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

3.2.10 · D3 · Physics › Orbital Mechanics & Astrodynamics › Vis-viva equation v² = GM(2 - r − 1 - a) — derivation

Shuru karne se pehle, teen simple-language reminders taaki koi bhi symbol anjana na lage:

Hum hamesha km aur seconds mein kaam karenge, isliye $GM$ ko $km^{3} / s^{2}$ mein rakhne se har answer km/s mein aata hai. Metres aur km mix karna #1 silent error hai — hum isse ek baar km choose karke aur kabhi switch na karke bachate hain.

Scenario matrix

Har vis-viva problem in cells mein se kisi ek mein hoti hai. Niche ke examples tagged hain us cell se jo woh cover karte hain.

Cell	Orbit type	Sign / value of $a$	Kya special hai	Example
C1	Circular	$a = r$ (finite $> 0$ )	$\frac{2}{r} - \frac{1}{a} = \frac{1}{r}$ ; speed constant	Ex 1
C2	Elliptical, perigee par	$a > 0$ , $r = r_{p} < a$	sabse fast point	Ex 2
C3	Elliptical, apogee par	$a > 0$ , $r = r_{a} > a$	sabse slow point	Ex 2
C4	Elliptical, general point	$a > 0$ , $r_{p} < r < r_{a}$	beech ki speed	Ex 3
C5	Parabolic (escape boundary)	$a \to \infty$ , $\frac{1}{a} \to 0$	$ε = 0$ exactly	Ex 4
C6	Hyperbolic (unbound)	$a < 0$	$v^{2} > 2 GM / r$ , excess speed	Ex 5
C7	Ulta solve karein $a$ ke liye	given $v, r$ → orbit size dhundho	orbit classify karo	Ex 6
C8	Real-world word problem	mission $Δ v$	Hohmann transfer	Ex 7
C9	Exam twist / degenerate	$r \to r$ at $a \to r$ limit, sign trap	galti pakdo	Ex 8

Worked Examples

Worked example Ex 1 — Circular orbit (Cell C1)

Ek GPS satellite Earth ke around $r = 26560$ km radius par circle kar rahi hai. Woh kitni fast ja rahi hai?

Forecast: Pehle guess karo — kya yeh low satellite ki $\sim 7.7$ km/s speed se fast hogi ya slow? (Zyada upar orbit → zyada slow.)

Step 1. Pehchano ki circular hai ⇒ $a = r$ . Yeh step kyun? Ek circle ek aisi orbit hai jis ki "sabse lambi width" uska diameter hi hai, toh uska aadha, $a$ , radius $r$ ke barabar hota hai. Yeh sirf yahi case hai jahan $r$ aur $a$ same number hote hain.

Step 2. $a = r$ ko vis-viva mein substitute karo: $v^{2} = GM (\frac{2}{r} - \frac{1}{r}) = \frac{GM}{r} .$ Yeh step kyun? Dono fractions collapse ho jaate hain kyunki unka same denominator hai — yeh "circular matlab constant speed" ka algebraic fingerprint hai.

Step 3. Numbers plug karo: $v = \frac{398600}{26560} = 15.008 \approx 3.874 km/s .$

Verify: Units: $km^{3} / s^{2} \div km = km^{2} / s^{2} = km/s$ ✓. Aur $3.87 < 7.7$ km/s — LEO se slow, exactly jaisa ek badi orbit mein hona chahiye. ✓

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

Step 1. Apsides se $a$ nikalo. Yeh step kyun? Vis-viva ko hamesha $a$ chahiye, aur dono apsides long axis ke opposite ends par hote hain, toh unki distances ka sum poori length $2 a$ hota hai: $a = \frac{r _{p} + r _{a}}{2} = \frac{7000 + 42000}{2} = 24500 km .$

Step 2. Perigee par speed, $r = r_{p} = 7000$ km: $v_{p}^{2} = 398600 (\frac{2}{7000} - \frac{1}{24500}) = 398600 (2.8571 \times 1 0^{- 4} - 4.0816 \times 1 0^{- 5}) .$ Yeh step kyun? Hum sirf $r = r_{p}$ aur fixed $a$ formula mein daalte hain — kuch aur nahin badalta. $v_{p}^{2} = 398600 \times 2.4490 \times 1 0^{- 4} = 97.62 \Rightarrow v_{p} \approx 9.88 km/s .$

Step 3. Apogee par speed, $r = r_{a} = 42000$ km, same $a$ : $v_{a}^{2} = 398600 (\frac{2}{42000} - \frac{1}{24500}) = 398600 (4.7619 \times 1 0^{- 5} - 4.0816 \times 1 0^{- 5}) .$ $v_{a}^{2} = 398600 \times 6.803 \times 1 0^{- 6} = 2.712 \Rightarrow v_{a} \approx 1.647 km/s .$

Verify: Angular momentum match karna chahiye: $r_{p} v_{p} = 7000 \times 9.88 = 6.92 \times 1 0^{4}$ aur $r_{a} v_{a} = 42000 \times 1.647 = 6.92 \times 1 0^{4}$ . Equal ✓ (yeh Conservation of Angular Momentum action mein hai). Aur $v_{p} > v_{a}$ jaise forecast tha. ✓

Worked example Ex 4 — Parabolic escape,

$a \to \infty$ limit (Cell C5) Ek probe $r = 7000$ km par hai. Isse Earth se just barely escape karne ke liye (infinity tak pahunchne ke liye zero speed ke saath) kitni speed chahiye?

Forecast: Escape speed famously circular speed ki $2$ times hoti hai — toh is radius par circular speed ka lagbhag $1.41 \times$ guess karo.

Step 1. "Just barely escape" matlab orbit parabolic hai, yaani $a \to \infty$ . Yeh step kyun? Ek escaping-but-not-overshooting path ka total specific energy $ε = 0$ hota hai (bound aur free ke beech ka exact boundary). Kyunki $ε = - GM / (2 a)$ , $ε = 0$ set karne se $a \to \infty$ force hota hai, toh $\frac{1}{a} \to 0$ .

Step 2. $\frac{1}{a}$ term drop karo: $v_{esc}^{2} = GM (\frac{2}{r} - 0) = \frac{2 GM}{r} = \frac{2 \times 398600}{7000} = 113.89.$ $v_{esc} \approx 10.67 km/s .$

Verify: Yahan circular speed hai $398600/7000 = 56.94 = 7.546$ km/s, aur $2 \times 7.546 = 10.67$ km/s ✓ — Escape Velocity se exactly match karta hai.

Worked example Ex 5 — Hyperbolic orbit,

$a < 0$ trap (Cell C6) Ek interstellar object Sun ke paas se hyperbolic orbit par guzarta hai jisme $a = - 50 \times 1 0^{6}$ km (negative — yeh unbound hai). $r = 150 \times 1 0^{6}$ km (1 AU) par yeh kitni fast move kar raha hai? Use karo $G M_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ .

Forecast: Hyperbola ke liye $v^{2} > 2 GM / r$ (escape speed aur usse bhi zyada). Toh local escape speed se upar ek speed expect karo.

Step 1. $a$ ko negative rakhho — uska sign mat badlo. Yeh step kyun? Convention " $a < 0$ for hyperbolas" exactly wohi hai jo $- \frac{1}{a}$ ko $+ \frac{1}{∣ a ∣}$ bana deti hai, energy add karti hai taaki speed escape se zyada ho. Agar aapne sign "fix" kiya toh physics destroy ho jaati hai.

Step 2. $\frac{1}{a} = \frac{1}{- 50 \times 1 0 ^{6}} = - 2.0 \times 1 0^{- 8}$ ke saath substitute karo: $v^{2} = 1.327 \times 1 0^{11} (\frac{2}{150 \times 1 0 ^{6}} - (- 2.0 \times 1 0^{- 8})) .$ $= 1.327 \times 1 0^{11} (1.3333 \times 1 0^{- 8} + 2.0 \times 1 0^{- 8}) = 1.327 \times 1 0^{11} \times 3.3333 \times 1 0^{- 8} .$ $v^{2} = 4423.3 \Rightarrow v \approx 66.51 km/s .$

Verify: Local escape speed hai $2 GM / r = 2 \times 1.327 \times 1 0^{11} / (150 \times 1 0^{6}) = 1769.3 = 42.06$ km/s. Hamara $66.5 > 42.06$ ✓ — genuinely hyperbolic, jaise forecast tha.

Worked example Ex 6 —

$a$ ke liye ulta solve karo (Cell C7) Ek spacecraft $r = 8000$ km par hai aur $v = 9.0$ km/s move karta hua measure kiya gaya. Semi-major axis $a$ kya hai, aur kya orbit bound hai ya unbound?

Forecast: $9.0$ km/s — kya yeh 8000 km par escape se upar hai ya niche? Escape hai $2 \times 398600/8000 = 9.98$ km/s. Hum usse niche hain, toh orbit bound hona chahiye (positive $a$ ).

Step 1. Vis-viva ko $\frac{1}{a}$ ke liye rearrange karo: $\frac{1}{a} = \frac{2}{r} - \frac{v ^{2}}{GM} .$ Yeh step kyun? Humein $v$ aur $r$ pata hai; sirf unknown $a$ hai, toh ise isolate karo. Yeh "orbit diagnose karo" wala move hai.

Step 2. Plug in karo: $\frac{1}{a} = \frac{2}{8000} - \frac{9. 0 ^{2}}{398600} = 2.5000 \times 1 0^{- 4} - 2.0321 \times 1 0^{- 4} = 4.679 \times 1 0^{- 5} .$ $a = \frac{1}{4.679 \times 1 0 ^{- 5}} \approx 21372 km .$

Step 3. Classify karo: $a > 0$ aur finite ⇒ bound ellipse.

Verify: Is $a$ se $v$ recompute karo: $v^{2} = 398600 (2/8000 - 1/21372) = 398600 (2.5 \times 1 0^{- 4} - 4.679 \times 1 0^{- 5}) = 398600 \times 2.032 \times 1 0^{- 4} = 81.00$ , toh $v = 9.00$ km/s ✓ round-trips. Aur $a > 0$ bound forecast se match karta hai. Dekho Orbital Elements.

Worked example Ex 7 — Real-world word problem: Hohmann

$Δ v$ (Cell C8) Ek satellite $r_{1} = 7000$ km par circular orbit mein hai. Hum ise Hohmann Transfer Orbit use karke $r_{2} = 42000$ km par circular orbit mein raise karna chahte hain. Pehla burn $Δ v_{1}$ dhundho — low circle chhodne aur transfer ellipse mein enter karne ke liye needed speed boost.

Forecast: Transfer ellipse ka perigee low circle par baithta hai, aur uski perigee speed (Ex 2 se, $9.88$ km/s) low circular speed se fast hai. Toh $Δ v_{1}$ ek modest positive number hai.

Step 1. $r_{1} = 7000$ par low circular speed (Cell C1 machinery): $v_{c 1} = \frac{398600}{7000} = 56.943 = 7.546 km/s .$ Yeh step kyun? Burn se pehle satellite circular speed par move kar rahi hai; yeh hamara starting point hai.

Step 2. Transfer-ellipse perigee speed. Transfer ellipse ka $r_{p} = 7000$ , $r_{a} = 42000$ ⇒ $a = 24500$ km hai — Ex 2 ke same. Toh uski perigee speed $v_{p} = 9.883$ km/s hai (Ex 2 se). Yeh step kyun? Burn transfer ellipse ke perigee par hota hai, jo low circle ke saath coincide karta hai, toh hume us ellipse ki perigee speed chahiye.

Step 3. Burn sirf difference poori karta hai: $Δ v_{1} = v_{p} - v_{c 1} = 9.883 - 7.546 = 2.337 km/s .$

Verify: $Δ v_{1} > 0$ (upar climb karne ke liye speed-up) ✓, aur itne transfer ke liye yeh realistic few-km/s figure hai ✓. Units km/s ✓.

Worked example Ex 8 — Exam twist: degenerate / sign trap (Cell C9)

"Ek student $r = 30000$ km par ek satellite ki speed compute karta hai jisme orbit ka $a = 24500$ km hai aur paata hai $v^{2} = 398600 (2/30000 - 1/24500)$ . Woh panic karta hai kyunki dono fractions close hain. Phir ek doosra student, jise ek hyperbolic orbit diya gaya hai $a = 24500$ km ke saath (woh sign bhool gaya), same tarah compute karta hai. Kya galat hai, aur pehle (legitimate) case mein correct speed kya hai?"

Forecast: Pehli calculation theek hai (bound ellipse, $r < r_{a} = ?$ ). Doosra classic sign mistake hai. Guess karo: legit answer couple of km/s hai.

Step 1. Pehla student — check karo ki orbit real hai. Yahan $a = 24500$ Ex 2 orbit hai, jiska apogee $r_{a} = 2 a - r_{p} = 2 (24500) - 7000 = 42000$ km hai. Kyunki $7000 \leq 30000 \leq 42000$ , point $r = 30000$ actually orbit par exist karta hai. Achha. Yeh step kyun? Vis-viva silently ek value return karta hai even un $r$ ke liye jo orbit kabhi reach nahin karta; aapko pehle confirm karna chahiye $r_{p} \leq r \leq r_{a}$ , warna "speed" fictitious hai.

Step 2. Compute karo: $v^{2} = 398600 (\frac{2}{30000} - \frac{1}{24500}) = 398600 (6.6667 \times 1 0^{- 5} - 4.0816 \times 1 0^{- 5}) = 398600 \times 2.585 \times 1 0^{- 5} .$ $v^{2} = 10.305 \Rightarrow v \approx 3.210 km/s .$

Step 3. Doosre student ki galti: hyperbola ke liye $a$ negative hona chahiye. $a = + 24500$ use karna ek bound ellipse describe karta hai, woh unbound path nahin jो unka matlab tha — sign cosmetic nahin hai, yeh $- \frac{1}{a}$ ko energy add karne se energy remove karne mein flip kar deta hai. Fix hai $a = - 24500$ , jo bahut bada $v^{2} = 398600 (2/30000 + 1/24500)$ deta.

Verify: Pehla case: $3.21$ km/s perigee $9.88$ aur apogee $1.65$ ke beech hai? Nahi — ruko, $3.21 < 1.65$ false hai; $1.65 < 3.21 < 9.88$ ✓ (yeh beech mein hai, apogee ke kareeb kyunki $30000$ far side ke paas hai). Consistent ✓.

Recall Main kaun se cell mein hoon? (quick reflex)

Ek orbit diya gaya, pehle uska type batao ::: circle→ $a = r$ ; ellipse→ $0 < a$ finite; parabola→ $a \to \infty$ ; hyperbola→ $a < 0$ . $v$ aur $r$ diya gaya, main type kaise dhundhunga ::: $1/ a = 2/ r - v^{2} / GM$ compute karo; $a$ ka sign batata hai (+ bound, ∞ parabolic, − hyperbolic). Ellipse par computed $v$ par trust karne se pehle kya check karna zaroori hai ::: ki $r_{p} \leq r \leq r_{a}$ ho, warna point orbit par nahin hai.

Connections

Vis-viva equation — derivation — woh formula jo har example use karta hai.
Conservation of Energy — isliye $a$ akela har cell mein total energy set karta hai.
Conservation of Angular Momentum — Ex 2 mein perigee/apogee cross-check.
Kepler's Laws — third law $a$ ko period se link karta hai jab $a$ pata ho.
Hohmann Transfer Orbit — Ex 7 uska ek burn hai.
Escape Velocity — Ex 4 ka parabolic Cell C5 limit.
Orbital Elements — $a$ , $e$ , $r_{p}$ , $r_{a}$ definitions jo poori jagah use hoti hain.