Orbit equation r = p - (1 + e·cos θ) — derivation from equations of motion

3.2.3 · Physics › Orbital Mechanics & Astrodynamics

Hum kya derive kar rahe hain

KAISE: first principles se derivation

Step 1 — Relative motion ke liye Newton's law

Newton's gravity orbiting body ka acceleration central mass ki taraf deta hai: $\ddot{r} = - \frac{μ}{r ^{2}} \overset{r}{^}, μ \equiv GM .$

Yeh step kyun? Yeh bas $F = m a$ hai jisme $F = GM m / r^{2}$ andar ki taraf point karta hai ( $- \overset{r}{^}$ ); $m$ se divide karne par orbiting mass hat jaati hai, isliye sab bodies ek hi path follow karti hain diye gaye $μ$ ke liye.

Step 2 — Angular momentum conserved hai (central force)

Specific angular momentum hai $h = r \times \dot{r}$ . Differentiate karo: $\frac{d h}{d t} = \dot{r} \times \dot{r} + r \times \ddot{r} = 0 + r \times (- \frac{μ}{r ^{2}} \overset{r}{^}) = 0.$

Yeh step kyun? $r$ aur $\overset{r}{^}$ parallel hain, isliye unka cross product zero hai. Isliye $h$ constant hai ⇒ motion planar hai, aur iska magnitude hai $h = r^{2} \dot{θ} = const.$

Step 3 — Substitution $u = 1/ r$ (woh trick jo ise linearize karti hai)

Plane mein polar $(r, θ)$ ke saath kaam karte hue, radial acceleration hai $\overset{r}{¨} - r \dot{θ}^{2} = - \frac{μ}{r ^{2}} .$

Hum $r$ ko $t$ ki jagah $θ$ ka function banana chahte hain. $h = r^{2} \dot{θ}$ use karo, to $\dot{θ} = h u^{2}$ jahan $u = 1/ r$ .

Time derivatives ko $θ$ derivatives mein convert karo. Kyun? Kyunki $θ$ monotonic hai (hamesha increase karta hai), isliye yeh ek clean independent variable hai. $\overset{r}{˙} = \frac{d r}{d θ} \dot{θ} = \frac{d}{d θ} (\frac{1}{u}) h u^{2} = - \frac{1}{u ^{2}} \frac{d u}{d θ} \cdot h u^{2} = - h \frac{d u}{d θ} .$ Dobara differentiate karo: $\overset{r}{¨} = \frac{d}{d t} (- h \frac{d u}{d θ}) = - h \frac{d ^{2} u}{d θ ^{2}} \dot{θ} = - h \frac{d ^{2} u}{d θ ^{2}} h u^{2} = - h^{2} u^{2} \frac{d ^{2} u}{d θ ^{2}} .$

Yeh step kyun? $u$ -substitution nasty $1/ r^{2}$ force ko ek constant mein badal deta hai, ek mushkil nonlinear ODE ko ek simple linear ODE mein convert kar deta hai.

Step 4 — Linear oscillator equation

$\overset{r}{¨}$ aur $r \dot{θ}^{2} = \frac{1}{u} (h u^{2})^{2} = h^{2} u^{3}$ ko radial equation mein plug karo: $- h^{2} u^{2} \frac{d ^{2} u}{d θ ^{2}} - h^{2} u^{3} = - μ u^{2} .$ $- h^{2} u^{2}$ se divide karo: $\frac{d ^{2} u}{d θ ^{2}} + u = \frac{μ}{h ^{2}}$

Yeh step kyun? Yeh ek simple harmonic oscillator ki equation hai jisme constant driving term $μ / h^{2}$ hai. Hum iska solution zubani yaad rakhte hain.

Step 5 — Solve karo aur interpret karo

General solution = particular + homogeneous: $u (θ) = \frac{μ}{h ^{2}} + A cos (θ - θ_{0}) .$ $θ_{0} = 0$ choose karo (angle wahan se measure karo jahan $r$ sabse chhota ho, yaani perihelion). Tab $\frac{1}{r} = \frac{μ}{h ^{2}} (1 + e cos θ), e \equiv \frac{A h ^{2}}{μ} .$ Invert karo: $r = \frac{h ^{2} / μ}{1 + e cos θ} = \frac{p}{1 + e cos θ}$

Yeh step kyun? $A$ ek integration constant hai jo initial conditions se set hota hai; ise dimensionless $e$ mein bundle karne se standard conic form milti hai. $p = h^{2} / μ$ automatically nikal aata hai.

Figure — Orbit equation r = p - (1 + e·cos θ) — derivation from equations of motion

Worked examples

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (answers cover kar lo!)

Kaun sa conserved quantity orbit ko planar banata hai? → angular momentum $h$ .
Equation ko linearize karne ke liye kaun sa substitution use hota hai? → $u = 1/ r$ .
Resulting ODE kya hai? → $u^{''} + u = μ / h^{2}$ .
$θ = 9 0^{\circ}$ par $r$ kya hai? → $r = p$ .

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho tum ek stretchy string par ek ball apne sar ke upar ghumate ho, lekin string sirf andar ki taraf khinchti hai, kabhi sideways nahi. Kyunki yeh sirf tumhare haath ki taraf khinchti hai, ball kabhi apni "spin amount" gain ya lose nahi kar sakti (wahi angular momentum hai) — isliye woh motion ke ek flat circle mein rehti hai. Gravity exactly wahi andar-sirf pull hai. Jab hum math karte hain, ball ka path ek perfectly smooth oval nikalta hai (ya circle, ya open swoosh). Formula $r = p / (1 + e cos θ)$ bas ek recipe hai: mujhe batao $θ$ angle tum kahan ho, aur yeh batata hai $r$ Sun se kitni door ho. Number $e$ decide karta hai oval mota hai ya patla; $p$ decide karta hai sab kitna bada hai.

Flashcards

Kaun si force property orbital angular momentum ko constant banati hai?

Yeh ek central force hai (

F ∥ \overset{r}{^}

), isliye

r \times \ddot{r} = 0

aur

h

conserved hai.

Semi-latus rectum

p

h

aur

μ

ke terms mein define karo.

p = h^{2} / μ

Radial equation of motion ko linearize karne ke liye kaun sa substitution use hota hai?

u = 1/ r

, jisse

u^{''} + u = μ / h^{2}

milta hai.

Full orbit equation kya hai?

r = p / (1 + e cos θ)

Is convention mein perihelion (closest approach) kahan hai?

θ = 0

par, jahan

r_{m i n} = p / (1 + e)

True anomaly

θ = 9 0^{\circ}

par

r

kya hai?

r = p

(semi-latus rectum).

p

aur semi-major axis

a

ko relate karo.

p = a (1 - e^{2})

Kaun si eccentricity parabola deti hai?

e = 1

Constant

h = r^{2} \dot{θ}

physically kya encode karta hai?

Areal velocity ka do guna (Kepler's 2nd law: equal areas in equal times).

Orbit equation mein

r

kis point se measure hoti hai?

Focus se (central mass ki location), center se nahi.

Connections

Kepler's Laws — orbit equation hi first law hai; $h =$ const hi second law hai.
Conic Sections — $r = p / (1 + e cos θ)$ focus ke baare mein polar conic hai.
Specific Angular Momentum h — $p$ aur area-sweep rate set karta hai.
Vis-viva Equation — energy version jo is geometry version ke saath pair hoti hai.
Eccentricity and Orbital Energy — $e = 1 + 2 E h^{2} / μ^{2}$ shape ko energy se link karta hai.
Central Force Motion — woh general framework jiska yeh ek special case hai.

$e$	shape
$0$	circle
$0 < e < 1$	ellipse
$1$	parabola
$> 1$	hyperbola