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

3.2.10 · D5 · Physics › Orbital Mechanics & Astrodynamics › Vis-viva equation v² = GM(2 - r − 1 - a) — derivation

Ye vis-viva derivation ka child hai. Us parent page mein woh key fact prove hoti hai jis par hum yahan rely karte hain: $ε = \frac{1}{2} v^{2} - \frac{GM}{r} = - \frac{GM}{2 a}$ Beech wala piece total specific energy hai; daayaa piece uski constant value hai. Hum ise dobara derive nahi karte — lekin is page ke figures dikhate hain ki kyun yeh sirf $a$ par depend karta hai.

Shuru karne se pehle, symbols ki list (taaki kuch bhi bina explanation ke use na ho):

$r$ = central body ke centre se aapki instantaneous distance (har second badlti rehti hai).
$a$ = semi-major axis, poori orbit ke liye ek fixed size-label.
$v$ = aapki current speed. $GM$ = central body ki gravity ki strength (the "standard gravitational parameter").
$ε = \frac{1}{2} v^{2} - \frac{GM}{r}$ = specific (per-kilogram) total energy — ek hi orbit par har jagah same number.
$e$ = eccentricity (ellipse kitna squash hai), $r_{p} = a (1 - e)$ = sabse close point, $r_{a} = a (1 + e)$ = sabse door point.

True or false — justify

Har line: claim ::: verdict + ek-sentence reason.

Ek circular orbit ke liye

r

aur

a

same number hote hain.

True — circle ka "semi-major axis" bas uska radius hota hai, isliye

r = a

har jagah aur vis-viva collapse hokar

v = GM / r

ban jaati hai.

Specific energy

ε

tab change hoti hai jab satellite perigee se apogee tak move karti hai.

False —

ε = - GM / (2 a)

sirf

a

par depend karta hai, jo ek fixed orbit property hai, isliye yeh har point par same hai; sirf kinetic aur potential ke beech split badlta hai (figure s01 mein flat line).

Same

a

lekin alag eccentricity wale do orbits ki total energy same hoti hai.

True — energy sirf

a

par depend karti hai, shape par nahi, isliye same

a

wala near-circular aur bahut elongated orbit equal

ε

carry karte hain (exactly figure s02 ke teen curves).

Zyada eccentric orbit hamesha higher-energy orbit hoti hai.

False — eccentricity

ε = - GM / (2 a)

mein appear nahi hoti; sirf

a

energy set karta hai, isliye akela eccentricity energy ke baare mein kuch nahi batata.

Kisi point par apni speed double karne se orbital energy double ho jaati hai.

False — energy mein

\frac{1}{2} v^{2}

kinetic term hai, isliye

v

double karne se kinetic part quadruple ho jaata hai, aur potential term unchanged rehta hai, isliye energy simply double nahi hoti.

Elliptical orbit par satellite perigee par sabse fast move karti hai.

True — perigee mein

r

sabse chhota hota hai, isliye

2/ r

sabse bada hota hai, jisse

v^{2}

wahan sabse bada hota hai.

Bound orbit ke liye specific energy

ε

negative hoti hai.

True — bound matlab aap infinity tak nahi pahunch sakte, isliye total energy zero-energy escape threshold se neeche rehti hai;

ε = - GM / (2 a) < 0

Kisi body se escape karne ke liye exactly local circular speed se do guni speed chahiye.

False —

2

guni circular speed chahiye, kyunki

v_{esc}^{2} = 2 GM / r

jabki

v_{c i r c}^{2} = GM / r

Parabolic escape trajectory ki

ε = 0

hoti hai.

True — yeh barely infinity tak pahunchti hai zero leftover speed ke saath, isliye kinetic plus potential energy exactly boundary value zero par sum hoti hai.

Hyperbolic orbit ke liye semi-major axis

a

negative hota hai.

True — convention ke mutabik unbound orbits ke liye

a < 0

hota hai, jisse

- 1/ a > 0

ho jaata hai taaki

v^{2} > 2 GM / r

, yaani speed escape se upar.

Spot the error

Har line ek flawed reasoning batati hai; answer mein bataao kya toot raha hai.

"Kyunki

r = a

circle ke liye kaam karta hai, main hamesha

r

ki jagah

a

use kar sakta hoon."

r = a

ki equality har jagah sirf circle ke liye hold karti hai; ellipse par

r

r_{p}

r_{a}

tak sweep karta hai jabki

a

fixed rehta hai, isliye yeh sirf exactly do points par

a

ke barabar hota hai (jahan orbit radius

a

cross karti hai) aur baaki jagah alag hota hai — ek ko doosre ki jagah substitute karna generally galat hai.

"Apogee par orbit slow hai, isliye wahan ki energy perigee se kam hai."

Energy conserved hai aur dono apsides par identical hai; apogee par kinetic energy kam hai lekin potential energy correspondingly high hai, isliye sum unchanged rehta hai.

a

ek length hai, aur lengths positive hoti hain, isliye

a > 0

hamesha."

Hyperbolic (unbound) orbits ke liye

a

negative define hota hai; ise positive treat karna galat se predict karega

v^{2} < 2 GM / r

ek escaping body ke liye.

"Main

ε = \frac{1}{2} v^{2} - GM / r

use karunga jisme full potential energy

- GM m / r

hai."

Yeh inconsistent hai —

ε

specific (per unit mass) hai, isliye uska potential term

- GM / r

hai (koi

m

nahi); per-mass aur total quantities mix karne se units

m

ke factor se off ho jaate hain.

"Derivation sirf energy conservation use karti hai."

Yeh do conservation laws use karti hai — energy aur angular momentum (

h = r_{p} v_{p} = r_{a} v_{a}

); sirf energy se aap dono apsis speeds separately solve nahi kar sakte.

"Escape speed is par depend karti hai ki aap kis direction mein launch karte ho."

Energy threshold ke liye sirf speed matter karti hai, direction nahi, kyunki kinetic energy

\frac{1}{2} v^{2}

direction ignore karti hai;

v_{esc}

kisi bhi launch angle ke liye same hai (atmosphere/ground ignore karte hue).

"Angular momentum

h

bhi final vis-viva formula mein appear karta hai, isliye mujhe ise use karne ke liye

e

chahiye."

Nahi — poora point yahi hai ki

h

(aur isliye

e

) energy step mein cancel ho jaata hai, leaving

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

sirf

r

aur

a

ke saath; vis-viva apply karne ke liye aapko kabhi

e

ki zaroorat nahi.

"Apogee par velocity mein radial part hota hai, isliye

h = r_{a} v_{a}

galat hai."

Apsides par radial velocity exactly zero hoti hai (distance momentarily ek turning point par hoti hai), isliye

v ⊥ r

aur

h = r_{a} v_{a}

cleanly hold karta hai.

Why questions

Orbital energy sirf

a

par kyun depend karti hai, eccentricity par kyun nahi?

Derivation mein apogee speed angular momentum se fix hoti hai, aur jab aap ise energy expression mein substitute karte ho to eccentricity-carrying factor

(r_{a} - r_{p})

upar-neeche cancel ho jaata hai (difference-of-squares cancellation), leaving pure

- GM / (2 a)

— visually, figure s02 dikhata hai ki flat energy line wahi rehti hai jab

e

badlta hai.

Gravitational potential energy negative kyun li jaati hai?

Ise infinity par zero se measure kiya jaata hai; kyunki aapko infinity tak climb karne ke liye energy add karni padti hai, ek bound body ek energy "well" mein zero se neeche baithti hai, isliye negative (figure s01 mein bowl).

Derivation energy specifically perigee aur apogee par kyun evaluate karti hai?

Un do points par velocity purely tangential hoti hai, awkward radial component khatam ho jaata hai, isliye

v

cleanly appear karta hai aur angular momentum simply

r v

padhta hai.

a \to \infty

set karne se escape condition kyun milti hai?

Infinite semi-major axis

- 1/ a \to 0

banata hai isliye

ε \to 0

, exactly bound (negative energy) aur free (positive energy) motion ke beech boundary — well ka rim.

Escape speed circular speed se

2

guna kyun hai, koi aur factor kyun nahi?

Circular energy

ε = - GM / (2 r)

hai jabki escape ke liye

ε = 0

chahiye; kinetic energy ko

GM / (2 r)

GM / r

tak raise karna

v^{2}

ko double karta hai, aur

2

speed ki doubling hai.

Vis-viva ko "conservation of energy in disguise" kyun kaha ja sakta hai?

Yeh literally

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

rearranged hai; left side total specific energy hai aur right side uski constant value

ε

hai.

Slow apogee pass aur fast perigee pass same energy kyun carry karte hain?

Satellite ke climb karne par (rising

r

) jo kinetic energy lose hoti hai woh exactly potential energy ke roop mein gain ho jaati hai, conserved sum constant rakhte hue.

Edge cases

Ellipse par kin points par

r

actually

a

ke barabar hota hai?

Exactly do points par — jahan orbit radius

a

ke circle ko cross karti hai (semi-minor axis ke ends); wahan

v^{2} = GM (2/ a - 1/ a) = GM / a

, same speed jo radius

a

ki circular orbit ki hogi.

Exact parabolic boundary

a \to \infty

par vis-viva ka kya hota hai?

1/ a

term vanish ho jaata hai, leaving

v^{2} = 2 GM / r

— escape-speed profile, parabola ke har

r

par valid.

Hyperbola ke liye body se door (

r \to \infty

) speed kya hai?

2/ r \to 0

, isliye

v^{2} \to GM (- 1/ a) = - GM / a

; kyunki

a < 0

yeh positive hai, woh leftover "hyperbolic excess speed" jo body infinity par rakhti hai.

Kya vis-viva se

v^{2}

negative aa sakta hai, aur uska kya matlab hoga?

Haan, agar aap orbit ke apogee se bada (ya bound orbit ki reach se bahar)

r

plug karte ho; negative

v^{2}

ek physically forbidden point signal karta hai — body actually us orbit par woh

r

tak kabhi nahi pahunchti.

Degenerate

e = 1

radial (seedha line drop) case kya hai, aur kya vis-viva tab bhi kaam karta hai?

Jab

e \to 1

finite

a

ke saath, ellipse focus se guzarne wali line mein collapse ho jaata hai (

r_{p} \to 0

), aur body seedha andar girke wapas aati hai; energy conservation

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

har

r

par correct speed deta rehta hai, lekin angular momentum zero hai isliye

h = r_{p} v_{p}

degenerate ho jaata hai — orbit shape toot jaata hai, bhalehee vis-viva speed valid rehti hai.

Perigee versus apogee par

ε

ka kaun sa term dominate karta hai?

Perigee par kinetic term

\frac{1}{2} v^{2}

bada hota hai aur potential deeply negative; apogee par kinetic chhota hai aur potential shallower — unka sum

- GM / (2 a)

par fixed rehta hai.

Kya koi aise orbit hai jahan

r

kabhi nahi badlta?

Haan — circle, jahan

r = a

constant hai, isliye

v

bhi constant hai; yeh ek aisa case hai jahan "instantaneous distance" aur "orbit size" permanently coincide karte hain.

Low circular orbit ki specific energy high orbit se compare karne par kaisi hoti hai?

Low orbit ka

a

chhota hota hai, isliye

ε = - GM / (2 a)

zyada negative hai (deeper well); higher orbits kam negative hain, yaani escape ke paas.