Specific orbital energy ε = −GM - 2a — Hinglish notes

3.2.11 · Physics › Orbital Mechanics & Astrodynamics

WHAT is specific orbital energy?

"Specific" kyun? Hum $m$ se divide karte hain taaki satellite ki apni mass cancel ho jaaye — ek pankh aur ek patthar, agar same orbit mein hain, toh dono ka same $ε$ hoga. Yahi wajah hai ki sab cheezein ek hi rate se girती hain.

WHY is total energy constant? (first principles)

Gravity ek conservative force hai: kiya gaya kaam sirf start/end positions par depend karta hai, path par nahi. Isliye kinetic + potential energy conserved rehti hai.

Kinetic energy per mass: $\frac{1}{2} v^{2}$ .
Potential energy per mass: $- \frac{GM}{r}$ (negative, infinity par zero — dekho Mistake 1).

Inका sum $ε$ free orbital motion mein kabhi nahi badalta. Yeh kyun matter karta hai: jab satellite andar girta hai ( $r ↓$ ), PE girti hai, toh KE badhti hai ( $v ↑$ ) — perigee par speed up hota hai, apogee par slow down, dono ko exchange karta hua jabki total same rehta hai.

DERIVATION: kyun $ε = - GM /2 a$

Humein do cheezein chahiye. Dono ko hum derive karenge.

Step 1 — Perigee aur apogee par energy. Perigee ( $r_{p}$ ) aur apogee ( $r_{a}$ ) par velocity radius ke perpendicular hoti hai, isliye saari speed "useful" hai. Energy dono jagah same hai: $\frac{v _{p}^{2}}{2} - \frac{GM}{r _{p}} = \frac{v _{a}^{2}}{2} - \frac{GM}{r _{a}} .$ Yeh step kyun? $ε$ constant hai, isliye main do convenient points par isko equate kar sakta hoon.

Step 2 — Angular momentum conserve karo. Un dono points par $v ⊥ r$ hai, isliye $L / m = h = r_{p} v_{p} = r_{a} v_{a}$ . Isse: $v_{a} = v_{p} \frac{r _{p}}{r _{a}} .$ Yeh step kyun? Central force se torque zero hota hai, isliye $h$ conserved hai — speeds ko link karne ke liye ek aur equation milti hai.

Step 3 — $v_{p}^{2}$ solve karo. $v_{a}$ substitute karo: $\frac{v _{p}^{2}}{2} - \frac{GM}{r _{p}} = \frac{v _{p}^{2}}{2} \frac{r _{p}^{2}}{r _{a}^{2}} - \frac{GM}{r _{a}} .$ $v_{p}^{2}$ collect karo: $\frac{v _{p}^{2}}{2} (1 - \frac{r _{p}^{2}}{r _{a}^{2}}) = GM (\frac{1}{r _{p}} - \frac{1}{r _{a}}) .$ Left bracket factor hota hai: $1 - \frac{r _{p}^{2}}{r _{a}^{2}} = \frac{( r _{a} - r _{p} ) ( r _{a} + r _{p} )}{r _{a}^{2}}$ . Right bracket $= \frac{r _{a} - r _{p}}{r _{p} r _{a}}$ . $(r_{a} - r_{p})$ cancel karo: $\frac{v _{p}^{2}}{2} \cdot \frac{r _{a} + r _{p}}{r _{a}^{2}} = \frac{GM}{r _{p} r _{a}} \Rightarrow v_{p}^{2} = \frac{2 GM r _{a}}{r _{p} ( r _{a} + r _{p} )} .$ Yeh step kyun? Algebraic cleanup se perigee speed sirf geometry ke terms mein alag ho jaati hai.

Step 4 — $ε$ mein plug karo. $ε = \frac{v _{p}^{2}}{2} - \frac{GM}{r _{p}} = \frac{GM r _{a}}{r _{p} ( r _{a} + r _{p} )} - \frac{GM}{r _{p}} = \frac{GM}{r _{p}} (\frac{r _{a} - ( r _{a} + r _{p} )}{r _{a} + r _{p}}) = \frac{GM}{r _{p}} \cdot \frac{- r _{p}}{r _{a} + r _{p}} .$ $ε = - \frac{GM}{r _{a} + r _{p}} .$ Yeh step kyun? Substitute karne se $v_{p}$ khatam ho jaata hai aur sirf pure geometry bachti hai.

Step 5 — $a$ ki definition use karo. Ellipse ki geometry se, perigee $+$ apogee distances major axis span karti hain: $r_{p} + r_{a} = 2 a$ . Isliye $ε = - \frac{GM}{2 a} . ■$

Figure — Specific orbital energy ε = −GM - 2a

Vis-viva equation (ek free bonus)

$ε$ ke dono formulas ko kisi bhi point $r$ par speed $v$ ke saath equate karo: $\frac{v ^{2}}{2} - \frac{GM}{r} = - \frac{GM}{2 a} \Rightarrow v^{2} = GM (\frac{2}{r} - \frac{1}{a}) .$ Yeh useful kyun hai: sirf $r$ aur $a$ se kisi bhi jagah speed milti hai — angle track karne ki zaroorat nahi.

$ε$ ke sign se orbit type

Worked examples

Common mistakes (steel-manned)

Recall Feynman: ek 12-saal ke bacche ko explain karo

Ek satellite ka ek "energy bank account" hota hai. Kuch paisa motion (speed) hai aur kuch height (Earth se kitna door) hai. Jab woh paas aake ghooma, usne height-money kharch kiya lekin speed-money kamaya; door jaate waqt ulta hua. Total balance kabhi nahi badalta. Aur yahan magic hai: woh total balance sirf satellite kitna bada loop maarta hai uss par depend karta hai — bada loop matlab zyada ameer (kam negative) account. Agar balance kabhi zero ho jaaye, toh satellite ke paas bilkul itni energy hai ki woh Earth ko hamesha ke liye chhod sake.

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Specific orbital energy kya hoti hai (words mein)?

Orbiting body ki total mechanical energy per unit mass; orbit mein constant rehti hai.

ε

ka formula

a

ke terms mein?

ε = - GM /2 a

ε

ka formula

v

aur

r

ke terms mein?

ε = \frac{v ^{2}}{2} - \frac{GM}{r}

ε

orbit mein constant kyun rehti hai?

Gravity conservative hai, isliye total energy conserved hoti hai.

ε

kis par depend karti hai (aur kis par NAHI)?

Sirf

a

par (aur

GM

par); position, kisi ek point ki speed, ya body ki mass par nahi.

Vis-viva equation batao.

v^{2} = GM (\frac{2}{r} - \frac{1}{a})

ε < 0

= 0

> 0

se kaunsa orbit type milta hai?

Ellipse (bound), parabola (escape), hyperbola (unbound).

Radius

r

se escape speed?

v_{esc} = 2 GM / r

(

ε = 0

set karo).

Circular orbit mein

a

aur

r

ka kya relation hai?

a = r

, aur

v = GM / r

Hyperbola ke liye

a < 0

kyun hona chahiye?

a = - GM /2 ε

aur

ε > 0

, isliye

a

negative hota hai.

ε

ko perigee aur apogee se express karo.

ε = - GM / (r_{p} + r_{a})

kyunki

r_{p} + r_{a} = 2 a

Mass se divide karke "specific" energy kyun lete hain?

Body ki mass cancel ho jaati hai, isliye same orbit mein saari bodies ek hi number share karti hain.

Connections

Vis-viva equation — seedha $ε$ se derive hota hai.
Kepler's Third Law — $T^{2} \propto a^{3}$ , wahi $a$ jo energy govern karta hai.
Conservation of angular momentum — derivation mein use hua.
Escape velocity — $ε = 0$ ka limiting case.
Hohmann transfer orbit — burns size karne ke liye energy/vis-viva use karta hai.
Conic sections in orbits — $ε$ ka sign ↔ ellipse/parabola/hyperbola.
Gravitational potential energy −GM/r — PE term ka origin.