Orbital velocity for circular orbit — derivation: Exercises

1.2.24 · D4 · Physics › Newton's Laws & Dynamics › Circular orbit ke liye Orbital velocity — derivation

Constants jo poore note mein use honge: $g = 9.8 m/s^{2}$ , $R_{\oplus} = 6.4 \times 1 0^{6} m$ , $G = 6.67 \times 1 0^{- 11} N\cdotm^{2} / kg^{2}$ , $M_{\oplus} = 6.0 \times 1 0^{24} kg$ .

Level 1 — Recognition

L1.1

Orbital-velocity formula batao aur words mein explain karo ki har symbol ka kya matlab hai.

Recall Solution

$v_{o} = \frac{GM}{r}$

$v_{o}$ — circular orbit ke liye constant tangential speed.
$G$ — universal gravitational constant (poore cosmos mein same rehta hai).
$M$ — central body ki mass jiske around orbit ho raha hai (e.g. Earth).
$r$ — $M$ ke centre se satellite tak ki distance, $r = R + h$ . Dhyan do: satellite ki apni mass $m$ formula mein nahi hai — woh derivation ke dauran cancel ho gayi.

L1.2

Ek satellite $r$ radius par orbit kar raha hai. Uski mass double kar di jaati hai, baaki kuch nahi badla. Kya $v_{o}$ change hoga? Ek sentence mein explain karo.

Recall Solution

Nahi. Satellite ki mass $m$ , $\frac{GM m}{r ^{2}} = \frac{m v _{o}^{2}}{r}$ mein cancel ho jaati hai, isliye $v_{o}$ sirf $M$ aur $r$ par depend karta hai. Ek feather aur ek truck same radius par identical speed se orbit karenge.

Level 2 — Application

L2.1

Earth ki surface ke bilkul upar skim karte satellite ki orbital speed nikalo ( $h = 0$ ).

Recall Solution

Surface ke paas $r \approx R$ hai, toh shortcut $v_{o} = g R$ use karo: $v_{o} = 9.8 \times 6.4 \times 1 0^{6} = 6.272 \times 1 0^{7} \approx 7.9 \times 1 0^{3} m/s = 7.9 km/s .$

L2.2

Ek satellite $h = 3.6 \times 1 0^{6} m$ altitude par orbit kar raha hai. Pehle $r$ nikalo, phir $v_{o}$ . $GM = g R^{2}$ use karo.

Recall Solution

$r = R + h = 6.4 \times 1 0^{6} + 3.6 \times 1 0^{6} = 1.0 \times 1 0^{7} m .$ $v_{o} = \frac{g R ^{2}}{r} = \frac{9.8 \times ( 6.4 \times 1 0 ^{6} ) ^{2}}{1.0 \times 1 0 ^{7}} = \frac{9.8 \times 4.096 \times 1 0 ^{13}}{1.0 \times 1 0 ^{7}} .$ $= 4.014 \times 1 0^{7} \approx 6.34 \times 1 0^{3} m/s \approx 6.3 km/s .$

L2.3

$r = 1.0 \times 1 0^{7} m$ par $G$ aur $M$ directly use karke $v_{o}$ calculate karo, aur check karo ki L2.2 se match karta hai.

Recall Solution

$v_{o} = \frac{GM}{r} = \frac{6.67 \times 1 0 ^{- 11} \times 6.0 \times 1 0 ^{24}}{1.0 \times 1 0 ^{7}} = \frac{4.002 \times 1 0 ^{14}}{1.0 \times 1 0 ^{7}} .$ $= 4.002 \times 1 0^{7} \approx 6.33 \times 1 0^{3} m/s \approx 6.3 km/s . ✓$ Dono routes agree karte hain kyunki $GM = g R^{2}$ hai — same physics, do alag data sets.

Level 3 — Analysis

L3.1

Ek satellite ko radius $r$ se $9 r$ par le jaaya jaata hai. Uski speed kis factor se change hogi?

Recall Solution

$v_{o} \propto 1/ r$ , toh $r$ ko $9$ se multiply karne par $v_{o}$ $1/ 9 = 1/3$ se multiply ho jaata hai. $v_{o} (9 r) = \frac{GM}{9 r} = \frac{1}{3} \frac{GM}{r} = \frac{1}{3} v_{o} (r) .$ Speed one-third ho jaati hai.

L3.2

Do satellites Earth orbit kar rahe hain. Satellite A $r_{A} = 7 \times 1 0^{6} m$ par hai, B $r_{B} = 4.2 \times 1 0^{7} m$ par. Dono speeds puri calculate kiye bina $v_{A} / v_{B}$ ratio nikalo.

Recall Solution

Kyunki $v_{o} \propto 1/ r$ : $\frac{v _{A}}{v _{B}} = \frac{r _{B}}{r _{A}} = \frac{4.2 \times 1 0 ^{7}}{7 \times 1 0 ^{6}} = 6 \approx 2.449.$ A, B se lagbhag 2.45 times faster hai. (Figure dekho — inner orbit race karta hai, outer dheeray chalti hai.)

Figure — Orbital velocity for circular orbit — derivation

L3.3

Dikhao ki same radius par $v_{esc} / v_{o}$ ka ratio ek constant hai, aur woh value nikalo.

Recall Solution

Escape velocity hai $v_{esc} = 2 GM / r$ (dekho Escape Velocity). $\frac{v _{esc}}{v _{o}} = \frac{2 GM / r}{GM / r} = 2 \approx 1.414.$ $G, M, r$ par independent — hamesha exactly $2$ . Circle karne se nikalne ke liye $41.4%$ zyada speed chahiye.

Level 4 — Synthesis

L4.1

$v_{o} = GM / r$ ko circumference relation $v_{o} = \frac{2 π r}{T}$ ke saath combine karke Kepler's Third Law $T^{2} \propto r^{3}$ derive karo.

Recall Solution

$v_{o}$ ke dono expressions barabar set karo: $\frac{2 π r}{T} = \frac{GM}{r} .$ Dono sides square karo: $\frac{4 π ^{2} r ^{2}}{T ^{2}} = \frac{GM}{r} \Rightarrow T^{2} = \frac{4 π ^{2}}{GM} r^{3} .$ Toh $T^{2} \propto r^{3}$ — exactly Kepler's Third Law, proportionality constant $4 π^{2} / GM$ ke saath.

L4.2

$T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ use karke ek geostationary satellite ( $T = 86400 s$ , ek sidereal-ish din) ka orbital radius $r$ nikalo. Phir $v_{o}$ nikalo.

Recall Solution

$r$ ke liye solve karo: $r = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3} .$ Numerator: $GM T^{2} = 4.002 \times 1 0^{14} \times (86400)^{2} = 4.002 \times 1 0^{14} \times 7.465 \times 1 0^{9} = 2.988 \times 1 0^{24}$ . $4 π^{2} = 39.48$ se divide karo: $7.568 \times 1 0^{22}$ . Cube root: $r = (7.568 \times 1 0^{22})^{1/3} \approx 4.23 \times 1 0^{7} m$ ( $\approx 42, 300 km$ centre se). Phir: $v_{o} = \frac{GM}{r} = \frac{4.002 \times 1 0 ^{14}}{4.23 \times 1 0 ^{7}} = 9.46 \times 1 0^{6} \approx 3.08 \times 1 0^{3} m/s \approx 3.1 km/s .$ Yeh Geostationary Orbit radius hai — check karo: sirf $3.1$ km/s, LEO ke $7.9$ km/s se kaafi slow, jaisa $1/ r$ predict karta hai.

L4.3

LEO mein ek satellite ( $v_{o} \approx 7.9$ km/s) ko same radius se Earth escape karne ke liye speed boost chahiye. Kitna extra speed $Δ v$ chahiye, aur woh escape speed ka kya fraction hai?

Recall Solution

$v_{esc} = 2 v_{o} = 1.414 \times 7.9 = 11.17 km/s$ . $Δ v = v_{esc} - v_{o} = 11.17 - 7.9 = 3.27 km/s .$ Fraction: $Δ v / v_{esc} = 3.27/11.17 \approx 0.293$ , lagbhag 29% zyada.

Level 5 — Mastery

L5.1

Moon Earth ke around $r = 3.84 \times 1 0^{8} m$ par orbit karta hai. $v_{o} = GM / r$ se uski orbital speed predict karo, aur actual $\approx 1.02$ km/s se compare karo.

Recall Solution

$v_{o} = \frac{4.002 \times 1 0 ^{14}}{3.84 \times 1 0 ^{8}} = 1.042 \times 1 0^{6} \approx 1.021 \times 1 0^{3} m/s \approx 1.02 km/s .$ Real value se excellent match — treated circular orbit Moon ki motion ko ek percent ke andar capture kar leta hai.

L5.2 (Different planet)

Ek planet jis par surface gravity $g_{p} = 3.7 m/s^{2}$ (Mars-jaisi) aur radius $R_{p} = 3.4 \times 1 0^{6} m$ hai, uske low-orbit speed $g_{p} R_{p}$ nikalo.

Recall Solution

$v_{o} = g_{p} R_{p} = 3.7 \times 3.4 \times 1 0^{6} = 1.258 \times 1 0^{7} \approx 3.55 \times 1 0^{3} m/s \approx 3.5 km/s .$ Kam gravity aur chhota radius — dono milkar surface-orbit speed ko Earth ke 7.9 km/s se neeche le aate hain.

L5.3 (Full-chain synthesis)

Ek satellite Earth ko $T = 5.4 \times 1 0^{3} s$ period ke saath orbit kar raha hai. (a) orbital radius $r$ , (b) orbital speed $v_{o}$ , aur (c) Earth ki surface se altitude $h$ nikalo.

Recall Solution

(a) Kepler se: $r = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3}$ . $GM T^{2} = 4.002 \times 1 0^{14} \times (5.4 \times 1 0^{3})^{2} = 4.002 \times 1 0^{14} \times 2.916 \times 1 0^{7} = 1.167 \times 1 0^{22}$ . $39.48$ se divide karo: $2.956 \times 1 0^{20}$ . Cube root: $r = (2.956 \times 1 0^{20})^{1/3} \approx 6.66 \times 1 0^{6} m$ . (b) $v_{o} = \frac{2 π r}{T} = \frac{2 π \times 6.66 \times 1 0 ^{6}}{5.4 \times 1 0 ^{3}} = \frac{4.184 \times 1 0 ^{7}}{5.4 \times 1 0 ^{3}} \approx 7.75 \times 1 0^{3} m/s \approx 7.7 km/s .$ (c) $h = r - R = 6.66 \times 1 0^{6} - 6.4 \times 1 0^{6} = 2.6 \times 1 0^{5} m \approx 260 km$ — ek realistic LEO hai.

Active recall

Recall Quick self-check

$v_{o}$ formula? ⇒ $v_{o} = GM / r$ .
Same radius par $v_{esc} / v_{o}$ ? ⇒ $2$ .
$T^{2}$ aur $r^{3}$ ko link karne wala Kepler constant? ⇒ $4 π^{2} / GM$ .
Geostationary radius (centre se)? ⇒ $\approx 4.23 \times 1 0^{7}$ m.
Moon ki orbital speed? ⇒ $\approx 1.02$ km/s.

Connections

Orbital velocity for circular orbit — derivation — parent derivation.
Newton's Law of Universal Gravitation
Centripetal Force and Uniform Circular Motion
Escape Velocity
Kepler's Third Law
Acceleration due to gravity g and GM = gR²
Geostationary Orbit