Circular orbit — velocity, period, energy: Foundations

3.2.13 · D1 · Physics › Orbital Mechanics & Astrodynamics › Circular orbit — velocity, period, energy

Ye page har letter aur squiggle build karti hai jo parent note use karta hai, bilkul scratch se. Agar koi symbol parent topic mein bina explanation ke aaya ho, toh usse yahan explain kiya gaya hai — ek picture ke saath.

Characters ki cast (build order mein)

Hum symbols ek ek karke introduce karte hain. Har naya symbol sirf unhi symbols par lean kar sakta hai jo pehle aa chuke hain.

1. $r$ — radius (center se distance)

Picture karo. Neeche ke figure mein bada body middle mein baitha hai aur satellite ek circle par chal rahi hai. Center se satellite tak ka seedha spoke $r$ hai.

Figure — Circular orbit — velocity, period, energy

Orbit map: center mein bada central body, chhota orbiting body, radius-spoke $r$ , aur circle ke saath-saath speed arrow. (Masses aur speed ke naam hum agle do sections mein denge.)

Topic ko iska kyu zaroorat hai. Gravity distance ke saath kamzor hoti hai, aur bada circle ek lambi trip hai. Dono effects is ek number $r$ se control hote hain, isliye ye har formula mein appear karta hai.

2. $m$ aur $M$ — do masses

Picture karo. Upar ke orbit map par wapas jaao: center mein badi disk ab $M$ (Earth, Sun, ...) naam se jaani jaati hai aur ghoomne wala chhota dot $m$ hai (satellite, planet, moon). Hum assume karte hain $M ≫ m$ (padho: "bahut, bahut bada"), isliye bada body muskil se hilta hai.

Topic ko iska kyu zaroorat hai. Gravity ki strength dono masses ke product par depend karti hai. Is chapter ka ek famous punchline ye hai ki $m$ eventually speed se cancel out ho jaata hai — ek pankh aur ek bus ek jaisi orbit karte hain — lekin aap ise cancel hote tabhi dekh sakte hain jab pehle jaante hon ki ye kya hai.

3. $v$ — orbital speed

Picture karo. Orbit map mein circle ke tangent par orange arrow $v$ hai. Ye hamesha travel ki direction mein point karta hai — spoke $r$ ke relative sideways, kabhi center ki taraf ya center se door nahi.

Topic ko iska kyu zaroorat hai. $v$ headline answer hai: "yahan orbit karne ke liye mujhe kitna fast jaana chahiye?" Baaki sab (period, energy) $v$ se build hote hain.

4. Gravity ka pull — $F_{g} = \frac{GM m}{r ^{2}}$

Ab hum upar ke symbols se apna pehla formula assemble karte hain, saath mein ek naya constant.

Isse piece by piece padho.

Upar $GM m$ : kisi bhi body ki zyada mass → zyada strong pull. $G$ bas ise scale karta hai.
Neeche $r^{2}$ : distance double karo aur pull quarter ho jaata hai (kyunki $2^{2} = 4$ ). Ye inverse-square law hai.

Inverse-square law: jaise jaise satellite bahar jaata hai gravity arrow $1/ r^{2}$ ki tarah shrink hota hai — double distance par ye sirf quarter strong hoti hai.

$r^{2}$ kyun aur $r$ nahi? Socho gravity center se paint ki tarah baahar failis jaati hai, hamesha bade-bade spheres par. Jo paint chhote sphere par thi ushe ab aisa sphere coat karna hai jiska area $r^{2}$ ki tarah badh raha hai, isliye ye $1/ r^{2}$ ki tarah patli ho jaati hai.

Topic ko iska kyu zaroorat hai. Ye orbit ka supply side hai: satellite ko circle par rokne ke liye available actual force. Dekho Newton's Law of Universal Gravitation.

5. Centripetal requirement — $F_{c} = \frac{m v ^{2}}{r}$

Ye koi nayi force nahi hai — ye ek demand hai. Ye kehta hai: "agar tum $r$ radius par $v$ speed se orbit chahte ho, toh itni inward force kahi se provide karni hogi."

Centripetal demand: har point par orange velocity arrow turn kar raha hai, aur turning ke liye center ki taraf plum arrow of inward force $m v^{2} / r$ chahiye.

Formula padho.

Zyada fast ( $v$ bada) → arrow zyada hard swing karta hai → zyada force chahiye. Aur $v$ squared hai, isliye speed double karne se demand quadruple ho jaati hai.
Tighter circle ( $r$ chhota) → sharper turn → zyada force. Isliye $r$ neeche hai.

Topic ko iska kyu zaroorat hai. Ye demand side hai. Dekho Centripetal Force and Uniform Circular Motion.

6. Balance — step by step $v$ solve karna

Poora topic ek sentence hai: supply = demand. Gravity hi ek available force hai, isliye use exactly centripetal amount provide karna hoga: $gravity supplies \frac{GM m}{r ^{2}} = circle demands \frac{m v ^{2}}{r}$

Ab algebra step by step dekhte hain — ye derivation hai jis par parent note lean karta hai, slowly ki gayi.

Step 1 — satellite mass $m$ cancel karo. Ye dono sides par ek ek baar appear karta hai, isliye dono sides ko $m$ se divide karo: $\frac{GM}{r ^{2}} = \frac{v ^{2}}{r}$ Isliye answer $m$ ki parwah nahi karega: ye khatam hone se pehle hi chala jaata hai.

Step 2 — neeche ka $r$ clear karo. Dono sides ko $r$ se multiply karo. Left mein $\frac{r}{r ^{2}} = \frac{1}{r}$ ho jaata hai; right mein $r$ poori tarah cancel ho jaata hai: $\frac{GM}{r} = v^{2}$

Step 3 — square undo karo. $v^{2}$ matlab " $v$ times itself"; $v$ wapas paane ke liye dono sides ka square root lete hain (square root hai ye sawaal ki "kaunsa number, khud se multiply hoke, ye deta hai?"). Squaring ek sign lose karta hai, isliye algebraically do answers hain, $+ GM / r$ aur $- GM / r$ . Lekin yahan $v$ ek speed hai — velocity arrow ki length — aur length kabhi negative nahi hoti. Isliye hum sirf positive root rakhte hain aur negative ko discard karte hain: $v = + \frac{GM}{r}$

7. $T$ — period, aur $2 π r$ — circumference

Picture karo. Orbit map ke circle ko $2 π r$ length ki seedhi line mein unroll karo. Satellite poori line $v$ speed se chalti hai, isliye time = distance ÷ speed: $T = \frac{2 π r}{v} .$

Topic ko iska kyu zaroorat hai. Ye ek definition, plus $v = GM / r$ , Kepler's Third Law $T^{2} \propto r^{3}$ deta hai — dekho Kepler's Three Laws.

8. $K$ , $U$ , $E$ — kinetic, potential, aur total energy

$U$ negative kyun hai? Hum choose karte hain $U = 0$ infinitely door (koi pull nahi bacha). Wahan se andar girne se energy release hoti hai, isliye koi bhi bound position zero ke neeche hoti hai — isliye minus sign aur $1/ r$ : andar close hona matlab aur negative.

$E$ explicitly build karo. Ek orbiting satellite ke liye humne pehle hi paya $v^{2} = \frac{GM}{r}$ , isliye kinetic energy ban jaati hai: $K = \frac{1}{2} m v^{2} = \frac{1}{2} m \cdot \frac{GM}{r} = \frac{GM m}{2 r} .$ Ab potential energy $U = - \frac{GM m}{r}$ add karo: $E = K + U = \frac{GM m}{2 r} - \frac{GM m}{r} = \frac{GM m}{2 r} - \frac{2 GM m}{2 r} = - \frac{GM m}{2 r} .$

Energy ladder: deep teal level $U = - GM m / r$ hai, orange arrow kinetic energy $K = GM m /2 r$ hai jo tumhe bottom se utha raha hai, aur unka sum $E = - GM m /2 r$ (plum) halfway upar baitha hai — phir bhi $E = 0$ escape line ke neeche.

Equipment checklist

Khud test karo — right side cover karo. Agar koi bhi answer fuzzy lage, upar woh section dobara padho.

r

kya measure karta hai, aur kahan se?

Center of the big body se satellite tak ki distance — surface ke upar ki altitude nahi.

r

kyun central body ke physical radius se zyada hona chahiye?

Warna "orbit" underground hogi; ek real orbit sirf central body ke bahar exist karti hai, jahan ye formulas hold karte hain.

m

aur

M

kya hain?

m

= chhota orbiting mass,

M

= bada central mass (kg mein),

M ≫ m

ke saath.

Circle mein speed aur velocity mein kya difference hai?

Speed (arrow ki length) constant hai; velocity (arrow + direction) badal jaati hai kyunki direction ghoomti rehti hai.

G

kya hai, uske units kya hain, aur kya aap kabhi iske liye solve karte hain?

Gravitational constant

6.674 \times 1 0^{- 11} m^{3} kg^{- 1} s^{- 2}

; nature ka ek fixed number jise aap look up karte hain, kabhi solve nahi karte.

F_{g} = GM m / r^{2}

mein

r

kyun squared hai?

Gravity

r^{2}

ki tarah badhte spheres ki area par faili hoti hai, isliye ye distance ke inverse square ki tarah patli hoti hai.

Kya centripetal force ek nayi tarah ki force hai?

Nahi — ye ek requirement (

m v^{2} / r

) hai jo kisi real force (yahan, gravity) ko provide karna hoga.

v^{2} = GM / r

solve karte waqt sirf positive root kyun rakhte hain?

v

ek speed (ek length) hai, jo kabhi negative nahi ho sakti, isliye negative root discard kar diya jaata hai.

Kaunsa mass cancel hota hai

v

solve karte waqt, aur kis step mein?

Satellite mass

m

, Step 1 mein,

GM m / r^{2} = m v^{2} / r

ke dono sides ko

m

se divide karke.

r

radius ke circle ki circumference kya hoti hai?

2 π r

, jahan

π \approx 3.14159

Speed se period kaise milta hai?

T = \frac{2 π r}{v}

= (ek lap ki distance) ÷ (speed).

E = - GM m / (2 r)

K

aur

U

se kaise aata hai?

K = GM m /2 r

aur

U = - GM m / r

add karo:

\frac{1}{2 r} - \frac{1}{r}

(times

GM m

)

= - \frac{1}{2 r}

U

negative kyun hai?

Humne

U = 0

infinity par set kiya; andar girne se energy release hoti hai, isliye bound positions zero ke neeche hoti hain.

Total energy

E

ka sign tumhe kya batata hai?

E < 0

bound (trapped),

E = 0

just escape karta hai,

E > 0

free hokar chala jaata hai.

Connections

Newton's Law of Universal Gravitation — $F_{g} = GM m / r^{2}$ kahan se aata hai.
Centripetal Force and Uniform Circular Motion — $m v^{2} / r$ kahan se aata hai.
Kepler's Three Laws — in symbols se build hua $T^{2} \propto r^{3}$ .
Gravitational Potential Energy — $U = - GM m / r$ ki origin.
Escape Velocity — $E = 0$ boundary.
Elliptical Orbits and the Vis-viva Equation — speed aur energy ko generalize karta hai.
Virial Theorem — explain karta hai $U = 2 E = - 2 K$ .