Circular orbit — velocity, period, energy: Worked examples

3.2.13 · D3 · Physics › Orbital Mechanics & Astrodynamics › Circular orbit — velocity, period, energy

Jo tools hum baar baar use karte hain (sab parent note mein bane hain): $v = \frac{GM}{r}, T = 2 π \frac{r ^{3}}{GM}, K = \frac{GM m}{2 r}, U = - \frac{GM m}{r}, E = - \frac{GM m}{2 r}$

Yahaan $G = 6.674 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}$ gravitational constant hai, $M$ central body ki mass hai (woh badi wali jise tum orbit karte ho), $m$ orbiting body ki mass hai (chhoti wali), $r$ badi body ke center se distance hai (surface ke upar altitude NAHI — yeh ek trap hai jo hum jaanboojhkar hit karte hain). Yahaan $K$ kinetic energy hai (motion ki energy), $U$ gravitational potential energy hai (position ki energy, negative kyunki orbit bound hai), aur $E = K + U$ total hai. Hum teeno ko explicitly Example 5 mein compute karte hain.

The scenario matrix

Circular orbits ke baare mein har problem in cells mein se kisi ek mein aata hai. Neeche ke examples cell ke hisaab se label kiye gaye hain.

Cell	Kya vary karta hai / kya poochha gaya hai	Example
A. Forward, Earth	diya $r$ → dhundho $v, T$	Ex 1
B. Backward	diya $T$ → dhundho $r$ (Kepler ko invert karo)	Ex 2
C. Altitude trap	diya altitude, planet radius add karna zaroori hai	Ex 3
D. Different central mass	Sun / kisi aur star ko orbit karo	Ex 4
E. Energy budget	diya ek maneuver → dhundho $K, U, E, Δ E$	Ex 5
F. Limiting case $r \to \infty$	$v, T, E$ edge par kya karte hain?	Ex 6
G. Degenerate: surface-grazing	sabse chhota possible orbit, $r = R_{planet}$	Ex 7
H. Mass-independence check	do bahut alag $m$ , same $r$	Ex 8
I. Exam twist	ratio problem, koi numbers plug nahi	Ex 9

Constants jo hum baar baar use karte hain: $M_{\oplus} = 5.97 \times 1 0^{24} kg, R_{\oplus} = 6.37 \times 1 0^{6} m, M_{⊙} = 1.989 \times 1 0^{30} kg$

Example 1 — Cell A: forward, Earth

Worked example Ek satellite Earth ko

$r = 7.00 \times 1 0^{6}$ m radius par orbit karta hai. $v$ aur $T$ dhundho. Forecast: Aage padhne se pehle speed ka andaza lagao. Kya yeh 1 km/s ke kareeb hai, 8 km/s, ya 30 km/s? Aur period minutes mein hoga ya hours mein?

Step 1 — Speed. $v = GM / r$ mein plug karo: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{7.00 \times 1 0 ^{6}} = 5.691 \times 1 0^{7} \approx 7.54 \times 1 0^{3} m/s$ Yeh step kyun? Yeh WOHI speed hai jis par gravity exactly centripetal need $m v^{2} / r$ ke barabar hoti hai is radius par — woh balance jo orbit ko define karta hai.

Step 2 — Period. $T = 2 π r / v$ use karo: $T = \frac{2 π ( 7.00 \times 1 0 ^{6} )}{7540} \approx 5.83 \times 1 0^{3} s \approx 97 min$ Yeh step kyun? Ek chakkar = circumference divided by speed. Pure definition hai, koi naya physics nahi.

Verify: $v \approx 7.5$ km/s (famous ISS ~7.7 km/s ke kareeb — theek hai, kyunki yeh radius ISS ki radius ke paas hai). Units: $(m^{3} s^{- 2}) / m = m^{2} s^{- 2} = m/s$ ✓. Period ~1.6 h low orbit ke liye sensible hai.

Example 2 — Cell B: backward, invert Kepler

Worked example Earth ke around ek satellite ka period

$T = 2.00$ hours $= 7200$ s hai. Uska orbital radius $r$ kya hai? Forecast: ISS (~1.5 h) se zyada period — toh kya $r$ , $7 \times 1 0^{6}$ m se bada hona chahiye ya chhota?

Step 1 — Yaad karo Kepler law ka two-form kahan se aata hai. Parent note ne likha $T = 2 π r^{3} / GM$ . Root hataane ke liye dono sides square karo: $T^{2} = (2 π)^{2} \cdot \frac{r ^{3}}{GM} = \frac{4 π ^{2}}{GM} r^{3}$ Yeh step kyun? Squaring se awkward square-root form ek clean $T^{2} \propto r^{3}$ form mein badal jaata hai jo $r$ ke liye invert karna aasaan hai. Dono forms ek hi law hain.

Step 2 — $r$ isolate karne ke liye rearrange karo: $r = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3}$ Yeh step kyun? Hume $T$ diya gaya aur $r$ maanga gaya; Kepler's third law wohi ek equation hai jo unhe link karti hai, isliye hum ise $r$ ke liye solve karte hain.

Step 3 — Plug in karo: $r = (\frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 7200 ) ^{2}}{4 π ^{2}})^{1/3} = (5.230 \times 1 0^{20})^{1/3} \approx 8.06 \times 1 0^{6} m$ Yeh step kyun? Cube root $r^{3}$ ko undo karta hai; hum ise last mein lete hain taaki andar ka arithmetic clean rahe.

Verify: $8.06 \times 1 0^{6} > 7 \times 1 0^{6}$ ✓ — zyada period matlab bada, dheema orbit, jo " $T$ grows with $r$ " se match karta hai. Sanity: cube karo wapas, $(8.06 \times 1 0^{6})^{3} = 5.24 \times 1 0^{20}$ ✓.

Example 3 — Cell C: the altitude trap

Worked example Ek weather satellite Earth ki surface ke upar

altitude 800 km par orbit karta hai. $v$ dhundho. Forecast: Naive galti hai $r = 800$ km $= 8 \times 1 0^{5}$ m use karna. Andaza lagao woh answer kitna galat hoga.

Step 1 — Altitude ko orbital radius mein convert karo. Formulas Earth ke center se distance use karte hain: $r = R_{\oplus} + h = 6.37 \times 1 0^{6} + 8.00 \times 1 0^{5} = 7.17 \times 1 0^{6} m$ Yeh step kyun? Gravity ka $1/ r^{2}$ center of mass se measure hota hai, ground se nahi. $R_{\oplus}$ skip karna sabse common orbit error hai.

Step 2 — Sahi $r$ ke saath speed: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{7.17 \times 1 0 ^{6}} = 5.557 \times 1 0^{7} \approx 7.45 \times 1 0^{3} m/s$ Yeh step kyun? Ex 1 jaisa hi master formula; subtlety sirf $r$ sahi karne mein thi.

Verify: Agar tum galti se $r = 8 \times 1 0^{5}$ m use karte toh $v = 4.98 \times 1 0^{8} \approx 2.23 \times 1 0^{4}$ m/s milta — teen guna bada aur LEO ke liye physically impossible. Sahi 7.45 km/s low-orbit band mein fit baithta hai ✓.

Example 4 — Cell D: different central mass (orbit the Sun)

Worked example Earth Sun ko

$r = 1.496 \times 1 0^{11}$ m par orbit karti hai. Earth ki orbital speed predict karo aur check karo ki period ek saal hai. Forecast: Sun ke around Earth ki orbital speed — satellite jaisi 8 km/s, ya bahut zyada?

Step 1 — $M = M_{⊙}$ use karo, $M_{\oplus}$ nahi. Central body ab Sun hai: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 1.989 \times 1 0 ^{30} )}{1.496 \times 1 0 ^{11}} = 8.872 \times 1 0^{8} \approx 2.98 \times 1 0^{4} m/s$ Yeh step kyun? $M$ hamesha wohi mass hai jise tum orbit karte ho. Sun ko swap karo aur same formula heliocentric speeds deta hai.

Step 2 — Period: $T = \frac{2 π ( 1.496 \times 1 0 ^{11} )}{2.98 \times 1 0 ^{4}} \approx 3.15 \times 1 0^{7} s$ Yeh step kyun? Same distance-over-speed. Convert karo: $3.15 \times 1 0^{7} s \div (86400 s/day) \approx 365$ days.

Verify: $v \approx 29.8$ km/s textbook Earth orbital speed hai ✓, aur $T \approx 365$ days = ek saal ✓. Wohi formula jo ISS ki 92 minutes deta hai, Earth ka saal bhi deta hai — same physics, alag $M$ aur $r$ .

Example 5 — Cell E: energy budget for a maneuver (K, U, E sab dikhaye)

Worked example Ek

$m = 1200$ kg satellite Earth ko $r_{1} = 6.80 \times 1 0^{6}$ m par orbit karta hai. (a) Uska $K$ , $U$ , aur $E$ dhundho. (b) Use $r_{2} = 1.02 \times 1 0^{7}$ m tak raise karo — kitni energy $Δ E$ supply karni padegi? Forecast: Zyada upar ka orbit = zyada total energy, isliye raise mein energy lagti hai. Lekin kya satellite end mein faster hoga ya slower?

Step 1 — $r_{1}$ par kinetic energy. $K = \frac{GM m}{2 r _{1}}$ use karo: $K_{1} = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{2 ( 6.80 \times 1 0 ^{6} )} \approx 3.514 \times 1 0^{10} J$ Yeh step kyun? $K$ motion ki energy hai; $v^{2} = GM / r$ ko $\frac{1}{2} m v^{2}$ mein plug karne se yeh compact form milta hai — pehle $v$ compute karne ki zaroorat nahi.

Step 2 — $r_{1}$ par potential energy. $U = - \frac{GM m}{r _{1}}$ use karo: $U_{1} = - \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{6.80 \times 1 0 ^{6}} \approx - 7.029 \times 1 0^{10} J$ Yeh step kyun? $U$ position ki energy hai, negative isliye kyunki orbit bound hai (infinity tak pahunchne ke liye energy add karni padegi).

Step 3 — $r_{1}$ par total energy. Dono add karo: $E_{1} = K_{1} + U_{1} \approx 3.514 \times 1 0^{10} - 7.029 \times 1 0^{10} = - 3.514 \times 1 0^{10} J$ Yeh step kyun? Dhyaan do $E_{1} = - K_{1}$ aur $U_{1} = 2 E_{1}$ — parent note ke virial relations, ek built-in check.

Step 4 — Orbit raise karne ki energy = totals ka difference. $Δ E = E_{2} - E_{1} = \frac{GM m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ $= 2.390 \times 1 0^{17} \times (1.4706 \times 1 0^{- 7} - 9.8039 \times 1 0^{- 8}) = 2.390 \times 1 0^{17} \times 4.902 \times 1 0^{- 8} \approx 1.17 \times 1 0^{10} J$ Yeh step kyun? Tum exactly do total energies ke beech ka gap supply karte ho; $K$ aur $U$ ke pieces orbit ke andar automatically rearrange ho jaate hain.

Verify: $E_{1} = - K_{1}$ ✓ aur $U_{1} = 2 E_{1}$ ✓ (virial theorem hold karta hai). $Δ E > 0$ ✓ (raise karne mein energy lagti hai). Twist: $v_{1} = GM / r_{1} = 7.65$ km/s, $v_{2} = GM / r_{2} = 6.25$ km/s — satellite end mein total energy gain karne ke bawajood slower ho jaata hai ✓ ( $K$ , $3.51 \times 1 0^{10}$ se $2.34 \times 1 0^{10}$ J tak girta hai; baaki $U$ mein chali gayi).

Example 6 — Cell F: the limit $r \to \infty$

Worked example Jab orbital radius bina bound ke barhta hai toh

$v$ , $T$ , aur $E$ ka kya hota hai? Forecast: Teeno mein se kaunsa zero ki taraf jaata hai, kaunsa blow up karta hai, aur kaunsa ek ceiling ki taraf creep karta hai?

Step 1 — Speed. $v = GM / r$ . Jab $r \to \infty$ , fraction $\to 0$ , isliye $lim_{r \to \infty} v = 0.$ Yeh step kyun? Formula se limit padhna: denominator mein $r$ ke neeche jo bhi hai woh $r$ barhne par vanish ho jaata hai.

Step 2 — Period. $T = 2 π r^{3} / GM \to \infty$ (numerator bina bound ke barhta hai). Yeh step kyun? $r^{3}$ dominate karta hai; chakkar forever lagta hai kyunki orbit enormous bhi hai aur speed bhi bahut dheemi hai.

Step 3 — Energy. $E = - GM m /2 r \to 0^{-}$ (zero ke neeche se approach karta hai). Yeh step kyun? $E$ negative hai aur uska magnitude $\propto 1/ r$ shrink hota hai. Neeche ke figure mein, violet curve $E$ dashed orange line $E = 0$ ki taraf climb karti hai lekin kabhi cross nahi karti, jabki magenta curve $v$ zero ki taraf sag karti hai.

Figure: horizontal axis orbital radius $r$ hai (millions of metres mein). Magenta curve orbital speed $v$ hai km/s mein, $r$ barhne par zero ki taraf girta hai. Violet curve total energy per kilogram $E$ hai MJ mein, dashed orange line $E = 0$ ki taraf rise karta hai — jo escape boundary hai — lekin kabhi nahi pahunchta.

Verify: Limit $E \to 0^{-}$ exactly escape boundary hai. $E = 0$ matlab body barely unbound hai (infinity par rest mein) — Escape Velocity ka darwaza. Bahut door ka orbit almost free hai ✓.

Example 7 — Cell G: degenerate surface-grazing orbit

Worked example Ek hypothetical satellite Earth ki surface ko skim karta hai,

$r = R_{\oplus} = 6.37 \times 1 0^{6}$ m (hawa ignore karo). Uski speed aur period kya hai? Yeh sabse chhota possible circular orbit hai. Forecast: ISS se faster hai ya slower? Yeh ek circular orbit ki sabse fast speed hai — kyun?

Step 1 — Minimum radius par speed: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{6.37 \times 1 0 ^{6}} = 6.254 \times 1 0^{7} \approx 7.91 \times 1 0^{3} m/s$ Yeh step kyun? Sabse chhota $r$ sabse bada $v$ deta hai (kyunki $v \propto r^{- 1/2}$ ). Koi orbit surface-grazer se faster nahi ho sakta.

Step 2 — Period: $T = \frac{2 π ( 6.37 \times 1 0 ^{6} )}{7910} \approx 5.06 \times 1 0^{3} s \approx 84 min$ Yeh step kyun? Sabse chhota orbit → sabse chhota period. Yeh ~84 min Earth ke liye famous "minimum orbital period" hai.

Verify: $v \approx 7.9$ km/s ≈ "first cosmic velocity" ✓. Yeh Ex 1 ke 7.54 km/s (jo bade $r$ par tha) se bada hai ✓, confirm karta hai ki chhota $r$ ⇒ faster. Koi bhi real orbit ( $r > R_{\oplus}$ ) is bound se slower aur longer-period hoga.

Example 8 — Cell H: mass-independence

Worked example Same radius

$r = 8.0 \times 1 0^{6}$ m par, ek $1$ kg cubesat aur ek $4.2 \times 1 0^{5}$ kg space station compare karo. Kya woh same speed par orbit karte hain? Forecast: Station 420,000× bhaari hai. Zaroor woh alag move karta hoga?

Step 1 — Balance likho aur $m$ cancel karo: $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow \frac{GM}{r} = v^{2}$ Yeh step kyun? Gravity $m$ ke saath scale karti hai, aur centripetal need bhi $m$ ke saath scale karti hai; dono $m$ exactly cancel ho jaate hain.

Step 2 — Woh ek speed compute karo jo dono share karte hain: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{8.0 \times 1 0 ^{6}} = 4.981 \times 1 0^{7} \approx 7.06 \times 1 0^{3} m/s$ Yeh step kyun? Kyunki $m$ drop out ho gaya, is $r$ par ek single orbital speed hai — feather aur freighter dono ke liye same.

Verify: Dono 7.06 km/s dete hain ✓. Unki energies zaroor alag hain: $E = - GM m /2 r$ , $m$ ke saath scale karta hai, isliye station ka $∣ E ∣$ , cubesat se $4.2 \times 1 0^{5}$ guna hai — lekin trajectory (speed, period) identical hai.

Example 9 — Cell I: the exam twist (ratio, no plugging)

Worked example Satellite B, satellite A ke radius se

chaar guna radius par same planet ko orbit karta hai. $v_{B} / v_{A}$ aur $T_{B} / T_{A}$ dhundho bina koi numbers compute kiye. Forecast: Agar radius chaar guna ho jaaye, toh period 4×, 8×, ya kuch beech mein hoga?

Step 1 — $v \propto r^{- 1/2}$ se speed ratio: $\frac{v _{B}}{v _{A}} = \frac{r _{A}}{r _{B}} = \frac{1}{4} = \frac{1}{2}$ Yeh step kyun? Ratio mein constants $G, M$ cancel ho jaate hain; sirf $r$ ki power bachti hai. $v_{B}$ , $v_{A}$ ka half hai.

Step 2 — $T \propto r^{3/2}$ se period ratio: $\frac{T _{B}}{T _{A}} = (\frac{r _{B}}{r _{A}})^{3/2} = 4^{3/2} = 8$ Yeh step kyun? Kepler's $T^{2} \propto r^{3}$ matlab $T \propto r^{3/2}$ ; $4$ ko $3/2$ power par raise karne se $8$ aata hai.

Verify: $T = 2 π r / v$ se consistency check: $T_{B} / T_{A} = (r_{B} / r_{A}) \div (v_{B} / v_{A}) = 4 \div \frac{1}{2} = 8$ ✓. Ratios exam ka sabse fast route hain — calculator ki zaroorat nahi.

Recall Kaun sa cell kaun sa hai — self-test

Given only the radius of a satellite around Earth, which formulas do you reach for? ::: $v = GM / r$ , then $T = 2 π r / v$ (Cell A). Given a period and asked for radius, what do you invert? ::: $r = (GM T^{2} /4 π^{2})^{1/3}$ (Cell B). A problem gives "altitude 500 km" — what must you do first? ::: Add the planet radius: $r = R + h$ (Cell C). How do you get $K$ , $U$ , $E$ once you know $r$ ? ::: $K = GM m /2 r$ , $U = - GM m / r$ , $E = - GM m /2 r$ (Cell E). As $r \to \infty$ , what do $v$ , $T$ , $E$ approach? ::: $v \to 0$ , $T \to \infty$ , $E \to 0^{-}$ (Cell F). The smallest possible circular orbit has $r = ?$ and is the fastest or slowest? ::: $r = R_{planet}$ ; it's the fastest circular orbit (Cell G). Does doubling the satellite's mass change $v$ or $T$ ? ::: No — $m$ cancels; only $E, K, U$ scale with $m$ (Cell H). If $r$ quadruples, $T$ multiplies by? ::: $4^{3/2} = 8$ (Cell I).

Connections

Parent: Circular orbit — velocity, period, energy — yahaan use kiye gaye chaar master formulas.
Newton's Law of Universal Gravitation — $GM m / r^{2}$ jisse hum har balance shuru karte hain.
Centripetal Force and Uniform Circular Motion — balance ki $m v^{2} / r$ wali side.
Kepler's Three Laws — $T^{2} \propto r^{3}$ ko invert karna Cells B aur I ko power deta hai.
Gravitational Potential Energy — Cell E ke peeche $U = - GM m / r$ .
Escape Velocity — $E = 0$ ceiling jis ki taraf Cell F creep karta hai.
Elliptical Orbits and the Vis-viva Equation — jahaan non-circular maneuvers rehte hain.
Virial Theorem — $E = - K$ , $U = 2 E$ bookkeeping jo energies ke peeche hai.