Specific angular momentum h = √(GMp)

3.2.12 · Physics › Orbital Mechanics & Astrodynamics

$h$ KYA hai?

$m$ se divide kyun karte hain? Kyunki two-body problem mein chhoti mass equation of motion se cancel ho jaati hai ($\ddot{\vec{r}}=-\frac{GM}{r^2}\hat{r}$). Dynamics sirf $\vec{r},\vec{v}$ par depend karti hai, isliye natural conserved quantities bhi per-mass hoti hain.

$h$ conserved kyun hai?

Scratch se derivation. $\frac{d\vec{h}}{dt}=\frac{d}{dt}(\vec{r}\times \vec{v})=\underset{=0}{\underbrace{\vec{v}\times \vec{v}}}+\vec{r}\times \vec{a}$ Yeh step kyun? Cross products ke liye product rule; $\vec{v}\times \vec{v}=0$ kyunki ek vector apne aap ke parallel hota hai.

Acceleration purely radial hai: $\vec{a}=-\frac{GM}{r^2}\hat{r}$. Toh $\vec{r}\times \vec{a}=-\frac{GM}{r^2}(\vec{r}\times \hat{r})=0$ Yeh step kyun? $\vec{r}=r\hat{r}$, aur $\hat{r}\times \hat{r}=0$. Isliye $\boxed{\frac{d\vec{h}}{dt}=0}\Rightarrow \vec{h}=\text{const.}$

Constant $\vec{h}$ ⟹ orbit ek fixed plane mein rehti hai (motion $\vec{h}$ ke perpendicular rehti hai) aur Kepler's 2nd law hold karta hai ($\frac{dA}{dt}=\frac{h}{2}$).

$h=\sqrt{GMp}$ kaise milta hai?

Humein orbit equation chahiye. Two-body problem ke conic-section solution se shuru karte hain (neeche ek baar derive kiya gaya hai).

$p=h^2/GM$ ki derivation (aur isliye $h=\sqrt{GMp}$)

Polar coordinates use karte hain. Radial equation of motion hai $\ddot{r}-r{\dot{\theta }}^2=-\frac{GM}{r^2}$ Kyun? Polar form mein standard acceleration; RHS gravity hai.

Angular momentum deta hai $h=r^2\dot{\theta }$ ($\vec{r}\times \vec{v}$ ka $z$-component). Toh $\dot{\theta }=h/r^2$.

Trick: $u=1/r$ substitute karo aur variable $\theta$ mein change karo. $\dot{r}=\frac{dr}{d\theta }\dot{\theta }=-\frac{1}{u^2}\frac{du}{d\theta }\cdot h{u}^2=-h\frac{du}{d\theta }$ Yeh step kyun? $r=1/u\Rightarrow \frac{dr}{d\theta }=-u^{-2}\frac{du}{d\theta }$, aur $\dot{\theta }=h{u}^2$.

Dobara differentiate karo: $\ddot{r}=-h\frac{d^{2}u}{d{\theta }^2}\dot{\theta }=-h^2{u}^2\frac{d^{2}u}{d{\theta }^2}$

Radial equation mein substitute karo (jahan $r{\dot{\theta }}^2=\frac{1}{u}(h{u}^2{)}^2=h^2{u}^3$): $-h^2{u}^2{u}^{\prime \prime }-h^2{u}^3=-GM{u}^2$ $-h^2{u}^2$ se divide karo: $u^{\prime \prime }+u=\frac{GM}{h^2}$ Yeh step kyun? Yeh ek clean linear ODE hai — Binet's equation.

Iska solution: particular $u_p=GM/h^2$, homogeneous $A\cos (\theta -{\theta }_0)$: $u=\frac{GM}{h^2}(1+e\cos \theta )\Rightarrow r=\frac{h^2/GM}{1+e\cos \theta }$

$r=\frac{p}{1+e\cos \theta }$ se compare karo: $p=\frac{h^2}{GM}⟹\boxed{h=\sqrt{GMp}}$

$p$ ka Geometric Meaning

Key checkpoints (sab $r=\frac{p}{1+e\cos \theta }$ se):

• Periapsis ($\theta =0$): $r_p=\frac{p}{1+e}$
• Apoapsis ($\theta =180^{\circ }$): $r_a=\frac{p}{1-e}$
• Harmonic-mean relation: $\frac{2}{p}=\frac{1}{r_p}+\frac{1}{r_a}$.

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• Two-body problem — jahan se orbit equation $r=p/(1+e\cos \theta )$ aati hai.
• Kepler's Second Law — $dA/dt=h/2$ constant $h$ ka direct consequence hai.
• Specific orbital energy — $h$ ke saath pair karta hai orbit ko fully fix karne ke liye ($a$ energy se, $e$ dono se).
• Eccentricity vector — $\vec{e}=\frac{\vec{v}\times \vec{h}}{GM}-\hat{r}$ directly $\vec{h}$ use karta hai.
• Flight-path angle — $h=rv\sin \varphi$ mein $\sin \varphi$ explain karta hai.
• Vis-viva equation — $v^2=GM(2/r-1/a)$ orbit ke along speeds dene ke liye $h$ se combine hota hai.

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