Circular orbit — velocity, period, energy: Exercises

3.2.13 · D4 · Physics › Orbital Mechanics & Astrodynamics › Circular orbit — velocity, period, energy

Figure — Circular orbit — velocity, period, energy

Upar ki picture poora toolkit hai: har quantity ( $v, T, K, U, E$ ) ek single number $r$ , yaani center se distance, ka function hai. Isko aise padho: " $r$ choose karo, sab kuch read kar lo."

Level 1 — Recognition

Goal: sahi formula shelf se uthao aur plug in karo. Koi algebraic rearranging nahi.

L1.1

Ek satellite Earth ko $r = 7.00 \times 1 0^{6} m$ radius par orbit karta hai. Uski orbital speed $v$ nikalo.

Recall Solution L1.1

KYA KARNA HAI: Seedha $v = GM / r$ use karo — $r$ pata hai, $v$ chahiye. $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{7.00 \times 1 0 ^{6}} = 5.691 \times 1 0^{7}$ $v \approx 7.54 \times 1 0^{3} m/s = 7.54 km/s .$ Satellite ki mass kyun nahi chahiye? Mass $m$ , $GM m / r^{2} = m v^{2} / r$ mein cancel ho jaati hai, isliye speed kabhi $m$ par depend nahi karti.

L1.2

Usi satellite ki mass $m = 1200 kg$ hai. Uski total mechanical energy $E$ nikalo.

Recall Solution L1.2

KYA KARNA HAI: $E = - \frac{GM m}{2 r}$ use karo. Yahan $m$ matter karta hai (energy mass ke saath scale hoti hai). $E = - \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{2 ( 7.00 \times 1 0 ^{6} )}$ $E \approx - 3.42 \times 1 0^{10} J .$ Negative kyun? Negative total energy ka matlab bound hai — satellite phansa hua hai, isse free karne ke liye energy add karni padegi.

L1.3

$r = 7.00 \times 1 0^{6} m$ par satellite ki period $T$ nikalo.

Recall Solution L1.3

KYA KARNA HAI: $T = 2 π r^{3} / GM$ use karo — radius se time chahiye. $T = 2 π \frac{( 7.00 \times 1 0 ^{6} ) ^{3}}{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )} = 2 π 8.610 \times 1 0^{6}$ $T \approx 2 π (2934) \approx 5.83 \times 1 0^{3} s \approx 97 min .$ Check: $T = 2 π r / v = 2 π (7.00 \times 1 0^{6}) /7540 \approx 5.83 \times 1 0^{3} s$ ✓ — dono tarike agree karte hain.

Level 2 — Application

Goal: ek formula rearrange karo, ya do formulas ko ek saath chain karo.

L2.1

Ek satellite Moon ( $M = 7.35 \times 1 0^{22} kg$ , radius $1.74 \times 1 0^{6} m$ ) ko $100 km$ altitude par orbit karta hai. Uski speed nikalo.

Recall Solution L2.1

Step 1 — radius banao. $r = 1.74 \times 1 0^{6} + 1.00 \times 1 0^{5} = 1.84 \times 1 0^{6} m$ . Step 2 — $v = GM / r$ mein Moon ka $M$ plug karo: $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 7.35 \times 1 0 ^{22} )}{1.84 \times 1 0 ^{6}} = 2.666 \times 1 0^{6} \approx 1.63 \times 1 0^{3} m/s .$ Lagbhag $1.63 km/s$ — Earth orbit se bahut slow, kyunki Moon ka $M$ bahut chota hai.

L2.2

Kepler ulta karo: ek Earth satellite ki period $T = 2.00 hours = 7200 s$ hai. Uski orbital radius $r$ nikalo.

Recall Solution L2.2

KYA KARNA HAI: $T$ pata hai, $r$ chahiye. $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ ko $r$ ke liye solve karo: $r = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3} .$ Cube-root kyun? $T^{2} \propto r^{3}$ hai, isliye cube undo karne ke liye $1/3$ power lete hain. $r = (\frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 7200 ) ^{2}}{4 π ^{2}})^{1/3} = (5.229 \times 1 0^{20})^{1/3}$ $r \approx 8.06 \times 1 0^{6} m .$

L2.3

Dikhao ki kisi bhi circular orbit ki kinetic energy $- E$ ke barabar hoti hai, aur L1 satellite ( $m = 1200 kg$ , $r = 7.00 \times 1 0^{6}$ ) ke liye numerically verify karo.

Recall Solution L2.3

Algebra: $K = \frac{GM m}{2 r}$ aur $E = - \frac{GM m}{2 r}$ hai, isliye $K = - E$ exactly (Virial Theorem for $1/ r$ gravity). Numbers: $K = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{2 ( 7.00 \times 1 0 ^{6} )} \approx 3.42 \times 1 0^{10} J$ . Aur L1.2 se, $E \approx - 3.42 \times 1 0^{10} J$ , isliye sach mein $K = - E$ ✓.

Level 3 — Analysis

Goal: do orbits compare karo, ratios aur change ki direction ke baare mein sochо.

L3.1

Satellite A, $r$ par orbit karta hai; satellite B, usi planet ke around $4 r$ par orbit karta hai. Ratios $v_{B} / v_{A}$ aur $T_{B} / T_{A}$ nikalo.

Recall Solution L3.1

Speed ratio. $v \propto r^{- 1/2}$ hai, isliye $\frac{v _{B}}{v _{A}} = (\frac{4 r}{r})^{- 1/2} = 4^{- 1/2} = \frac{1}{2}$ . B aadhi speed se chalta hai. Period ratio. $T \propto r^{3/2}$ hai, isliye $\frac{T _{B}}{T _{A}} = 4^{3/2} = 8$ . B ko ek lap mein 8 guna zyada waqt lagta hai. Sanity check: B, 4 guna door hai lekin sirf aadhi speed se, isliye time $\propto r / v = 4/ (1/2) = 8$ ✓.

L3.2

Jab satellite ko $r$ se $4 r$ par le jaate ho, total energy $E$ kitne factor se change hoti hai? Ye upar jaati hai ya neeche?

Recall Solution L3.2

$E \propto - 1/ r$ hai. Isliye $\frac{E _{B}}{E _{A}} = \frac{- 1/ ( 4 r )}{- 1/ r} = \frac{1}{4}$ . $E_{B}$ , magnitude mein $E_{A}$ ka $\frac{1}{4}$ hai, lekin dono negative hain, isliye $E_{B}$ kam negative hai — energy badh gayi hai (zero ki taraf gayi). Uuchi orbit mein total energy zyada hai, bhale hi woh slower move kare.

L3.3

Ek satellite circular orbit mein hai. Uski speed us radius par local circular speed se exactly $2$ guni maapi gayi hai. Uski total energy kya hai, aur uska kya hoga?

Recall Solution L3.3

KYA KARNA HAI: Usi $r$ par, circular speed deti hai $v_{c}^{2} = GM / r$ . Yahan $v^{2} = 2 v_{c}^{2} = 2 GM / r$ . Kinetic: $K = \frac{1}{2} m v^{2} = \frac{GM m}{r}$ . Potential (unchanged, sirf $r$ par depend karta hai): $U = - \frac{GM m}{r}$ . Total: $E = K + U = \frac{GM m}{r} - \frac{GM m}{r} = 0$ . $E = 0$ bilkul escape boundary hai — body ab bound nahi hai. Isliye Escape Velocity satisfy karta hai $v_{esc} = 2 v_{c i r c}$ .

Level 4 — Synthesis

Goal: energy bookkeeping, Kepler, aur definitions ko ek multi-step plan mein combine karo.

L4.1 — Orbit raise karne ke liye energy

Ek $2000 kg$ satellite ko Earth ke around $r_{1} = 7.00 \times 1 0^{6} m$ ki circular orbit se $r_{2} = 1.20 \times 1 0^{7} m$ ki circular orbit par le jao. Kitni energy add karni hogi?

Recall Solution L4.1

KYA KARNA HAI: Add ki gayi energy $= E_{2} - E_{1}$ , total orbital energies ka difference. $Δ E = - \frac{GM m}{2 r _{2}} - (- \frac{GM m}{2 r _{1}}) = \frac{GM m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}}) .$ Ye form kyun? Uuchi orbit mein $E$ kam negative hoti hai, isliye $Δ E > 0$ : orbit raise karna energy kharch karta hai. $Δ E = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 2000 )}{2} (\frac{1}{7.00 \times 1 0 ^{6}} - \frac{1}{1.20 \times 1 0 ^{7}}) .$ Prefactor $\frac{GM m}{2} = 3.984 \times 1 0^{17}$ . Bracket $= 1.4286 \times 1 0^{- 7} - 8.333 \times 1 0^{- 8} = 5.952 \times 1 0^{- 8}$ . $Δ E \approx (3.984 \times 1 0^{17}) (5.952 \times 1 0^{- 8}) \approx 2.37 \times 1 0^{10} J .$

L4.2 — Surface launch se orbit tak

Earth ki surface ( $R_{\oplus}$ radius) par rest se shuru karke, $r = 7.00 \times 1 0^{6} m$ ki circular orbit tak pahunchne ke liye har kilogram ke liye kitni energy chahiye? (Earth ki rotation aur air ignore karo.)

Recall Solution L4.2

KYA KARNA HAI: Dono states mein total energy per unit mass compare karo. Surface par rest mein: $E_{i} / m = - \frac{GM}{R _{\oplus}}$ (sirf potential, koi kinetic nahi). Orbit mein: $E_{f} / m = - \frac{GM}{2 r}$ . $\frac{Δ E}{m} = - \frac{GM}{2 r} + \frac{GM}{R _{\oplus}} = GM (\frac{1}{R _{\oplus}} - \frac{1}{2 r}) .$ $GM = (6.674 \times 1 0^{- 11}) (5.97 \times 1 0^{24}) = 3.984 \times 1 0^{14}$ . Bracket $= \frac{1}{6.37 \times 1 0 ^{6}} - \frac{1}{1.40 \times 1 0 ^{7}} = 1.5699 \times 1 0^{- 7} - 7.143 \times 1 0^{- 8} = 8.556 \times 1 0^{- 8}$ . $\frac{Δ E}{m} \approx (3.984 \times 1 0^{14}) (8.556 \times 1 0^{- 8}) \approx 3.41 \times 1 0^{7} J/kg .$ Lagbhag $34 MJ$ per kilogram — low Earth orbit ki fundamental energy cost yahi hai.

L4.3 — Grazing orbit se planet ki density

Ek satellite ek planet ki surface ke beechabech ghoomta hai ( $r \approx R$ , planet radius) period $T$ ke saath. Dikhao ki period sirf planet ki density $ρ$ par depend karti hai, aur $ρ = 5510 kg/m^{3}$ (Earth ki mean density) ke liye $T$ calculate karo.

Recall Solution L4.3

Step 1 — $r = R$ par Kepler: $T^{2} = \frac{4 π ^{2} R ^{3}}{GM}$ . Step 2 — $M$ ko density se replace karo. Sphere ke liye $M = ρ \cdot \frac{4}{3} π R^{3}$ . Substitute karo: $T^{2} = \frac{4 π ^{2} R ^{3}}{G \cdot ρ \frac{4}{3} π R ^{3}} = \frac{4 π ^{2}}{\frac{4}{3} π Gρ} = \frac{3 π}{Gρ} .$ $R$ kyun gayab ho gaya? Numerator ka $R^{3}$ aur mass ka $R^{3}$ dono cancel ho gaye — grazing period size ke baare mein kuch nahi jaanti, sirf density ke baare mein. $T = \frac{3 π}{Gρ} = \frac{3 π}{( 6.674 \times 1 0 ^{- 11} ) ( 5510 )} = 2.562 \times 1 0^{7} \approx 5.06 \times 1 0^{3} s \approx 84 min .$ Yahi famous "minimum orbital period" hai — Earth-jaise density wale har rocky planet ki grazing period 84 minutes ke paas hoti hai.

Level 5 — Mastery

Goal: ek naya relation derive karo, ya ek limiting/degenerate case carefully handle karo.

L5.1 — Virial theorem, verified

Circular orbit ke liye prove karo ki time-averaged kinetic energy satisfy karti hai $⟨ K ⟩ = - \frac{1}{2} ⟨ U ⟩$ , aur $E = - K$ . (Circle ke liye har instant same hai, isliye time-averages sirf constant values hain.)

Recall Solution L5.1

Constants (circular ⇒ koi averaging nahi chahiye): $K = \frac{GM m}{2 r}, U = - \frac{GM m}{r} .$ Virial check karo: $- \frac{1}{2} U = - \frac{1}{2} (- \frac{GM m}{r}) = \frac{GM m}{2 r} = K$ ✓. $E = - K$ check karo: $E = K + U = \frac{GM m}{2 r} - \frac{GM m}{r} = - \frac{GM m}{2 r} = - K$ ✓. Ye Virial Theorem ke relations hain $2 K + U = 0$ aur $E = \frac{1}{2} U = - K$ , jo inverse-square force ke liye exactly hold karte hain.

L5.2 — Vis-viva sanity check

General vis-viva equation semi-major axis $a$ (ellipse ka aadha lamba diameter) wali orbit par $r$ distance par speed deti hai: $v^{2} = GM (\frac{2}{r} - \frac{1}{a}) .$ Dikhao ki $a = r$ set karne par (ek circle, jahan semi-major axis constant radius ke barabar hai) $v = GM / r$ wapas milta hai.

Recall Solution L5.2

KYA KARNA HAI: Circle ke liye har point same distance par hota hai, isliye $a = r$ . Substitute karo: $v^{2} = GM (\frac{2}{r} - \frac{1}{r}) = GM \cdot \frac{1}{r} = \frac{GM}{r} .$ $v = \frac{GM}{r} . ✓$ Circular formula, vis-viva ka special case hai jab eccentricity zero ho. Isliye parent note ki speed seedha general orbit law se nikaali ja sakti hai.

L5.3 — Degenerate limit: $r \to \infty$

Dekho kya hota hai $v$ , $T$ , aur $E$ ka jab orbit radius bina bound ke badhta hai. Har limit ko physically interpret karo.

Recall Solution L5.3

Speed: $v = GM / r \to 0$ . Infinitely badi orbits infinitely slowly traverse hoti hain — gravity itni weak hai ki koi speed demand nahi karta.
Period: $T = 2 π r^{3} / GM \to \infty$ . Ek lap lene mein forever lagta hai.
Energy: $E = - GM m / (2 r) \to 0^{-}$ . Orbit, escape boundary $E = 0$ ko neeche se approach karti hai: infinitely door, barely bound, rest mein. Thodi si aur energy add karo aur orbit open (hyperbolic, $E > 0$ ) ho jaati hai aur body hamesha ke liye chali jaati hai.

Doosri extreme, $r \to R_{\oplus}$ (grazing): $v$ maximum hai, $T$ minimum (~84 min from L4.3), $E$ most negative — surface clear karne wali sabse tightly bound circular orbit.

Recall Master self-test (answers chhupaao)

$r = 7.00 \times 1 0^{6}$ m par Earth ke around circular speed ::: $\approx 7.54 km/s$ Us radius par period ::: $\approx 5.83 \times 1 0^{3} s \approx 97 min$ $r$ aur $4 r$ par orbits ke liye speed ratio ::: $v_{4 r} / v_{r} = \frac{1}{2}$ $r$ aur $4 r$ par orbits ke liye period ratio ::: $T_{4 r} / T_{r} = 8$ Jo speed $r$ par $E = 0$ banaye ::: $2 v_{c i r c}$ (escape) Grazing-orbit period sirf kis par depend karti hai ::: planet density $ρ$ par, $T = 3 π / Gρ$ se $a = r$ ke saath vis-viva deta hai ::: $v = GM / r$ , circular case

Connections

3.2.13 Circular orbit — velocity, period, energy (Hinglish) — parent topic (Hinglish companion).
Newton's Law of Universal Gravitation — woh force jo yahan har formula ke peeche hai.
Centripetal Force and Uniform Circular Motion — $m v^{2} / r$ requirement.
Kepler's Three Laws — L2.2, L4.3 mein $T^{2} \propto r^{3}$ ko ulta kiya.
Gravitational Potential Energy — har energy problem mein use hone wala $U = - GM m / r$ .
Escape Velocity — $E = 0$ boundary jo L3.3 aur L5.3 mein mili.
Elliptical Orbits and the Vis-viva Equation — L5.2 ka general law.
Virial Theorem — L2.3 aur L5.1 ka $E = - K$ , $U = 2 E$ .