Orbital velocity for circular orbit — derivation: Question bank

1.2.24 · D5 · Physics › Newton's Laws & Dynamics › Circular orbit ke liye Orbital velocity — derivation

Poore note mein, symbols wahi mean karte hain jo parent note ne build kiye hain:

$v_{o}$ = circular orbit ke liye constant tangential (sideways) speed.
$r$ = planet ke centre se satellite tak ki distance, yaani $r = R + h$ ( $R$ = planet radius, $h$ = altitude).
$M$ = planet ki mass, $m$ = satellite ki mass, $G$ = gravitational constant.
Master formula: $v_{o} = \frac{GM}{r}$ .

Spot the error

"400 km altitude par orbit karne ke liye,

r = 400

km ko

v_{o} = GM / r

mein plug karo."

Error

r

mein hai: ye Earth ke centre se measure hota hai, isliye

r = R + h = 6400 + 400 = 6800

km.

400

km use karne se bahut galat (bahut zyada badi) speed milti hai.

"Ek heavier satellite ko same orbit mein rehne ke liye zyada strong rocket chahiye, isliye use tezi se move karna hoga."

Gravitational pull aur required centripetal force dono exactly

m

ke proportion mein badhte hain, isliye

m

cancel ho jaata hai — kisi bhi mass ke liye zaroori speed identical hai.

v_{o} = GM / r

, isliye badi orbit ko badi speed chahiye kyunki

r

bada hai."

r

denominator mein root ke neeche baitha hai: bada

r

fraction ko chota banata hai, isliye

v_{o}

kam hoti hai. Is reasoning ne formula mein

r

ki jagah ko ignore kar diya.

"Gravity satellite ko uske path ke along aage push karti hai taaki use moving rakh sake."

Gravity inward point karti hai (planet ke centre ki taraf), motion ke perpendicular — ye path ko bend karti hai lekin kabhi tez nahi karti. Vacuum mein koi forward push nahi hota aur kisi ki zaroorat bhi nahi.

"Kyunki

g

altitude ke saath decrease karta hai, kisi bhi orbit ke liye surface value

g = 9.8

use karo."

Aap surface

g

tabhi use kar sakte hain jab identity

GM = g R^{2}

aur sahi

r = R + h

use karein;

v_{o} = g R

oochi altitude par likhna double-count karta hai kyunki ye ignore karta hai ki

r

kaise badla.

"Net inward force paane ke liye hum gravity aur centripetal force ko add karte hain."

Koi alag "centripetal force" add karne ke liye hai hi nahi — gravity yahan centripetal force hai. Derivation

F_{g}

ko required

m v_{o}^{2} / r

ke barabar set karti hai; ye ek hi single force hai jo do tarike se dekhi jaati hai.

"Escape velocity

2 v_{o}

hai kyunki aapko gravity ko poori tarah overcome karna padta hai."

2 v_{o}

hai,

2 v_{o}

nahi. Factor energy condition

\frac{1}{2} v_{esc}^{2} = GM / r

se aata hai, jo

2

deta hai,

2

nahi.

Why questions

Satellite ki apni mass final formula se kyun gayab ho jaati hai?

Kyunki

\frac{GM m}{r ^{2}} = \frac{m v _{o}^{2}}{r}

ke dono sides par

m

ka factor hai; use cancel karne se pata chalta hai ki orbit speed sirf

M

aur

r

par depend karti hai, bilkul waise jaise sabhi masses ek hi rate se girti hain.

Gravity exactly

m v_{o}^{2} / r

ke barabar kyun honi chahiye, zyada ya kam nahi?

Agar gravity zyada hoti toh path andar ki taraf spiral karta; agar kam hoti, toh satellite bahar drift karta. Sirf exact equality radius ko constant rakhti hai — circular orbit ki defining condition yahi hai.

Hum

r

surface se nahi, centre se kyun measure karte hain?

Newton's law

GM m / r^{2}

ek spherical planet ko aisa treat karta hai jaise uski saari mass centre mein ho, isliye relevant distance centre-to-satellite hai, jo

R + h

hai.

Low satellite itna fast (≈ 7.9 km/s) kyun hai jabki Moon itna slow (≈ 1 km/s)?

v_{o} \propto 1/ r

: Moon roughly 60 Earth-radii door hai, isliye

60 \approx 7.7

times slower — badi doori pull ko kamzor kar deti hai aur pace ko dheel kar deti hai.

Derivation specifically

a_{c} = v_{o}^{2} / r

kyun use karti hai?

Kyunki constant speed par turning hone par bhi acceleration hota hai, aur geometry dikhati hai ki radius

r

ke circle par speed

v_{o}

se chalne waali object exactly isi rate se inward curve karti hai — ye woh force hai jo orbit "demand" karti hai.

Substitution

GM = g R^{2}

practically kyun helpful hai?

Hume aksar

G

aur

M

alag alag pata nahi hote, lekin hum hamesha kisi planet ki surface gravity

g

aur radius

R

jaante hain; ye identity humein measurable quantities se

v_{o}

compute karne deti hai.

Edge cases

v_{o} = GM / r

kya predict karega jab

r \to \infty

Speed zero ki taraf jaati hai — infinitely door, gravity negligibly weak hai, isliye orbit karne waali body ko barely move karne ki zaroorat hogi. (Practically isko infinite time bhi lagta hai.)

Formula

r \to 0

par (orbit bilkul centre mein) kya "chahta" hai?

Ye infinity tak blow up ho jaata hai, jo ek signal hai ki model break down kar raha hai: aap planet ke andar orbit nahi kar sakte, aur

1/ r^{2}

point-mass law wahan valid nahi hai.

Kya surface par hi circular orbit (

h = 0

r = R

) physically real hai?

Mathematically

v_{o} = g R \approx 7.9

km/s, lekin physically air resistance aur mountains ise impossible banate hain — ye idealized limiting case hai, achievable orbit nahi.

Kya formula elliptical orbit par apply hota hai?

Nahi.

v_{o} = GM / r

constant speed aur constant radius assume karta hai; ellipse mein dono change hote hain (planet ke paas fast, door slow), isliye ye single-speed result wahan nahi chalta.

Agar radius

r

par satellite ki speed

v_{o}

se thodi kam hai, toh kya hoga?

Gravity ab centripetal requirement se zyada hai, isliye path andar ki taraf curve karta hai — orbit ek ellipse ban jaati hai jo planet ke aur paas jaati hai, stable circle nahi rehti.

Agar us radius par speed thodi zyada hai

v_{o}

se?

Gravity itni tez motion ko same circle mein bend karne ke liye kaafi weak hai, isliye satellite ek bade elliptical orbit mein bahar swing karta hai (aur agar kaafi fast ho, toh escape kar jaata hai).

Connections

Newton's Law of Universal Gravitation
Centripetal Force and Uniform Circular Motion
Escape Velocity
Kepler's Third Law
Acceleration due to gravity g and GM = gR²
Geostationary Orbit

Question bank — Orbital velocity for circular orbit — derivation

True or false — justify