Orbital velocity for circular orbit — derivation: Worked examples

1.2.24 · D3 · Physics › Newton's Laws & Dynamics › Orbital velocity for circular orbit — derivation

Yeh practice arena hai parent derivation ke liye. Humne already woh ek formula earn kar liya hai jo poora show chalata hai:

Neeche hum har tarah ka question tackle karte hain jo yeh ek akela formula produce kar sakta hai.

Scenario matrix

Kisi formula ke "scenarios" ko un alag-alag shapes of question ke roop mein socho jo ek teacher ya reality pose kar sakti hai. Yahan har input diya ja sakta hai, chhupaya ja sakta hai, double kiya ja sakta hai, zero bheja ja sakta hai, ya infinity bheja ja sakta hai — aur har aisa move ek cell hai:

Cell	Kya vary/test kiya jaata hai	Example
A. Surface orbit	$h = 0$ , isliye $r = R$ — $g R$ shortcut	Ex 1
B. High orbit	$h \neq = 0$ , $r = R + h$ use karna zaroori	Ex 2
C. Ratio/scaling	speeds compare karna, koi numbers plug nahi	Ex 3
D. Solve backwards	$v_{o}$ diya hai, $r$ (ya $h$ ) find karo	Ex 4
E. Different planet	$M, R$ Earth ke nahi — general $GM$	Ex 5
F. Limiting cases	$r \to \infty$ , $r \to R$ , " $r$ planet ke andar"	Ex 6
G. Real-world word problem	altitude quoted, speed + period + ground track find karo	Ex 7
H. Exam twist	escape velocity / mass cancellation trap ko mix karta hai	Ex 8
I. Kepler link	$v = 2 π r / T$ ke saath combine karo	Ex 9

Neeche ke nau examples saare nau cells cover karte hain. Figures geometry carry karti hain jahan geometry matter karti hai.

Ex 1 — Cell A: surface orbit (shortcut)

Worked example Statement

Ek satellite Earth ki surface ke bilkul upar skim karta hai (air ignore karo). $v_{o}$ find karo. Use $g = 9.8 m/s^{2}$ , $R = 6.4 \times 1 0^{6} m$ .

Forecast: Kya answer 1 km/s ke kareeb hoga, 8 km/s, ya 100 km/s? Padhne se pehle guess karo.

Step 1 — $h = 0$ recognize karo. Yeh step kyun? Surface par altitude $h$ zero hoti hai, isliye $r = R + h = R + 0 = R$ . Poora formula $g R^{2} / r$ collapse hokar $g R^{2} / R = g R$ ban jaata hai.

Step 2 — Numbers plug karo. $v_{o} = g R = 9.8 \times 6.4 \times 1 0^{6} = 6.272 \times 1 0^{7} m/s$ Yeh step kyun? Right side par sab kuch known hai; satellite ki mass $m$ ke baare mein kuch bhi nahi chahiye.

Step 3 — Root evaluate karo. $v_{o} \approx 7.92 \times 1 0^{3} m/s \approx 7.9 km/s$

Verify: Units: $(m/s^{2}) (m) = m^{2} / s^{2} = m/s$ ✓. Yeh ~28,500 km/h hai — ek Low Earth Orbit speed. Forecast ~8 km/s hona chahiye tha.

Ex 2 — Cell B: high orbit, $r = R + h$

Worked example Statement

Ek satellite altitude $h = 3.6 \times 1 0^{7} m$ par orbit karta hai (roughly geostationary height). $v_{o}$ find karo. Use $GM = g R^{2}$ with $g = 9.8$ , $R = 6.4 \times 1 0^{6}$ .

Forecast: Ex 1 se zyaada ya kam? Roughly kis factor se?

Step 1 — Centre se $r$ build karo. Yeh step kyun? Formula ko centre-distance chahiye. $r = R + h = 6.4 \times 1 0^{6} + 3.6 \times 1 0^{7} = 4.24 \times 1 0^{7} m$ .

Step 2 — $GM$ compute karo. Yeh step kyun? Hume $G$ aur $M$ alag-alag nahi diye gaye, lekin $GM = g R^{2}$ . $GM = 9.8 \times (6.4 \times 1 0^{6})^{2} = 9.8 \times 4.096 \times 1 0^{13} = 4.014 \times 1 0^{14} m^{3} / s^{2}$

Step 3 — Master formula apply karo. Yeh step kyun? Ab hamare paas formula ko chahiye dono cheezein hain — Step 2 se $GM$ aur Step 1 se $r$ — isliye yahan woh combine hokar actual speed deliver karte hain. $v_{o} = \frac{GM}{r} = \frac{4.014 \times 1 0 ^{14}}{4.24 \times 1 0 ^{7}} = 9.47 \times 1 0^{6} \approx 3.08 \times 1 0^{3} m/s$

Verify: ~3.1 km/s, LEO ke 7.9 km/s se kaafi kam, expected hai kyunki $r$ ~6.6× bada hai aur $v_{o} \propto 1/ r$ : $7.9/ 6.6 \approx 3.1$ ✓. Forecast: lower.

Ex 3 — Cell C: pure scaling, koi numbers nahi

Neeche ki figure woh $1/ r$ curve plot karti hai: P (yellow dot) ko radius $r$ par speed 1 ke saath dhundho, phir blue curve ke saath Q (red dot) tak trace karo $9 r$ par — curve par uski height exactly $1/3$ tak gir gayi hai. Dashed guide lines "9× out, 3× down" trade ko ek nazar mein visible banate hain.

Figure — Orbital velocity for circular orbit — derivation

Ex 4 — Cell D: solve backwards ( $v_{o}$ se $r$ find karo)

Worked example Statement

Ek probe $v_{o} = 5.0 \times 1 0^{3} m/s$ par Earth ke around circular orbit mein move karte hue measure kiya gaya hai. Woh kis altitude $h$ par hai? ( $GM = 3.986 \times 1 0^{14}$ , $R = 6.4 \times 1 0^{6}$ .)

Forecast: LEO se zyaada upar ya neeche (jo 7.9 km/s tha)? Kyunki yeh slower hai, high ya low altitude guess karo?

Step 1 — Master formula ko $r$ ke liye rearrange karo. Yeh step kyun? Ab unknown root ke andar hai; usse free karne ke liye dono sides square karo. $v_{o}^{2} = \frac{GM}{r} \Rightarrow r = \frac{GM}{v _{o}^{2}}$

Step 2 — Plug in karo. Yeh step kyun? $r$ ab left side par isolated hai, measured speed aur Earth ka $GM$ substitute karne se rearranged formula ek actual centre-distance number ban jaata hai. $r = \frac{3.986 \times 1 0 ^{14}}{( 5.0 \times 1 0 ^{3} ) ^{2}} = \frac{3.986 \times 1 0 ^{14}}{2.5 \times 1 0 ^{7}} = 1.594 \times 1 0^{7} m$

Step 3 — Altitude mein convert karo. Yeh step kyun? $r$ centre se hai; altitude $R$ subtract karta hai. $h = r - R = 1.594 \times 1 0^{7} - 6.4 \times 1 0^{6} = 9.54 \times 1 0^{6} m \approx 9, 540 km$

Verify: LEO se slower ⇒ higher orbit ⇒ positive altitude ✓. Back-check: $GM / r = 3.986 \times 1 0^{14} /1.594 \times 1 0^{7} = 2.5 \times 1 0^{7} = 5000$ m/s ✓.

Ex 5 — Cell E: alag planet (Mars)

Worked example Statement

Mars ki $M = 6.42 \times 1 0^{23} kg$ , radius $R = 3.39 \times 1 0^{6} m$ hai. Low Mars orbit ( $h \approx 0$ ) ki orbital speed find karo. $G = 6.67 \times 1 0^{- 11}$ .

Forecast: Mars Earth se chhota aur halka hai — Earth ke 7.9 km/s se faster ya slower?

Step 1 — $GM$ directly use karo. Yeh step kyun? Yahan hume $M$ aur $G$ alag-alag diye gaye hain (koi $g$ nahi diya), isliye $GM$ compute karo. $GM = 6.67 \times 1 0^{- 11} \times 6.42 \times 1 0^{23} = 4.282 \times 1 0^{13} m^{3} / s^{2}$

Step 2 — Low orbit ⇒ $r = R$ , phir formula apply karo. Yeh step kyun? "Low orbit" matlab altitude $h \approx 0$ , isliye $r = R + 0 = R$ — exactly wahi surface-orbit condition jo Ex 1 mein tha, lekin Mars ke liye. Mars ka apna $R$ master formula mein feed karne se uski low-orbit speed milti hai. $v_{o} = \frac{GM}{R} = \frac{4.282 \times 1 0 ^{13}}{3.39 \times 1 0 ^{6}} = 1.263 \times 1 0^{7} \approx 3.55 \times 1 0^{3} m/s$

Verify: ~3.6 km/s, Earth se aadhe se bhi kam — Mars ki kamzor gravity ke saath consistent ✓. Units Ex 1 ki tarah. Forecast: slower.

Ex 6 — Cell F: limiting & degenerate cases

Worked example Statement

Discuss karo ki $v_{o} = GM / r$ kya predict karta hai jab (a) $r \to \infty$ , (b) $r \to R$ upar se, (c) agar koi naively $r < R$ set kar de (planet ke andar).

Forecast: Kya speed zero, infinity, ya finite value tak jaati hai jab orbit bahut badi ho jaaye?

Step 1 — Case $r \to \infty$ . Yeh step kyun? Root ke andar denominator bina bound ke badh jaata hai. $lim_{r \to \infty} \frac{GM}{r} = 0$ Infinitely door orbit ko essentially zero speed chahiye — wahan gravity vanishingly weak hai, isliye required centripetal force tiny hai.

Step 2 — Case $r \to R^{+}$ . Yeh step kyun? Yeh fastest possible real orbit hai — surface ko hug karta hua. Surface relation $GM = g R^{2}$ yaad karo (Earth ki surface gravity uske radius squared se multiply); usse substitute karne se hum dono equivalent forms ke beech swap kar sakte hain: $v_{o} \to \frac{GM}{R} = \frac{g R ^{2}}{R} = g R \approx 7.9 km/s (Earth)$ Yeh Earth ke liye maximum circular orbital speed hai; tum ground skim karne se zyaada fast circle nahi kar sakte (aur woh bhi air ignore karke).

Step 3 — Case $r < R$ (planet ke andar). Yeh step kyun? Domain limit test karo. Bare formula abhi bhi ek number deta hai, lekin woh physically meaningless hai: surface ke neeche, sirf radius $r$ ke andar wali mass tumhe pull karti hai (shell theorem), isliye $M$ poora planet nahi raha. Jaise likha hua hai formula assume karta hai ki saari $M$ tumhare orbit ke andar hai. Isliye $r \geq R$ valid domain hai.

Verify: Monotonic decrease confirm hua: $\frac{d v _{o}}{d r} = - \frac{1}{2} GM r^{- 3/2} < 0$ saare $r > 0$ ke liye, isliye speed strictly girती hai jaise $r$ badhta hai, apni ceiling $r = R$ par hit karti hai aur infinity par $\to 0$ jaati hai ✓.

Neeche ki figure exactly in teen regions ko $v_{o} (r)$ curve par draw karti hai: green dot surface ceiling ko $r = R$ par mark karta hai (fastest orbit), yellow arrow curve ko zero ki taraf slide karte hue follow karta hai jaise $r \to \infty$ , aur red shaded band $r < R$ forbidden interior ko flag karta hai jahan whole-mass assumption toot jaati hai.

Ex 7 — Cell G: real-world word problem (speed, period, ground track)

Worked example Statement

International Space Station altitude $h = 4.0 \times 1 0^{5} m$ (400 km) par orbit karti hai. Uski speed, ek orbit kitne time mein complete hoti hai, aur equator ke saath har orbit mein ground track kitna aage badhta hai yeh find karo ( $R = 6.4 \times 1 0^{6}$ m). Use $GM = 3.986 \times 1 0^{14}$ .

Forecast: Log kehte hain "ISS har 90 minutes mein Earth ka chakkar lagaati hai." Sach? Haan/nahi guess karo.

Step 1 — Centre-distance. Yeh step kyun? Classic trap: 400 km altitude $h$ hai, $r$ nahi. $r = R + h = 6.4 \times 1 0^{6} + 4.0 \times 1 0^{5} = 6.8 \times 1 0^{6} m$ .

Step 2 — Speed. Yeh step kyun? Speed master formula ka direct output hai, aur baad ki saari cheezein (period, ground track) usi se build hoti hain, isliye hum pehle ise compute karte hain. $v_{o} = \frac{3.986 \times 1 0 ^{14}}{6.8 \times 1 0 ^{6}} = 5.862 \times 1 0^{7} \approx 7.66 \times 1 0^{3} m/s$

Step 3 — Geometry se Period. Yeh step kyun? Ek lap circumference $2 π r$ cover karta hai constant speed par, isliye $T = \frac{2 π r}{v _{o}}$ . $T = \frac{2 π \times 6.8 \times 1 0 ^{6}}{7.66 \times 1 0 ^{3}} \approx \frac{4.273 \times 1 0 ^{7}}{7.66 \times 1 0 ^{3}} \approx 5.58 \times 1 0^{3} s$

Step 4 — Ground track shift. Yeh step kyun? Jab ISS ek loop complete karta hai, Earth khud neeche eastward spin karti rehti hai. Equator ki surface $v_{eq} = \frac{2 π R}{T _{day}}$ par move karti hai jahan $T_{day} = 86400$ s, isliye time $T$ mein neeche ki ground $Δ x = v_{eq} T$ slide karti hai — yahi reason hai ki ISS har orbit mein different longitudes ke upar se guzarta hai. $v_{eq} = \frac{2 π \times 6.4 \times 1 0 ^{6}}{86400} \approx 465 m/s, Δ x = 465 \times 5580 \approx 2.60 \times 1 0^{6} m \approx 2600 km$

Verify: $5580 s \div 60 \approx 93$ minutes ✓ — famous "~90 minute" ISS orbit se match karta hai. Ground track har pass mein ~2600 km westward relative to ground shift karta hai, yahi reason hai ki ISS tracking maps par woh tilted sine-like paths trace karta hai ✓. Forecast: haan.

Ex 8 — Cell H: exam twist (escape-velocity + mass trap)

Worked example Statement

Do satellites Earth ke same radius $r$ par orbit karte hain: satellite X ki mass $1000 kg$ hai, satellite Y ki mass $2000 kg$ . (a) Kaun faster hai? (b) Agar X ki orbital speed $v_{o}$ hai, toh us radius se escape karne ke liye use kitni speed chahiye?

Forecast: Kya bhaari wala slower, faster, ya same orbit karta hai?

Step 1 — Mass cancellation. Yeh step kyun? Derivation mein $\frac{GM m}{r ^{2}} = \frac{m v _{o}^{2}}{r}$ mein, satellite mass $m$ cancel ho jaata hai. Isliye $v_{o}$ sirf $M$ (Earth) aur $r$ par depend karta hai — X ya Y ki mass par nahi. $v_{X} = v_{Y} = \frac{GM}{r}$ Woh exactly same speed par travel karte hain. (a) answer ho gaya: koi bhi faster nahi.

Step 2 — Same radius se escape. Yeh step kyun? Escape velocity ek alag result hai: $v_{esc} = 2 GM / r = 2 v_{o}$ . $v_{esc} = 2 v_{o} \approx 1.414 v_{o}$

Verify: Agar, maan lo, $v_{o} = 7.66 km/s$ (Ex 7 radius), escape ke liye $1.414 \times 7.66 \approx 10.8 km/s$ chahiye ✓. "Bhaari ⇒ faster" trap false hai; "orbital = escape" trap exactly $2$ se off hai. Dekho Escape Velocity.

Ex 9 — Cell I: Kepler link (period vs radius)

Worked example Statement

Dikhao ki $v_{o} = GM / r$ ko $v_{o} = 2 π r / T$ ke saath combine karne se Kepler's Third Law $T^{2} \propto r^{3}$ milta hai, phir ise use karo: agar orbit A ka period $T_{A}$ radius $r$ par hai, toh radius $8 r$ par period kya hoga?

Forecast: Bade radius par period longer ya shorter? $8 r$ ke liye kis factor se?

Step 1 — Dono speed expressions ko equate karo. Yeh step kyun? Dono $v_{o}$ ke barabar hain, isliye woh ek doosre ke barabar hain, hume $v$ eliminate karne aur $T$ – $r$ relationship expose karne deta hai. $\frac{2 π r}{T} = \frac{GM}{r}$

Step 2 — $T$ ke liye solve karo. Dono sides square karo: $\frac{4 π ^{2} r ^{2}}{T ^{2}} = \frac{GM}{r} \Rightarrow T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ Yeh step kyun? Yeh exactly Kepler's Third Law hai: $T^{2} \propto r^{3}$ . Dekho Kepler's Third Law.

Step 3 — $8 r$ ke liye ratio apply karo. Yeh step kyun? Kyunki $T \propto r^{3/2}$ , $r$ ko 8 se multiply karne par $T$ ko $8^{3/2}$ se multiply kiya jaata hai. $\frac{T _{B}}{T _{A}} = 8^{3/2} = (8)^{3} = (2.828)^{3} \approx 22.6$

Verify: $8^{3/2} = 8^{3} = 512 \approx 22.6$ ✓. Bada orbit ⇒ bahut lamba period — door Moon ko ~27 days lagte hain jabki LEO ko ~90 min. Forecast: longer, factor ~22.6.

Active recall

Recall Scenarios test karo

Earth ke liye $h = 0$ shortcut speed? ⇒ $g R \approx 7.9$ km/s.
$v_{o}$ diya ho toh $r$ kaise nikaalte ho? ⇒ $r = GM / v_{o}^{2}$ .
Fastest possible circular orbit kis $r$ par hoti hai? ⇒ $r = R$ (surface-skimming).
$r \to \infty$ par, $v_{o} \to$ ? ⇒ $0$ .
Kya satellite mass $m$ double karne se $v_{o}$ change hoti hai? ⇒ Nahi — $m$ cancel ho jaata hai.
Formula se $T$ vs $r$ relation? ⇒ $T^{2} \propto r^{3}$ .

Connections

Orbital velocity for circular orbit — derivation — woh parent jise yeh drill karta hai.
Newton's Law of Universal Gravitation
Centripetal Force and Uniform Circular Motion
Escape Velocity — $2$ wala cousin (Ex 8).
Kepler's Third Law — Ex 9 mein derive kiya gaya.
Acceleration due to gravity g and GM = gR² — $GM = g R^{2}$ swap jo poore mein use hua.
Geostationary Orbit — Ex 2 ka $h \approx 3.6 \times 1 0^{7}$ m case.