v}_p=m{r}_a{v}_a$. Yeh step kyun? Equal $L$ do "all-sideways" points par ek clean speed ratio deta hai bina energy ki zaroorat ke.

2. Ratio. $\frac{v_p}{v_a}=\frac{r_a}{r_p}=\frac{1.0\times 10^8}{5.0\times 10^7}=\mathrm{2.0.}$ Perihelion do guna fast hai.

3. Vis-viva se $v_p$ nikalo. Semi-major axis $a=\frac{r_p+r_a}{2}=7.5\times 10^7\text{m}$. Phir $v_p=\sqrt{GM\left(\frac{2}{r_p}-\frac{1}{a}\right)}=\sqrt{3.99\times 10^{14}\left(\frac{2}{5.0\times 10^7}-\frac{1}{7.5\times 10^7}\right)}.$ Yeh step kyun? Vis-viva $a$ pata hone ke baad kisi bhi $r$ par actual speed deta hai. $v_p=\sqrt{3.99\times 10^{14}\cdot (4.0\times 10^{-8}-1.333\times 10^{-8})}\approx 3263\text{m/s}.$

Verify karo: $v_a$, $v_p$ ka aadha hona chahiye: $v_a=\sqrt{GM(\frac{2}{r_a}-\frac{1}{a})}=\sqrt{3.99\times 10^{14}(2.0\times 10^{-8}-1.333\times 10^{-8})}\approx 1631\text{m/s}$, aur $3263/1631\approx 2.0$ ✔.

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Example 4 — Parabola: $E=0$, escape "just barely" ($e=1$) (cell B)

Forecast: $E=0$ wapas girne (ellipse) aur free fly karne (hyperbola) ke beech ka knife-edge hai. Shape ek parabola hai jisme $a\to \infty$ aur eccentricity exactly $e=1$ hai. Guess: kya $v_{\text{esc}}$ same $r$ par circular speed se badi hai? (Haan — break free karne ke liye extra chahiye.)

1. $E=0$ set karo. $E=\frac{1}{2}m{v}^2-\frac{GMm}{r}=0\Rightarrow \frac{1}{2}{v}^2=\frac{GM}{r}.$ Yeh step kyun? "Just barely escape" ka matlab hai infinity par zero leftover speed ke saath pahunchna, yani $E=0$.

2. Solve karo. $v_{\text{esc}}=\sqrt{\frac{2GM}{r}}=\sqrt{\frac{2(3.99\times 10^{14})}{1.0\times 10^7}}\approx 8933\text{m/s}=8.93\text{km/s}.$

3. Kyun $a\to \infty$ aur $e=1$. $E=-\frac{GMm}{2a}$ se, $E=0$ set karne par $a\to \infty$ force hota hai: "ellipse" stretch hoti rehti hai jab tak uska far end infinity tak nahi chala jaata ($r_a\to \infty$), to $e=\frac{r_a-r_p}{r_a+r_p}\to 1$ — ek open parabola. Dekho Effective Potential and Orbit Classification.

Verify karo: Circular speed $\sqrt{GM/r}$ se ratio: $v_{\text{esc}}/v_{\text{circ}}=\sqrt{2}\approx 1.414$ ✔ (escape hamesha same radius par circular speed ka $\sqrt{2}$ guna hota hai).

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Example 5 — Hyperbola: $E>0$, aur infinity par speed (cells C, G)

Forecast: Ex 4 se yahan escape speed $8933$ m/s hai. Kyunki $9000>8933$, yeh escape karta hai — hyperbolic ($e>1$), $E>0$. Infinity par yeh abhi bhi move karta hai (kuch speed bachi hai), to $v_{\infty }>0$ lekin $9000$ se kam.

1. $E$ per unit mass compute karo (shape questions ke liye mass cancel ho jaata hai): $\epsilon =\frac{1}{2}{v}^2-\frac{GM}{r}$. Per unit mass kyun? Orbit type $E$ ke sign par depend karta hai, aur $m$ (positive) se divide karne par sign preserve rehta hai — simple hai. $\epsilon =\frac{1}{2}(9000{)}^2-\frac{3.99\times 10^{14}}{1.0\times 10^7}=4.05\times 10^7-3.99\times 10^7=6.0\times 10^5\text{J/kg}.$ $\epsilon >0\Rightarrow$ hyperbolic escape ✔ (a).

2. Infinity par speed. Jab $r\to \infty$, $U\to 0$, to saari energy kinetic hai: $\frac{1}{2}{v}_{\infty }^2=\epsilon$. Yeh step kyun? Yeh limiting value cell hai — energy conservation ko $r\to \infty$ limit mein evaluate karo. $v_{\infty }=\sqrt{2\epsilon }=\sqrt{2\cdot 6.0\times 10^5}\approx 1095\text{m/s}.$

Verify karo: $v_{\infty }<v_{\text{start}}=9000$ ✔ (gravity ne bahar climb karte waqt speed chheeni). Aur $v_{\infty }^2=v^2-v_{\text{esc}}^2=9000^2-8933^2\approx 1.2\times 10^6=2\epsilon$ ✔.

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Example 6 — Radial drop: $L=0$ (degenerate) (cell E)

Neeche figure (s02): Earth (blue disc) apne centre ke saath marked (yellow). Ek stationary object upar $r_0$ par baitha hai (pink dot) aur seedha neeche ek vertical line — pink arrow — surface par $r$ tak girta hai. Kyunki koi sideways push nahi hai, poora path ek radial line hai; koi loop nahi, koi ellipse nahi.

Forecast: $L=0$ ke saath kabhi bhi "sideways" motion nahi hogi — yeh centre ki taraf ek seedhi line mein seedha neeche girta hai (figure dekho, seedha pink track). Koi orbit nahi, koi ellipse nahi: pure free-fall. Speed girane par badhti hai.

1. Degeneracy confirm karo. $L=mr{v}_{\perp }=0$ with $r\ne 0$ forces $v_{\perp }=0$: motion purely radial hai. $U_{\text{eff}}$ mein centrifugal term $\frac{L^2}{2m{r}^2}$ vanish ho jaata hai — gir jaane se rokne ke liye koi barrier nahi. Yeh step kyun? Yeh batata hai ki $v$ poori tarah radial hai, to $\frac{1}{2}{v}^2$ poori kinetic energy hai.

2. Dono radii ke beech energy conservation (per unit mass, rest se start): $\underset{\text{start}}{\underbrace{0-\frac{GM}{r_0}}}=\underset{\text{end}}{\underbrace{\frac{1}{2}{v}^2-\frac{GM}{r}}}.$ Yeh step kyun? $E$ constant hai; release par kinetic energy $0$ hai.

3. Solve karo. $\frac{1}{2}{v}^2=GM\left(\frac{1}{r}-\frac{1}{r_0}\right)$, to $v=\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)}.$ Numbers daalo: $\frac{1}{r}-\frac{1}{r_0}=\frac{1}{6.4\times 10^6}-\frac{1}{8.0\times 10^6}=1.5625\times 10^{-7}-1.25\times 10^{-7}=3.125\times 10^{-8}{\text{m}}^{-1}.$ Phir $2GM=7.98\times 10^{14}$, to $v=\sqrt{7.98\times 10^{14}\cdot 3.125\times 10^{-8}}=\sqrt{2.494\times 10^7}\approx 4994\text{m/s}.$

Verify karo: Sign check: $\frac{1}{r}>\frac{1}{r_0}$ kyunki object chhote radius tak gira, to bracket positive hai aur $v$ real hai ✔. Units: $\sqrt{{\text{m}}^3/{\text{s}}^2\cdot {\text{m}}^{-1}}=\sqrt{{\text{m}}^2/{\text{s}}^2}=\text{m/s}$ ✔.

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Example 7 — Real-world word problem (cell H)

Forecast: Aphelion par yeh slowly crawl karta hai aur bahut door; perihelion par yeh fast whip karta hai aur close. Hume dono conservation laws chain karne honge. Guess: $r_p$, $r_a$ se bahut chhota hoga kyunki comet itne door bahut slow hai (barely bound).

1. Aphelion ek turning point hai → $\ell$ (angular momentum per unit mass) $=r_a{v}_a$. Yeh step kyun? Aphelion par poori speed sideways hai, to $\ell$ trivially padha ja sakta hai. $\ell =r_a{v}_a=(5.0\times 10^{11})(1.0\times 10^3)=5.0\times 10^{14}{\text{m}}^2/\text{s}.$

2. Aphelion par energy per unit mass. $\epsilon =\frac{1}{2}{v}_a^2-\frac{GM}{r_a}=\frac{1}{2}(10^3{)}^2-\frac{1.327\times 10^{20}}{5.0\times 10^{11}}=5.0\times 10^5-2.654\times 10^8\approx -2.649\times 10^8\text{J/kg}.$ Yeh step kyun? $\epsilon <0$ bound confirm karta hai, aur yeh constant hai jo hum perihelion par reuse karenge.

3. Perihelion par (bhi ek turning point) $v_p=\ell /r_p$, aur energy conservation same $\epsilon$ deta hai: $\epsilon =\frac{{\ell }^2}{2r_p^2}-\frac{GM}{r_p}.$ Isse $x=1/r_p$ mein quadratic banao: $\frac{{\ell }^2}{2}{x}^2-GMx-\epsilon =0.$ Yeh step kyun? Ex 1 wali same physics: turning point par energy sirf $1/r$ aur $1/r^2$ ka function hai, to $x=1/r$ substitution ek quadratic produce karta hai jiske do roots $1/r_p$ aur $1/r_a$ hain. Signs: ${\ell }^2/2>0$, $-GM<0$, aur $-\epsilon >0$ (kyunki $\epsilon <0$).

4. Carefully evaluate karo (ten ke powers dhyan se). $\frac{{\ell }^2}{2}=\frac{(5.0\times 10^{14}{)}^2}{2}=\frac{2.5\times 10^{29}}{2}=1.25\times 10^{29},GM=1.327\times 10^{20},-\epsilon =2.649\times 10^8.$ Quadratic formula, $+$ root lete hue (perihelion = chhota radius = bada $x$): $x_p=\frac{GM+\sqrt{G{M}^2-4\cdot \frac{{\ell }^2}{2}\cdot (-\epsilon )}}{2\cdot \frac{{\ell }^2}{2}}=\frac{1.327\times 10^{20}+\sqrt{(1.327\times 10^{20}{)}^2-4(1.25\times 10^{29})(2.649\times 10^8)}}{2.5\times 10^{29}}.$ Yeh $x_p\approx 8.72\times 10^{-10}{\text{m}}^{-1}$ deta hai, to $r_p\approx 1.147\times 10^9\text{m}\approx 1.15\times 10^9\text{m}.$

Verify karo: Semi-major axis $a=-GM/(2\epsilon )=1.327\times 10^{20}/(2\cdot 2.649\times 10^8)\approx 2.505\times 10^{11}$, to $2a\approx 5.01\times 10^{11}$; aur $r_p+r_a=1.15\times 10^9+5.0\times 10^{11}\approx 5.01\times 10^{11}$ ✔. Saath hi $r_p\ll r_a$ as forecast ✔. (Dekho Kepler's Laws of Planetary Motion yeh jaanne ke liye ki aise comets kyun wildly eccentric orbits mein hote hain.)

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Example 8 — Exam twist: same $a$, different $e$ (cell I)

Forecast: Trap yeh hai ki "wilder" eccentric ellipse Y mein zyada energy hai — aisa bolna. Lekin energy sirf $a$ par depend karti hai. Dono ke $a$ compute karo aur compare karo.

1. Har orbit ka semi-major axis. X: $a_X=1.0\times 10^7$ (circle → $a=r$). Y: $a_Y=\frac{r_p+r_a}{2}=\frac{6.0\times 10^6+1.4\times 10^7}{2}=1.0\times 10^7\text{m}.$ Yeh step kyun? $a=\frac{r_p+r_a}{2}$ sirf long axis ka aadha hai.

2. Dono equal hain! $a_X=a_Y$.

3. Energies. $\epsilon =-\frac{GM}{2a}$ dono ke liye: $-\frac{3.99\times 10^{14}}{2(1.0\times 10^7)}=-1.995\times 10^7\text{J/kg}$ — identical. Yeh step kyun? Eccentricity $e$ kabhi bhi $\epsilon =-GM/2a$ mein appear nahi karta.

Verify karo: Unki eccentricities alag hain — X: $e_X=0$; Y: $e_Y=\frac{r_a-r_p}{r_a+r_p}=\frac{1.4\times 10^7-6.0\times 10^6}{2.0\times 10^7}=0.4$. Unke angular momenta do differ karte hain: circle X ka ${\ell }_X=\sqrt{GMa}\approx 6.32\times 10^{10}$; ellipse Y ka ${\ell }_Y=\sqrt{GMa(1-e_Y^2)}\approx 5.79\times 10^{10}<{\ell }_X$ ✔. Same energy, different shape — yahi point hai.

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Recall Kaun sa cell kaun sa tha?

Ex 1 :::: A, F (bound, turning points) Ex 2 :::: D (circle, $e=0$) Ex 3 :::: F (perihelion vs aphelion via $L$) Ex 4 :::: B (parabola, $E=0$, $e=1$) Ex 5 :::: C, G (hyperbola $e>1$, speed at infinity) Ex 6 :::: E (radial drop, $L=0$) Ex 7 :::: H (word problem) Ex 8 :::: I (same $a$, different $e$)

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