Visual walkthrough — Specific angular momentum h = √(GMp)

3.2.12 · D2 · Physics › Orbital Mechanics & Astrodynamics › Specific angular momentum h = √(GMp)

Hume sirf teen raw ideas chahiye, aur hum har ek ko draw karenge:

1. Ek bade mass ke paas move karta hua dot (position aur velocity arrows ke roop mein).
2. Mass se dot tak ki line ki "sweep rate."
3. Ek clever variable change jo messy orbit ko ek clean sine wave mein badal deta hai.

---

Step 1 — Do arrows draw karo: $\vec{r}$ aur $\vec{v}$

KYA. Ek heavy mass $M$ ko ek fixed point $O$ (the focus) par rakho. Ek tiny satellite point $P$ par baitha hai. $O$ se $P$ tak ka arrow draw karo — ise $\vec{r}$, yaani position vector kaho. Ek doosra arrow draw karo jo dikhaaye ki $P$ abhi kis taraf move kar raha hai — ise $\vec{v}$, yaani velocity vector kaho.

YEH DONo KYUN. Chhote mass ke liye Newton's law hai $\ddot{\vec{r}}=-\frac{GM}{r^2}\hat{r}$. Chhota mass $m$ already cancel ho chuka hai, isliye motion sirf $\vec{r}$ aur $\vec{v}$ par depend karta hai. Agar hum ek conserved number chahte hain, toh woh in do arrows aur kuch nahi se banana chahiye.

PICTURE. Figure dekho: amber arrow $\vec{v}$ seedha white arrow $\vec{r}$ ki taraf point nahi kar raha — un dono ke beech angle $\varphi$ hai. $\vec{v}$ ka woh bacha hua sideways part hi "$O$ ke around ghoomna" matlab hai.

---

Step 2 — Spin number $\vec{h}=\vec{r}\times \vec{v}$ banao

KYA. Hum do arrows ko cross product $\vec{r}\times \vec{v}$ se combine karte hain. Ise padhо: "$\vec{r}$ aur $\vec{v}$ lo, aur ek naya arrow produce karo jo seedha un ke plane se bahar nikle, jis ki length un ke banaye parallelogram ke area ke barabar ho."

Cross product kyun, dot product kyun nahi? Dot product $\vec{r}\cdot \vec{v}$ measure karta hai ki $\vec{v}$ kitna $\vec{r}$ ke saath point kar raha hai — yaani satellite kitna tezi se paas aa raha hai ya door ja raha hai. Yeh bilkul ulta hai jo hum chahte hain. Hum sideways motion chahte hain, woh part jo satellite ko $O$ ke around le jaata hai. Cross product exactly wohi pick karta hai: iski size $rv\sin \varphi$ hai, aur $\sin \varphi$ tab sabse bada hota hai jab $\vec{v}$ perpendicular ho (pure sideways) aur zero jab $\vec{v}$ seedha bahar point kare (pure radial, koi spin nahi).

$$\vec{h}\equiv \underset{\text{kahan hai}}{\underbrace{\vec{r}}}\times \underset{\text{kahan ja raha hai}}{\underbrace{\vec{v}}},h=|\vec{h}|=\underset{\text{length}}{\underbrace{r}}\underset{\text{speed}}{\underbrace{v}}\underset{\text{sideways fraction}}{\underbrace{\sin \varphi }}$$

PICTURE. Cyan parallelogram $\vec{r}$ aur $\vec{v}$ se bana hai. Iski area $h$ hai. Purple arrow $\vec{h}$ seedha page se bahar nikalta hai — yahi woh axis hai jiske around satellite ghoomta hai.

---

Step 3 — Dikhaao ki $\vec{h}$ kabhi nahi badalta

KYA. Hum check karte hain ki $\vec{h}=\vec{r}\times \vec{v}$ ko time mein differentiate karke $\vec{h}$ kitna tezi se badal raha hai.

$$\frac{d\vec{h}}{dt}=\underset{\vec{v}\times \vec{v}=0}{\underbrace{\frac{d\vec{r}}{dt}\times \vec{v}}}+\underset{\vec{r}\times \vec{a}}{\underbrace{\vec{r}\times \frac{d\vec{v}}{dt}}}$$

HAR PIECE KYUN KHATAM HOTI HAI.

• $\frac{d\vec{r}}{dt}=\vec{v}$, aur $\vec{v}\times \vec{v}=0$ kyunki koi bhi arrow khud se cross kiya hua zero deta hai — kisi arrow ka khud se parallelogram ek flat line hai, area $0$.
• $\vec{a}=-\frac{GM}{r^2}\hat{r}$ seedha $\vec{r}$ ke saath point karta hai (gravity $O$ ki taraf kheenchti hai). Aur $\vec{r}\times \hat{r}=0$ usi reason se — parallel arrows zero-area parallelogram banate hain.

Toh $\frac{d\vec{h}}{dt}=0$: $\vec{h}$ hamesha ke liye frozen hai. Isi liye orbit ek flat plane mein rehta hai aur equal areas equal times mein sweep hote hain.

PICTURE. Orbit ke do snapshots, pehle aur baad mein. Parallelogram tilt hota aur stretch hota hai jaise satellite move karta hai, lekin iski area bilkul same rehti hai aur purple $\vec{h}$ arrow kabhi nahi hilta.

---

Step 4 — $h$ ko polar coordinates mein likhte hain: $h=r^2\dot{\theta }$

KYA. Satellite ko $O$ se distance $r$ aur angle $\theta$ (true anomaly) se describe karo, jo ek fixed reference direction se measure hota hai. Sideways speed $r\dot{\theta }$ hai (distance $\times$ angular rate), aur outward speed $\dot{r}$ hai.

KYUN. Humne dikhaaya ki $h$ motion ka sideways part hai times $r$. Polar language mein sideways velocity exactly $r\dot{\theta }$ hai, toh

$$h=r\cdot (\text{sideways speed})=r\cdot r\dot{\theta }=r^2\dot{\theta }.$$

Dot ka matlab hai "per second rate of change": $\dot{\theta }=d\theta /dt$ yeh hai ki angle kitna tezi se badh raha hai.

PICTURE. Velocity arrow $\vec{v}$ ek white radial piece $\dot{r}$ ($\vec{r}$ ke saath) aur ek amber tangential piece $r\dot{\theta }$ (perpendicular) mein split hota hai. Sirf amber piece spin ko feed karta hai.

---

Step 5 — Clever swap: $u=1/r$ lo

KYA. Hum orbit ki shape $r(\theta )$ chahte hain, $r(t)$ nahi. Toh hum time $t$ ko angle $\theta$ ke liye $\dot{\theta }=h/r^2$ use karke trade karte hain, aur $r$ ko uske reciprocal $u=1/r$ se replace karte hain.

$u=1/r$ KYUN? Equation of motion mein $1/r^2$ terms hain (gravity se) aur $r$ terms hain (curving se). $r$ mein likhe jaayein toh yeh aapas mein fight karte hain aur nonlinear rehte hain. $u=1/r$ mein likhe jaayein toh magic hota hai: gravity term ek plain constant ban jaata hai, aur sab kuch physics ki sabse simple equation mein collapse ho jaata hai — ek spring equation. Tool isliye choose kiya jaata hai kyunki yeh problem ko linearise karta hai.

Step-by-step (har line ek substitution hai):

$$\dot{r}=\frac{dr}{d\theta }\dot{\theta }=\left(\underset{dr/d\theta ,\text{kyunki }r=1/u}{\underbrace{-\frac{1}{u^2}\frac{du}{d\theta }}}\right)\left(\underset{\dot{\theta }=h/r^2}{\underbrace{h{u}^2}}\right)=-h\frac{du}{d\theta }$$ $$\ddot{r}=\frac{d}{dt}\left(-h\frac{du}{d\theta }\right)=-h\frac{d^{2}u}{d{\theta }^2}\dot{\theta }=-h^2{u}^2\frac{d^{2}u}{d{\theta }^2}$$

PICTURE. Usi orbit ke do side-by-side plots: left mein real ellipse $r(\theta )$ (curvy, awkward); right mein $u(\theta )=1/r$, jo ek clean cosine wave trace karta hai. Swap literally problem ko ek wave mein straighten kar deta hai.

---

Step 6 — Binet's equation: orbit ek spring hai

KYA. $\ddot{r}$ aur $r{\dot{\theta }}^2$ ko radial equation of motion mein plug karo

$$\ddot{r}-r{\dot{\theta }}^2=-\frac{GM}{r^2}.$$

$r{\dot{\theta }}^2=\frac{1}{u}(h{u}^2{)}^2=h^2{u}^3$ aur $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$ use karke (jahan $u^{\prime \prime }=d^{2}u/d{\theta }^2$):

$$-h^2{u}^2{u}^{\prime \prime }-h^2{u}^3=-GM{u}^2.$$

Har term ko $-h^2{u}^2$ se divide karo:

$$\boxed{u^{\prime \prime }+u=\frac{GM}{h^2}}$$

YEH BEAUTIFUL KYUN HAI. Yeh Binet's equation hai, aur yeh form mein bilkul ek steady force se push hote hue mass on a spring jaisi hai: $x^{\prime \prime }+x=\text{const}$. Hum pehle se iski solution jaante hain — ek constant offset plus ek cosine wave.

$$u(\theta )=\underset{\text{constant offset (the push)}}{\underbrace{\frac{GM}{h^2}}}+\underset{\text{free oscillation}}{\underbrace{A\cos (\theta -{\theta }_0)}}$$

Reference angle choose karo taki ${\theta }_0=0$ ho aur $A=\frac{GM}{h^2}e$ likho (yeh sirf wave ki height ko ek number $e\ge 0$ ke terms mein name karta hai):

$$u=\frac{GM}{h^2}(1+e\cos \theta ).$$

PICTURE. Straight offset line $GM/h^2$ jiske upar ek cosine ride kar raha hai — sum hai $u(\theta )$. Ripple ki amplitude woh hai jise hum eccentricity $e$ kahenge.

---

Step 7 — $r$ par wapas jaate hain aur $p$ read karte hain

KYA. $u=1/r$ ko ulta karo orbit recover karne ke liye:

$$r=\frac{1}{u}=\frac{h^2/GM}{1+e\cos \theta }.$$

Two-body problem ki standard conic form se term-by-term compare karo:

$$r=\frac{p}{1+e\cos \theta }⟹\underset{\text{numerator}}{\underbrace{p}}=\frac{h^2}{GM}.$$

LAST SQUEEZE KYUN. Do fractions jo har $\theta$ ke liye equal hain unke numerators equal hone chahiye (denominators already identical hain). Toh numerator $h^2/GM$ hai semi-latus rectum $p$. $h$ ke liye solve karo:

$$p=\frac{h^2}{GM}⟹h^2=GMp⟹\boxed{h=\sqrt{GMp}}.$$

Hum positive square root lete hain kyunki $h=r^2\dot{\theta }$ ek magnitude hai — length² over time, kabhi negative nahi.

PICTURE. Finished ellipse jisme $O$ focus par hai. Amber segment seedha focus se upar draw kiya gaya hai ($\theta =90^{\circ }$ par, jahan $\cos \theta =0$) jis ki length exactly $p$ hai — "perpendicular peek."

---

Step 8 — Har case, checked

KYA. Hum un saare shapes se guzarte hain jo number $e$ produce kar sakta hai, taaki koi bhi reader kabhi un-drawn scenario se na mile.

$e$	shape	$p$ aur $h$ kya karte hain
$e=0$	circle	$p=a=r$, toh $h=\sqrt{GMr}=v_{\text{circ}}r$
$0<e<1$	ellipse	$p=a(1-e^2)$, toh $h=\sqrt{GMa(1-e^2)}$
$e=1$	parabola	$a\to \infty$ lekin $p$ finite; phir bhi $h=\sqrt{GMp}$
$e>1$	hyperbola	$p=a(e^2-1)$ ($a<0$ ke saath); phir bhi $h=\sqrt{GMp}$

Degenerate case $h=0$. Agar $\varphi =0$ (velocity dead-on radial), toh $h=rv\sin 0=0$, toh $p=h^2/GM=0$: "orbit" $O$ se guzarne wali seedhi upar-neeche line mein collapse ho jaati hai — bilkul koi going-around nahi. Formula phir bhi hold karta hai, bas ek fall describe karta hai.

YEH KYUN MATTER KARTA HAI. Single relation $h=\sqrt{GMp}$ har conic mein survive karta hai. Sirf link $p\leftrightarrow a$ shape ke saath badalta hai; $p$ khud hamesha $h$ carry karta hai.

PICTURE. Charon conics ek hi focus $O$ share karte hain, har ek ka apna semi-latus rectum amber vertical segment ke roop mein drawn — mota segment, tezi spin.

---

Ek-picture summary

Upar sab kuch ek diagram mein compress kiya gaya: do arrows $\to$ frozen spin $\vec{h}$ $\to$ wave $u(\theta )$ $\to$ conic apni half-width $p$ ke saath $\to$ boxed formula.

Recall Pure walkthrough ki Feynman retelling

Ek pathar ki kalpana karo jo ek heavy magnet se springy string se bandha hai, whirl kar raha hai. String sirf kabhi magnet ki taraf straight kheenchti hai — kabhi sideways nahi. Kyunki yeh kabhi sideways push nahi karti, stone ki "amount of going-around" kabhi nahi badal sakti. Hum ne us go-around amount ko ek naam diya, $h$, aur ise "pathar kahan hai" aur "kahan ja raha hai" se bane chhote parallelogram ke area ke roop mein banaya. Phir humne poocha: path kis shape mein banta hai? Pathar ko time mein track karne ke bajaaye, humne uska distance dekha jaise angle ghuma, aur humne cleverly distance ki jagah $1/\text{distance}$ track kiya. Isse ugly curve ek smooth wave mein flip ho gayi — ek wave jiska flat baseline pull $GM$ divided by $h^2$ hai, aur jis ka ripple size eccentricity hai. Wave ko ulta karne par orbit equation milti hai, aur uska numerator — orbit ki width jo seedhe magnet se measure ki gayi — $p$ hai. Numerator likhne ke do tarike line up karo, aur poori kahani ek breath mein nikal aati hai: $h=\sqrt{GMp}$. Zyada peek, zyada strong pull $\Rightarrow$ zyada spin.

---

Connections

• Two-body problem — Step 7 mein standard conic form provide karta hai jis se hum match karte hain.
• Kepler's Second Law — Step 3/4: constant $h$ exactly equal-areas-in-equal-times ke barabar hai.
• Flight-path angle — angle $\varphi$ jo Step 1 mein introduce kiya, $h=rv\sin \varphi$ mein use hota hai.
• Eccentricity vector — ripple amplitude $e$ jo Step 6 mein janam leti hai.
• Vis-viva equation — speed ko radius ke saath pair karta hai; orbit fully fix karne ke liye $h$ complement karta hai.
• Specific orbital energy — doosra conserved number; $h$ ke saath milkar $a$ aur $e$ determine karta hai.