Vis-viva equation v² = GM(2 - r − 1 - a) — derivation: Visual walkthrough

3.2.10 · D2 · Physics › Orbital Mechanics & Astrodynamics › Vis-viva equation v² = GM(2 - r − 1 - a) — derivation

Step 1 — Picture: ek chota body, ek bada body

Kisi bhi formula se pehle, setup dekho.

Figure — Vis-viva equation v² = GM(2 - r − 1 - a) — derivation

Bada orange dot central body hai, mass $M$ (socho Sun ya Earth). Hum ise fixed maante hain.
Chota magenta dot orbiting body hai, mass $m$ (ek satellite ya planet), itna chota ki woh bade ko hila nahi sakta.
Purple arrow jis par $r$ likha hai woh $M$ ke centre se $m$ ki taraf point karta hai. Iski length $r$ tumhari abhi ki distance hai.
Magenta arrow jis par $v$ likha hai woh velocity hai — $m$ abhi kis direction mein, aur kitni tez ja raha hai. Iski length $v$ speed hai.

Yahan se kyun shuru karein? Neeche ka har symbol ya to $r$ hai, $v$ hai, ya is picture se bana hua hai. Agar hum ek distance aur ek speed measure kar sakte hain, toh hamare paas sab kuch hai.

Step 2 — Loop ki shape: ek ellipse aur uski size $a$

Orbit generally circle nahi hoti — yeh ek dabba hua circle hota hai jise ellipse kehte hain. Humein iske parts ke names chahiye.

Central body ek focus par baitha hai (orange dot), beech mein nahi. Yeh Kepler's first law hai (dekho Kepler's Laws).
Woh point jahan $m$ sabse paas hota hai use periapsis kehte hain, distance $r_{p}$ (magenta).
Woh point jo sabse door hai use apoapsis kehte hain, distance $r_{a}$ (violet).
Ellipse ki lambi axis (periapsis seedha apoapsis tak) ki length $r_{p} + r_{a}$ hai. Iski aadhi length ==semi-major axis $a$ == hai.

$a$ ko is tarah define kyun karein? Kyunki — jaise hum Step 7 mein prove karenge — orbit ki total energy sirf is ek number par depend karti hai aur kisi cheez par nahi. $a$ woh akela knob hai jo set karta hai ki poori orbit mein kitni "oomph" hai.

Step 3 — Energy balance: kinetic plus potential constant hai

Ab do conservation laws mein se pehli. Dekho Conservation of Energy.

Gravity ek conservative force hai: yeh kabhi secretly energy add ya drain nahi karti. Toh neeche ka sum orbit ke har point par same number hota hai.

Figure mein do colored bars ki height swap hoti hai jab $m$ move karta hai: periapsis ke paas kinetic bar tall hai (tez), potential bar deep hai (well mein deep); apoapsis ke paas yeh swap ho jaate hain. Unka total (dashed line) kabhi nahi badlta. Woh flat dashed line $ε$ hai.

Yeh tool kyun? Energy conservation humein ek equation deta hai jo har jagah ek saath true hai. Yahi humein chahiye ek door wale point ko ek paas wale point se link karne ke liye.

Step 4 — Clever trick: do ends par energy padho

Energy har jagah same hai, toh hum ise wahan evaluate karne ke liye free hain jahan algebra sabse aasaan ho. Do sabse aasaan spots periapsis aur apoapsis hain.

Ends kyun? Periapsis aur apoapsis par velocity arrow exactly $r$ ke perpendicular hota hai — planet momentarily na climb kar raha hota hai na fall, bas sideways sweep kar raha hota hai. Toh speed purely sideways speed hai, koi confusing radial part nahi. Baki har jagah velocity slanted hoti hai aur zyada messy hoti hai.

Kyunki $ε$ dono ends par equal hai:

Ek equation, do unknown speeds. Humein ek aur relation chahiye — woh agla step hai.

Step 5 — Doosra balance: angular momentum speeds ko tie karta hai

Dekho Conservation of Angular Momentum. Angular momentum "ghoomne-ka-pan" measure karta hai: body kitna sideways sweep carry karti hai. Gravity seedha centre ki taraf kheenchti hai, toh yeh spin nahi badal sakti — yeh bhi conserved hai.

Do ends par, jahan $v ⊥ r$ , poori speed sideways speed hai, toh:

Yeh kyun chahiye? Equation (1) mein do unknowns the. Equation (2) $v_{a}$ ko $v_{p}$ se express karta hai. Do equations, ab sirf ek unknown ( $v_{p}$ ). Hum solve kar sakte hain.

Step 6 — Algebra ek clean speed par collapse hoti hai

(2) ko (1) mein substitute karo — $v_{a}$ ki jagah $v_{p} (r_{p} / r_{a})$ rakho — aur dekho yeh kaise simplify hota hai.

$\frac{1}{2} v_{p}^{2} - \frac{μ}{r _{p}} = \frac{1}{2} v_{p}^{2} \frac{r _{p}^{2}}{r _{a}^{2}} - \frac{μ}{r _{a}}$

$v_{p}^{2}$ ke pieces left mein, $μ$ ke pieces right mein gather karo:

$\frac{1}{2} v_{p}^{2} difference of two squares (1 - \frac{r _{p}^{2}}{r _{a}^{2}}) = μ (\frac{1}{r _{p}} - \frac{1}{r _{a}})$

Bracket $1 - r_{p}^{2} / r_{a}^{2}$ factor hota hai $\frac{( r _{a} - r _{p} ) ( r _{a} + r _{p} )}{r _{a}^{2}}$ ke roop mein, aur right side common denominator par $\frac{r _{a} - r _{p}}{r _{p} r _{a}}$ hai. Factor $(r_{a} - r_{p})$ dono sides par appear karta hai aur cancel ho jaata hai:

Figure mein dono $(r_{a} - r_{p})$ brackets dono sides par glowing dikhte hain, ek arrow unka cancel hona dikhata hai — wahi cancellation poori magic hai jo ise collapse karta hai.

Yeh miracle jaisa kyun lagta hai? Difference-of-squares mein right-hand side ke saath ek common factor chupta hua tha. Nature ne aise arrange kiya ki ek messy quadratic ek tidy fraction mein reduce ho jaaye.

Step 7 — Bada reveal: energy SIRF $a$ par depend karti hai

Ab Step 2 ka fact use karo: $r_{a} + r_{p} = 2 a$ . Ise $v_{p}^{2}$ ke result mein substitute karo:

$v_{p}^{2} = \frac{2 μ r _{a}}{r _{p} \cdot 2 a} = \frac{μ r _{a}}{a r _{p}}$

Ise periapsis par energy expression mein wapas rakho aur $r_{a} = 2 a - r_{p}$ use karo:

$ε = \frac{1}{2} v_{p}^{2} - \frac{μ}{r _{p}} = \frac{μ r _{a}}{2 a r _{p}} - \frac{μ}{r _{p}} = \frac{μ}{r _{p}} (\frac{2 a - r _{p}}{2 a} - 1) = \frac{μ}{r _{p}} (- \frac{r _{p}}{2 a})$

$r_{p}$ cancel ho jaata hai aur hum pahuunchte hain:

Figure mein teen alag-alag shape ke ellipses hain jo same $a$ share karte hain; side par ek bar dikhata hai ki woh sab same energy level par hain. Alag eccentricities, identical energy.

Step 8 — Kisi bhi point par vis-viva padho

Energy constant hai, toh kisi bhi point par (distance $r$ , speed $v$ ) balance wahi value ke barabar hoga jo humne abhi nikali:

$\frac{1}{2} v^{2} - \frac{μ}{r} = - \frac{μ}{2 a}$

2 se multiply karo aur rearrange karo:

Step 9 — Saare cases ek picture mein

Yeh akela formula secretly char alag tarah ke paths cover karta hai $a$ ke value ke hisaab se. Humein sab check karne chahiye.

Circle ( $r = a$ ): $v^{2} = μ (\frac{2}{r} - \frac{1}{r}) = \frac{μ}{r}$ . Constant speed — loop kabhi dip ya climb nahi karta. (Magenta path.)
Ellipse ( $a$ finite, $r$ varies): speed periapsis ke paas badhti hai, apoapsis ke paas ghatati hai. (Violet path.)
Parabola / escape ( $a \to \infty$ , toh $\frac{1}{a} \to 0$ ): $v^{2} = \frac{2 μ}{r}$ . Yeh boundary $ε = 0$ hai — bas barely free. Dekho Escape Velocity. (Orange path.)
Hyperbola ( $a < 0$ , toh $- \frac{1}{a} > 0$ ): $v^{2} > \frac{2 μ}{r}$ . Infinity par bhi extra "leftover" speed — ek flyby jo kabhi wapas nahi aata. (Navy dashed path.)

Ek-picture summary

Do conservation laws (energy + angular momentum), orbit ke do aasaan ends par evaluate kiye gaye, ek difference-of-squares cancellation se collapse hokar $ε = - μ /2 a$ bante hain — energy sirf size se set hoti hai — jo vis-viva mein unroll hoti hai.

Recall Feynman: poora walk seedhe alfazon mein

Socho ek ball tumhare around ek stretchy string par oval loop mein ghoom rahi hai. Uske baare mein do cheezein kabhi nahi badlti chahe woh kahaan bhi swing kare: uski total energy (motion-energy plus kitni-upar-ki energy) aur uski ghoomne-ki-ness. Main do extreme moments dekhta hoon — sabse paas aur sabse door — kyunki wahan ball purely sideways chalti hai, toh uski speed clean aur simple hai. "Closest par energy = farthest par energy" aur "closest par spin = farthest par spin" likhna mujhe do facts deta hai. Main unhe ek saath mash karta hoon, terms ka bada dhher cancel ho jaata hai, aur ek surprisingly simple truth nikalti hai: total energy sirf is par depend karti hai ki oval kitna bada hai — uski sabse lambi width ka aadha, jise hum $a$ kehte hain. Uski shape par nahi, ball abhi kahan hai uspar nahi — sirf uski size par. Jab main jaanta hoon ki energy $a$ se fixed hai, main loop par kahaan bhi khada ho sakta hoon, apni current distance $r$ plug in kar sakta hoon, aur turant apni speed jaan sakta hoon. Wahi recipe vis-viva equation hai.

Connections

Conservation of Energy — Step 3 mein flat dashed line.
Conservation of Angular Momentum — Step 5 mein equal-sweep rule.
Kepler's Laws — Sun ek focus par baitha hai (Step 2).
Escape Velocity — Step 9 mein $a \to \infty$ case.
Hohmann Transfer Orbit — orbits change karne ke liye ise do baar apply karta hai.
Orbital Elements — $a$ , $e$ , $r_{p}$ , $r_{a}$ ki formal definitions.

Recall Quick self-check

Hum energy sirf periapsis aur apoapsis par kyun evaluate karte hain? ::: Energy har jagah same hai, aur apsides par $v ⊥ r$ hota hai, toh speed purely sideways hai — algebra wahan sabse clean hoti hai. Kaunsi akeli quantity total orbital energy set karti hai? ::: Semi-major axis $a$ , $ε = - μ /2 a$ ke zariye. Formula mein hyperbola escape se tez kyun hai? ::: Negative semi-major axis, $a < 0$ , jo $- 1/ a > 0$ banata hai aur $v^{2}$ ko $2 μ / r$ se upar push karta hai. Specific angular momentum $h$ define karo. ::: $h = ∣ r \times v ∣$ — distance times velocity ka sideways part; constant kyunki gravity koi twist apply nahi karti.