Circular orbit — velocity, period, energy — Hinglish notes

3.2.13 · Physics › Orbital Mechanics & Astrodynamics

KYU ek circular orbit ko ek specific speed chahiye?

KYA hona chahiye? Constant speed par circular motion ke liye, net force ek centripetal force honi chahiye — center ki taraf point karte hue, magnitude $\frac{m v ^{2}}{r}$ ke saath. Available sirf gravity hai. Toh gravity ko exactly centripetal requirement supply karni padegi. Na zyada, na kam. Usi ek balance se speed fix ho jaati hai.

KAISE derive karte hain orbital velocity (scratch se)

Step 1 — Do forces likho jo equal honi chahiye.

$m$ par gravity: $F_{g} = \frac{GM m}{r ^{2}}$

Yeh step kyun? Newton ka law of gravitation — distance $r$ par inward pull.

Uniform circular motion ke liye centripetal requirement: $F_{c} = \frac{m v ^{2}}{r}$

Yeh step kyun? Koi bhi object jo radius $r$ ke circle mein speed $v$ se move kare usse yeh inward force chahiye; yeh kinematics hai, koi naya force nahi.

Step 2 — Inhe equal karo (gravity hi centripetal force hai): $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$

Step 3 — Cancel karo aur solve karo. $m$ cancel karo (toh answer satellite ke mass se independent ho jaata hai!) aur $r$ ki ek power: $\frac{GM}{r} = v^{2}$

KAISE period $T$ nikaalte hain

Step 1 — Distance / speed. Ek orbit mein satellite circumference $2 π r$ travel karta hai speed $v$ par: $T = \frac{2 π r}{v}$

Yeh step kyun? Period = ek chakkar ka time = (ek chakkar ki distance) / (speed). Bilkul definition hai.

Step 2 — $v = GM / r$ substitute karo: $T = \frac{2 π r}{GM / r} = 2 π r \cdot \frac{r}{GM} = 2 π \frac{r ^{3}}{GM}$

KAISE energies nikaalte hain

Kinetic energy. Bas $v^{2} = GM / r$ ko $K = \frac{1}{2} m v^{2}$ mein plug karo: $K = \frac{1}{2} m \frac{GM}{r} = \frac{GM m}{2 r}$

Potential energy. Gravitational PE (infinity par zero, bound hone par negative): $U = - \frac{GM m}{r}$ Negative kyun aur $1/ r$ kyun? $U$ woh kaam hai jo $m$ ko infinity se laane mein hota hai; andar kheenchne se energy release hoti hai, toh bound states zero ke neeche hoti hain. ( $U = - \int_{\infty}^{r} F_{g} d r$ se derive ho sakta hai.)

Total mechanical energy. Dono add karo: $E = K + U = \frac{GM m}{2 r} - \frac{GM m}{r} = - \frac{GM m}{2 r}$

Figure — Circular orbit — velocity, period, energy

Worked Examples

Common mistake Steel-manned misconceptions

"Higher orbit → faster, kyunki uski zyada energy hai." Yeh sahi kyun lagta hai: higher orbit ki total energy $E$ zyada hoti hai (kam negative) aur badi PE, toh "zyada energy = zyada speed" natural lagta hai. Fix: Extra energy potential energy mein gayi. Kinetic energy actually girती hai: $K = \frac{GM m}{2 r}$ $r$ badhne par ghaṭti hai, aur $v = GM / r$ bhi ghaṭti hai. Higher = slower lekin zyada total energy.

"Satellite ka mass speed ke liye matter karta hai." Yeh sahi kyun lagta hai: bhari cheezein zyada gravity feel karti hain. Fix: Unhe zyada centripetal force bhi chahiye, aur $m$ $GM / r^{2} = v^{2} / r$ mein cancel ho jaata hai. Speed aur period $m$ se independent hain. ( $K$ , $U$ , $E$ zaroor $m$ ke saath scale karte hain.)

" $U = m g h$ orbits ke liye kaam karta hai." Yeh sahi kyun lagta hai: yeh school physics ka formula hai. Fix: $m g h$ flat-Earth, constant- $g$ approximation hai. Orbital distances par $g$ vary karta hai, toh tumhe zaroor $U = - GM m / r$ use karna chahiye.

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek ball ko bahut, bahut tezi se sideways phenko. Normally woh girta hai aur zameen se takraata hai. Lekin Earth round hai, toh zameen curve karti hai. Agar ball ko bilkul sahi speed par phenko (lagbhag har second 8 km!), toh zameen exactly utni hi tezi se curve karti hai jitna ball girta hai — toh ball Earth ko "miss" karta rehta hai aur hamesha ke liye uske around girta rehta hai. Yahi orbit hai: kabhи land kiye bina girna. Ise upar se phenko aur yeh slower ja sakta hai, kyunki wahan gravity kamzor hoti hai. Tez-girne wale hisse aur gravity-kheenchne wale hisse ko perfectly balance karna hota hai — yahi ek speed hai jo har height par kaam karti hai.

Flashcards

What balance defines a circular orbit?

Gravity supplies exactly the centripetal force:

GM m / r^{2} = m v^{2} / r

Formula for circular orbital speed?

v = GM / r

(independent of orbiting mass

m

Why does satellite mass not affect the speed or period?

m

cancels in

GM m / r^{2} = m v^{2} / r

Formula for the period of a circular orbit?

T = 2 π r^{3} / GM

, i.e.

T^{2} \propto r^{3}

(Kepler III).

Kinetic energy of a circular orbit?

K = + GM m / (2 r)

Potential energy of a circular orbit?

U = - GM m / r

Total mechanical energy of a circular orbit?

E = - GM m / (2 r)

(negative ⇒ bound).

Relations between K, U, E?

E = - K

and

U = 2 E = - 2 K

(virial theorem for

1/ r

As orbit radius increases, what happens to v, T, E?

v decreases, T increases, E increases (less negative).

Does a higher orbit move faster or slower?

Slower (

v \sim r^{- 1/2}

), but it has more total energy.

Energy to move between circular orbits

r_{1} \to r_{2}

Δ E = \frac{GM m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}})

Why is

E < 0

physically meaningful?

Bound orbit; you'd need to add

+ GM m /2 r

to reach escape (

E = 0

Connections

Newton's Law of Universal Gravitation — woh force jo hum centripetal ke equal set karte hain.
Centripetal Force and Uniform Circular Motion — $m v^{2} / r$ supply karta hai.
Kepler's Three Laws — $T^{2} \propto r^{3}$ yahan ek special case hai.
Gravitational Potential Energy — $U = - GM m / r$ ka origin.
Escape Velocity — $E = 0$ boundary; $v_{esc} = 2 v_{c i r c}$ .
Elliptical Orbits and the Vis-viva Equation — $v$ , $E$ ko non-circular orbits tak generalize karta hai.
Virial Theorem — explain karta hai $U = 2 E = - 2 K$ .