Circular orbit — velocity, period, energy: Visual walkthrough

3.2.13 · D2 · Physics › Orbital Mechanics & Astrodynamics › Circular orbit — velocity, period, energy

Step 1 — "Circle mein move karna" secretly kya maangta hai

KYA. Ek ball ko string ke end pe imagine karo, apne haath ke around steady speed se ghumao. Woh ek circle trace karti hai. Ab poochho: kya koi isse kheench raha hai? Haan — string. Agar string kaat do, toh ball seedhi straight line mein ud jaati hai, circle ki tangent ki taraf.

KYUN. Yeh circular motion ki sabse gehri baat hai: circle mein move karne wali cheez lagaatar center ki taraf khichi ja rahi hoti hai. Us inward yank ke bina, woh seedhi jaati (yahi "no force" karta hai — Newton's first law). Toh circular motion "no force" nahi hai; yeh har instant ek specific inward force hai. Us inward direction ko hum ek naam dete hain: centripetal (Latin mein "center-seeking" yani center ki taraf jaana).

PICTURE. Blue ball circle ke saath move kar raha hai. Green arrow uski velocity hai — hamesha tangent, jis taraf woh ja raha hai. Orange arrow woh inward pull hai jo path ko bend karta hai. Pull kato (dashed gray) aur woh seedha bhaag jaata hai.

Step 2 — Ek tiny arc se inward acceleration $v^{2} / r$ banana

KYA. Pehle yeh kehne se pehle ki inward force kitni badi hai, humein yeh dekhna hoga ki inward acceleration kitni badi hai — kyunki turning kuch nahi balki velocity vector ka direction badalna hai, aur "acceleration" ka matlab sirf "velocity kitni tezi se badalti hai" hai. Chalo $v^{2} / r$ apne haathon se do triangles use karke construct karte hain.

Body ko dekho jo thodi si time $Δ t$ mein circle ka ek tiny slice travel karti hai (Greek letter delta, "ek chhoti change" ka matlab). Woh center par ek chota angle $Δ θ$ sweep karti hai.

Position triangle (figure ke left mein). Dono radii (har ek ki length $r$ ) jo start aur end points tak jaati hain, aur unke beech ki chhoti chord milke ek patla triangle banati hain jiska tip-angle $Δ θ$ hai. Ek tiny slice ke liye chord essentially arc length hai, aur arc $=$ radius $\times$ angle, toh chord $\approx r Δ θ$ .
Velocity triangle (figure ke right mein). Velocity arrow ki same length $v$ hai start aur end dono par (steady speed) lekin bilkul usi angle $Δ θ$ se rotate hui hai (kyunki woh tangent rehti hai, aur tangent exactly utna hi ghoom jaata hai jitna radius ghoomta hai). Dono velocity arrows ko tail-to-tail rakh do: woh ek patla triangle banate hain, jiska bhi tip-angle $Δ θ$ hai, jiske dono equal sides ki length $v$ hai. Chhoti teesri side velocity mein change $Δ v$ hai, aur woh inward, center ki taraf point karti hai. Wahi reasoning: $Δ v \approx v Δ θ$ .

YEH DO TRIANGLES KYUN. Woh same shape ke hain — dono isosceles with identical tip-angle $Δ θ$ — toh unke short-side / long-side ratios match karte hain:

$\frac{Δ v}{v} = Δ θ = \frac{chord}{r} = \frac{v Δ t}{r}$

(chord $\approx$ distance travelled $= v Δ t$ use karke). Rearrange karne par,

$acceleration a \frac{Δ v}{Δ t} = \frac{v ^{2}}{r}$

Woh left-hand ratio hi acceleration hai (velocity mein change per unit time), aur woh inward point karta hai kyunki $Δ v$ ne kiya tha. Bilkul tiny slice ki limit mein ( $Δ t \to 0$ ) approximations exact ho jaati hain:

$a_{c} = \frac{v ^{2}}{r} (center ki taraf directed)$

Term by term: $v^{2}$ upar (tezi se jaane wala body apni velocity ko zyada tezi se turn karta hai per second, aur arrows lambe aur tezi se swing karte hain — isliye square); $r$ neeche (wider circle velocity ko har metre travel pe chhote angle se turn karta hai).

PICTURE. Left: position triangle, do radii $r$ aur chord $v Δ t$ , tip-angle $Δ θ$ . Right: velocity triangle, length $v$ ke do arrows, tip-angle $Δ θ$ , chhoti inward side $Δ v$ . Matching angle dono par green mein highlighted hai.

Ab force. Newton kehta hai force $=$ mass $\times$ acceleration. Humara freshly built $a_{c}$ body ke mass $m$ se multiply karo:

$F_{c} = m a_{c} = \frac{m v ^{2}}{r}$

$m$ (kilograms) — heavier body ko equally turn karne ke liye zyada pull chahiye.
$v^{2}$ — speed double karo, required pull chaar guna ho jaati hai.
$r$ neeche — ek wider, gentle circle ko kam pull chahiye.

Step 3 — Bill kaun bharta hai? Gravity, aur sirf gravity

KYA. Space mein koi string nahi hai. Orbiting satellite par sirf ek hi force hai — gravity — bade body (mass $M$ , jaise Earth) ki chhote body (mass $m$ , satellite) par pull. Newton's law of universal gravitation distance $r$ par uski size deta hai:

$F_{g} = \frac{G M m}{r ^{2}}$

Term by term:

$G$ = universal gravitational constant, $6.674 \times 1 0^{- 11}$ (nature ka ek fixed number; yeh har jagah gravity ki strength set karta hai).
$M$ = bade central body ka mass (kg) — bada planet, zyada pull.
$m$ = chhote orbiting body ka mass (kg) — heavier satellite, us par zyada pull.
$r^{2}$ = distance squared, neeche — inverse-square law: do guna door jao aur gravity chauthai ho jaati hai. (Dekho Newton's Law of Universal Gravitation.)

KYUN. Sabse important baat yeh hai ki gravity center ki taraf point karti hai — straight at the big body. Yeh exactly Steps 1–2 ka centripetal direction hai. Toh gravity inward pull ke liye perfectly positioned hai jo circle maangta hai. Yahi ek candidate hai, toh yahi bill bharni chahiye.

PICTURE. Earth center mein; satellite apne circle par. Orange arrow (gravity) satellite se Earth ke center ki taraf point karta hai — wahi direction jaise Step 1 ka "center-seeking" arrow.

Step 4 — Bill ko payment ke barabar set karo

KYA. Circle maangta hai $\frac{m v ^{2}}{r}$ (Step 2). Gravity offer karta hai $\frac{GM m}{r ^{2}}$ (Step 3). Constant radius par steady circular orbit ke liye, yeh exactly barabar hone chahiye — na bache, na kami:

$gravity supply karta hai \frac{GM m}{r ^{2}} = circle maangta hai \frac{m v ^{2}}{r}$

BARABAR KYUN, SIRF "GREATER THAN" KYUN NAHI? Agar gravity demand se zyada hoti, toh surplus pull satellite ko andar kheench leti — radius shrink hoti (woh spiral down karta). Agar gravity kam hoti, toh body bahar swing hoti — radius badhti (woh wider ellipse mein ud jaata). Sirf exact balance $r$ ko constant rakhta hai. Woh akela equation ek circular orbit ki poori physics hai.

PICTURE. Ek balance scale: ek pan mein gravity, doosre mein centripetal requirement, level. Saath mein, do failure modes — bahut zyada pull spiral in karta hai (red), bahut kam fly out karta hai (blue).

Step 5 — Cancel karo aur speed ke liye solve karo

KYA. Ab sirf balance equation par algebra karo. Dono sides mein satellite mass $m$ hai — use divide kar do. Dono sides mein neeche $r$ hai (ek power); ek bachane ke liye multiply through karo:

\;\;\xrightarrow{\;\div\, m\;}\;\; \frac{GM}{r^2}=\frac{v^2}{r} \;\;\xrightarrow{\;\times\, r\;}\;\; \frac{GM}{r}=v^2$$ (Positive) square root lo: $$\boxed{\,v=\sqrt{\dfrac{GM}{r}}\,}$$ Answer mein term by term: - $G,M$ upar — bada/zyada strong central body matlab tezi orbit. - $r$ neeche, square root ke andar — bada orbit → **chhoti** speed, $1/\sqrt{r}$ jaisi girती hai. - **$m$ chala gaya.** Woh cancel ho gaya. Ek feather aur ek bus same altitude par *same* speed se orbit karte hain. **CANCELLATION KYUN MATTER KARTA HAI.** Ek heavier satellite zyada gravity feel karta hai (achha, zyada payment) lekin zyada centripetal force bhi maangta hai (bada bill) — aur dono $m$ ke saath identically scale karte hain, toh $m$ disappear ho jaata hai. Yahi reason hai ki saari cheezein same rate se girti hain. **PICTURE.** Curve $v=\sqrt{GM/r}$: chhote $r$ ke liye high aur steep, bade $r$ ke liye gently sink karta. Do dots — feather aur bus — same $r$ par *same* point par baithe hain. > [!mistake] "Higher orbit → tez" > *Lagta sahi hai:* zyada altitude, zyada energy. *Sach:* $r$ root ke andar **neeche** hai, toh higher = **slower**. Extra energy potential hai, kinetic nahi. (Parent note mein poori kahani.) --- ## Step 6 — Speed se period tak (ek free lap) **KYA.** Period $T$ ek poore loop ka time hai. Ek loop circle ki circumference $2\pi r$ (distance) cover karta hai speed $v$ se. Time = distance ÷ speed: $$T=\frac{2\pi r}{v}$$ Ab woh $v$ substitute karo jo humne abhi earn kiya: $$T=\frac{2\pi r}{\sqrt{GM/r}} =2\pi r\sqrt{\frac{r}{GM}} =2\pi\sqrt{\frac{r^3}{GM}}$$ - $2\pi r$ = lap distance (circumference). - $\sqrt{GM/r}$ se divide karne par aur theek karne par ek extra $r$ root ke andar aa jaata hai, $r^3$ deta hai. Dono sides square karo pattern dekhne ke liye: $$T^2=\frac{4\pi^2}{GM}\,r^3\quad\Longrightarrow\quad T^2\propto r^3$$ **YEH FAMOUS KYUN HAI.** Woh aakhri line ==Kepler's Third Law== hai — aur yahan woh Newton's gravity se seedha nikal aati hai, proportionality constant $4\pi^2/GM$ humein free mein mila. Dekho [[Kepler's Three Laws]]. **PICTURE.** Growing radius ki teen orbits; unke saath bar chart dikhata hai $T$ kitni tezi se $r$ se zyada badhta hai — $r^{3/2}$ ki steepness. --- ## Step 7 — Degenerate & limiting cases (jahan formula test hota hai) **KYA.** Ek formula tabhi trustworthy hai jab extremes par bhi sahi behave kare. Edges check karo: - **$r\to\infty$ (infinitely door):** $v=\sqrt{GM/r}\to 0$ aur $T=2\pi\sqrt{r^3/GM}\to\infty$. Door door, gravity kamzor hai, toh tum hamesha ke liye creep karte rehte ho. ✓ Sensible. - **$r\to$ surface radius (lowest possible orbit):** $v$ *sabse badi* hai — yeh woh minimum orbital speed hai jo ek planet allow karta hai, surface ko graze karte hue. Earth ke liye, lagbhag $7.9$ km/s. - **$r\to 0$:** $v\to\infty$ — unphysical, lekin honestly: tum center *par* orbit nahi kar sakte; math chillake warn karta hai. - **$M\to 0$ (koi central body nahi):** $v\to0$, $T\to\infty$ — koi pull nahi, koi bending nahi, "orbit" infinite radius par straight line mein degenerate ho jaata hai. ✓ - **[[Escape Velocity]] se compare karo:** total energy ko zero set karo aur tumhe milta hai $v_{\text{esc}}=\sqrt{2GM/r}=\sqrt{2}\,v_{\text{circ}}$. Toh circular speed exactly $1/\sqrt2\approx0.707$ hai escape speed ki same radius par — orbit escape line ke *neeche* rehti hai. **YEH KYUN DIKHAO.** Har reader eventually sochega "agar $r$ bahut bada / bahut chhota / planet gayab ho jaaye toh kya?" — yahan sab hain, kuch chance par nahi chhoda. **PICTURE.** $v(r)$ curve phir se, ab do lines ke saath: orange circular-speed curve aur, uske upar $\sqrt2$ factor se, red escape-speed curve. Unke beech ki gap bound elliptical orbits ki family hai (dekho [[Elliptical Orbits and the Vis-viva Equation]]). --- ## Ek-picture summary Upar sab kuch, do clean panels mein compress kiya. **Pehle**, woh ek idea jo poora show chalata hai — gravity *hi* string hai, toh uski inward pull circle ki demand ke barabar hai, aur $m$ cancel karke speed milti hai: **Doosra**, teen results jo phir free mein nikal aate hain, har ek radius $r$ ka ek simple function: > [!recall]- Feynman retelling — seedhe shabdon mein poora walkthrough > Ek ball ko string par ghuma: woh circle par tabhi rehti hai jab string use lagaatar inward yank karta rahe. Yank $mv^2/r$ kyun hona chahiye? Do tiny triangles draw karo — ek nearby do points ke radii se bana, ek un do points par velocity arrows se bana. Woh same skinny shape ke hain (same chhota corner angle), toh velocity ka inward "kink" per second exactly $v^2/r$ nikalta hai; mass se multiply karo aur circle ki zaroorat ki force milti hai. Space mein koi string nahi — *sirf ek hi* pull hai gravity, $GMm/r^2$, aur khush kismat yeh ki woh seedha center ki taraf point karti hai, exactly wahan jahan circle ko pull chahiye. Toh gravity string ka role play karti hai. "Jitna yank chahiye" ko "jitna yank gravity deti hai" ke barabar set karo, satellite ka apna weight $m$ cross out karo (woh cancel hota hai kyunki heavier satellite ko bigger yank chahiye *aur* equally bigger yank milta hai), aur tumhare paas $v=\sqrt{GM/r}$ bachta hai: bada orbit, slower speed. Poochho ek loop kitna time leta hai — circumference over speed — aur tumhe milta hai $T=2\pi\sqrt{r^3/GM}$, jo exactly Kepler's law $T^2\propto r^3$ hai. Edges check karo: infinitely door → creeping aur eternal; bahut tez (factor $\sqrt2$ se) → tum free ho jaate ho. Sab ek idea hai: *orbit ek circle mein girna hai, us ek speed par jahan gravity ki pull exactly woh pull hai jo circle maangta hai.* > [!mnemonic] Apni pocket mein rakh lo > **"Bill = Payment."** Circle $mv^2/r$ bill karta hai; gravity $GMm/r^2$ pay karta hai. Inhe equal karo, $m$ cancel karo, aur $v=\sqrt{GM/r}$ padh lo. --- ## Connections - [[Centripetal Force and Uniform Circular Motion]] — $mv^2/r$ demand ka source (Steps 1–2). - [[Newton's Law of Universal Gravitation]] — $GMm/r^2$ payment (Step 3). - [[Kepler's Three Laws]] — $T^2\propto r^3$ Step 6 mein emerge hota hai. - [[Escape Velocity]] — Step 7 mein $\sqrt2$ line. - [[Elliptical Orbits and the Vis-viva Equation]] — circular aur escape speeds ke beech kya rehta hai. - [[Gravitational Potential Energy]] · [[Virial Theorem]] — story ko energies mein le jaao (parent note).