Visual walkthrough — Orbital velocity for circular orbit — derivation

1.2.24 · D2 · Physics › Newton's Laws & Dynamics › Orbital velocity for circular orbit — derivation

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Step 1 — Kyun ek tez sideways throw ek orbit ban jaati hai

KYA. Socho tum ek bahut unche pahad par khade ho aur ball ko seedha sideways phenko. Gravity use neeche kheenchti hai; wo land ho jaati hai. Zyada zor se phenko — aur door land hogi. Bepanah zor se phenko, aur kuch ajeeb hota hai.

KYUN. Kisi bhi algebra se pehle, hume physical picture chahiye: ek orbit koi rahasya nahi hai — ye ek aisa projectile hai jo itni tez phenka gaya ho ki zameen neeche se curve karti jaaye bilkul utni hi tez jitna wo girta hai. Ye poora idea hai, aur baad ke har symbol ko bas yahi picture precise banati hai.

PICTURE. Figure dekho. Teen kaale arcs gentle, medium, aur zor se phenke gaye throws hain — wo sab neeche curve hote hain aur zameen se takraate hain. Red curve woh special throw hai: wo zameen se hamesha miss karti rehti hai aur ek circle mein close ho jaati hai.

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Step 2 — Circle mein ghoomna hi ek acceleration hai

KYA. Bhale hi special red ball constant speed se move kare, uski direction badlati rehti hai. Velocity arrow mein badlaav — chahe sirf direction ka ho — ek acceleration hai.

KYUN. Hum yahan acceleration introduce karte hain kyunki Newton ka 2nd law ($F=ma$) tab hi kaam karta hai jab koi acceleration ho. Toh pehle hume ye prove karna hoga ki orbiting ball accelerate kar rahi hai, aur kis taraf.

PICTURE. Figure mein do kareeb waqt par velocity arrow $\vec{v}$ dikhaya gaya hai (kaala). Doosre arrow ko pehle par slide karo: chota red arrow $\Delta \vec{v}$ jo unke tips ko jodta hai seedha centre ki taraf point karta hai. Woh inward direction hi acceleration ki direction hai.

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Step 3 — Koi cheez us inward force ko supply karni chahiye

KYA. Newton ka 2nd law kehta hai: mass $m$ ko $a_c$ se andar accelerate karne ke liye, ek real inward force honi chahiye jiska size $F_c=m{a}_c$ ho.

KYUN. Acceleration kabhi apne aap nahi hota — ek force use cause karti hai. Toh hum us force ko naam dete hain jo circle maangta hai. Abhi hum nahi jaante kaun provide karta hai; hum sirf itna jaante hain ki uska size kitna hona chahiye.

PICTURE. Ball par red arrow andar ki taraf point kar raha hai, $F_c$ label kiya hua. Ye woh "bill" hai jo circle humhe bhejta hai: "mujhe itni inward force do warna main circle rehna band kar dunga."

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Step 4 — Earth gravity se bill chukati hai

KYA. Ab hum poochte hain: kaunsi real force available hai? Vacuum mein satellite par sirf ek hi force hoti hai (na engine, na hawa) — Earth ki gravity, jo use andar ki taraf kheenchti hai.

KYUN. Hume demand (Step 3) ko ek real supply se connect karna hai. Newton ka law of gravitation us inward pull ka exact size deta hai. Dekho Newton's Law of Universal Gravitation.

PICTURE. Figure mein same ball hai lekin ab inward red arrow $F_g$ label kiya hua hai — Earth ke centre se gravitational pull. Dhyaan do ye Step 3 ke $F_c$ arrow ke same direction mein point karta hai. Ye alignment hi poora raaz hai.

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Step 5 — Gravity hi centripetal force hai (key claim)

KYA. Hum do forces add nahi kar rahe. Gravity sirf ek hi force hai, aur ye exactly wahan point karti hai jahan circle ko apni centripetal force chahiye. Toh gravity centripetal force ka role play karti hai: unhe equal set karo.

KYUN. Ye ek equality hi poori derivation ka physical core hai. Har pichla step sirf yeh ek line likhne ko justify karne ke liye tha.

PICTURE. Steps 3 aur 4 ke dono red arrows ek doosre ke upar draw kiye gaye hain — same length, same direction. Jab "bill" $F_c$ aur "payment" $F_g$ perfectly match karte hain, orbit stable hoti hai.

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Step 6 — Cancel karo, aur mass gayab ho jaata hai

KYA. Dono sides mein $m$ (ball ki mass) ka factor hai aur dono mein ek $r$ hai. Unhe cancel karo.

KYUN. Cancel karna sirf safaai nahi hai — $m$ ka gayab hona ek gehri sach le kar aata hai: orbit ko parwah nahi ki satellite kitna bhaari hai.

PICTURE. Figure mein ek feather aur ek truck same red orbit par, same radius par, same speed se move karte hue dikhaye gaye hain. Mass $m$ literally strike through ho kar cancel ho jaata hai.

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Step 7 — $1/\sqrt{r}$ law, drawn kiya hua

KYA. Formula $v_o=\sqrt{GM/r}$ kehta hai: jaise $r$ badhta hai, $v_o$ ghatta hai — lekin dheere, square-root ki tarah, seedha neeche nahi.

KYUN. Numbers abstract hote hain; ek curve yaadgar hota hai. Shape dekhne se tum compute kiye bina predict kar sakte ho: $r$ ko chaar guna karo, $v_o$ aadha ho jaayega.

PICTURE. Red curve chhote $r$ ke paas tezi se girta hai, phir bade $r$ ke liye flat ho jaata hai. Do marked points dikhate hain ki $r$ se $4r$ jaane par speed exactly aadhi ho jaati hai.

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Step 8 — Degenerate / edge cases

Har scenario cover honi chahiye taaki koi surprise na aaye.

KYA & PICTURE. Figure mein $1/\sqrt{r}$ curve par teen cases red markers ke roop mein stacked hain:

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Ek-picture summary

Poori kahaani ek frame mein: circle ek inward force maangta hai ($F_c$), gravity supply karti hai ($F_g$), unhe equal karna aur $m$ aur ek $r$ cancel karne par $v_o=\sqrt{GM/r}$ milta hai — red formula orbit ke centre mein glow karta hua.

Recall Feynman retelling — plain words mein walkthrough

Ek pahad se ball ko sideways phenko. Gravity use hamesha neeche kheenchti hai, toh ye curve hoti hai aur land ho jaati hai. Zyada zor se phenko aur ye aur door land hoti hai. Ise ek sahi bepanah speed par phenko — lagbhag 8 kilometre har second — aur yahan hai trick: gol Earth neeche se utni hi tezi se curve karti hai jitna ye girta hai, toh ye zameen miss karta rehta hai aur hamesha loop karta rehta hai. Ye hai orbit — itna tezi se sideways girna ki tum kabhi land nahi karte.

Ab kyun exactly woh speed? Circle mein jaana matlab direction baar baar badalti rehna, aur badalana ek tarah ka acceleration hai jo centre ki taraf point karta hai — iska size $v^2/r$ hai. Newton kehta hai acceleration ke liye force chahiye, toh circle ek inward force $m{v}^2/r$ maangta hai. Sirf ek hi cheez hamari ball ko andar dhakkelti hai — gravity, jo $GMm/r^2$ se kheenchti hai. Kyunki gravity sirf ek hi force hai aur exactly andar point karti hai, gravity hi woh centripetal force hai — toh hum dono ko equal set karte hain. Jab hum aisa karte hain, ball ki apni mass $m$ dono sides se cancel ho jaati hai (isliye feather aur truck same orbit karte hain), ek $r$ bhi cancel ho jaata hai, aur hum $v_o=\sqrt{GM/r}$ par pahunch jaate hain. Neeche ki orbits tez hoti hain, door ki orbits slow, aur tum kabhi centre orbit nahi kar sakte.

Recall Quick self-test

• Circle mein acceleration kis taraf point karta hai? ⇒ Centre ki taraf (centripetal).
• Satellite ke liye centripetal force kaun supply karta hai? ⇒ Gravity, aur sirf gravity.
• Final formula mein $m$ kahan jaata hai? ⇒ Cancel ho jaata hai — orbit mass-independent hai.
• $400$ km upar satellite ke liye $r$ kya hai? ⇒ $r=R+h=6400+400=6800$ km, Earth ke centre se.
• $4r$ par speed $r$ se compare mein? ⇒ Aadhi, kyunki $v_o\propto 1/\sqrt{r}$.

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Connections

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