Exercises — Specific angular momentum h = √(GMp)

WHAT: circle matlab $e=0$, isliye $p=a=r$. WHY: jab orbit perfect circle hoti hai, semi-latus rectum radius mein collapse ho jaata hai. $h=\sqrt{GMp}=\sqrt{GMr}=\sqrt{3.986\times 10^{14}\times 6.6\times 10^6}$ $=\sqrt{2.631\times 10^{21}}\approx 5.13\times 10^{10}{\mathrm{m}}^2/s.$

Units hain ${\mathrm{m}}^2/s$. Kyunki $h$ angular momentum per unit mass hai ($h=L/m$); humne deliberately orbiting body ki mass divide kar di hai, isliye $\mathrm{kg}$ cancel ho jaata hai.

Direct plug-in, $p$ ke aage koi shape info nahi chahiye — yahi to formula ki puri power hai. $h=\sqrt{3.986\times 10^{14}\times 1.30\times 10^7}=\sqrt{5.182\times 10^{21}}\approx 7.20\times 10^{10}{\mathrm{m}}^2/s.$

WHAT: periapsis par velocity purely tangential hoti hai, isliye $\vec{v}\perp \vec{r}$ aur $\varphi =90^{\circ }$. WHY: tab $\sin \varphi =1$, isliye $h=r_p{v}_p$ bina kisi angle loss ke. $h=r_p{v}_p=8.0\times 10^6\times 9.0\times 10^3=7.20\times 10^{10}{\mathrm{m}}^2/s.$ $h=\sqrt{GMp}$ ko invert karo: $p=\frac{h^2}{GM}=\frac{(7.20\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{5.184\times 10^{21}}{3.986\times 10^{14}}\approx 1.30\times 10^7\mathrm{m}.$

Pehle $p=a(1-e^2)$ use karo — ellipse ki half-width, $a$ khud nahi. $p=10^7\times (1-0.09)=9.10\times 10^6\mathrm{m}.$ $h=\sqrt{GMp}=\sqrt{3.986\times 10^{14}\times 9.10\times 10^6}=\sqrt{3.627\times 10^{21}}\approx 6.02\times 10^{10}{\mathrm{m}}^2/s.$

Orbit equation $r=\frac{p}{1+e\cos \theta }$ ko $p$ ke liye rearrange karo: $p=r(1+e\cos \theta )=9.0\times 10^6(1+0.20\times \cos 60^{\circ }).$ Kyunki $\cos 60^{\circ }=0.5$: $p=9.0\times 10^6\times (1+0.10)=9.90\times 10^6\mathrm{m}.$ $h=\sqrt{3.986\times 10^{14}\times 9.90\times 10^6}=\sqrt{3.946\times 10^{21}}\approx 6.28\times 10^{10}{\mathrm{m}}^2/s.$

Step 1 (WHAT): harmonic-mean relation $\frac{2}{p}=\frac{1}{r_p}+\frac{1}{r_a}$ use karo. WHY: yeh hume pehle $a$ ya $e$ nikale bina do apsis radii se seedha $p$ dila deta hai. $\frac{2}{p}=\frac{1}{7.0\times 10^6}+\frac{1}{2.1\times 10^7}=1.4286\times 10^{-7}+4.762\times 10^{-8}=1.9048\times 10^{-7}.$ $p=\frac{2}{1.9048\times 10^{-7}}=1.050\times 10^7\mathrm{m}.$ Step 2: eccentricity $r_p=\frac{p}{1+e}\Rightarrow e=\frac{p}{r_p}-1$ se: $e=\frac{1.050\times 10^7}{7.0\times 10^6}-1=1.500-1=\mathrm{0.500.}$ Step 3: $h=\sqrt{GMp}=\sqrt{3.986\times 10^{14}\times 1.050\times 10^7}=\sqrt{4.185\times 10^{21}}\approx 6.47\times 10^{10}{\mathrm{m}}^2/s.$

Dono apsides par $\vec{v}\perp \vec{r}$, isliye $h=rv$ aur $v=h/r$. $v_p=\frac{h}{r_p}=\frac{6.47\times 10^{10}}{7.0\times 10^6}\approx 9.24\times 10^{3}m/s.$ $v_a=\frac{h}{r_a}=\frac{6.47\times 10^{10}}{2.1\times 10^7}\approx 3.08\times 10^{3}m/s.$ Sanity check: $v_p/v_a=r_a/r_p=3.0$ — body wahan tez chalti hai jahan orbit tight hai, exactly jaisa constant $h$ demand karta hai.

WHAT: yahan $\vec{v}$, $\vec{r}$ ke perpendicular nahi hai, isliye hume poora $h=rv\sin \varphi$ use karna hoga. WHY: cross product $|\vec{r}\times \vec{v}|=rv\sin \varphi$ sirf $\vec{v}$ ka woh component pick karta hai jo $\vec{r}$ ke perpendicular hai. Dekho Flight-path angle. $h=rv\sin \varphi =10^7\times 6000\times \sin 30^{\circ }=10^7\times 6000\times 0.5=3.0\times 10^{10}{\mathrm{m}}^2/s.$ $p=\frac{h^2}{GM}=\frac{(3.0\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{9.0\times 10^{20}}{3.986\times 10^{14}}\approx 2.258\times 10^6\mathrm{m}.$

Step 1: $h$ se $p$ nikalo: $p=\frac{h^2}{GM}=\frac{(6.0\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{3.6\times 10^{21}}{3.986\times 10^{14}}=9.031\times 10^6\mathrm{m}.$ Step 2: $p=a(1-e^2)$ ko $e$ ke liye invert karo: $1-e^2=\frac{p}{a}=\frac{9.031\times 10^6}{1.2\times 10^7}=0.7526\Rightarrow e^2=0.2474\Rightarrow e\approx \mathrm{0.497.}$ Step 3: $r_p=a(1-e)=1.2\times 10^7\times (1-0.497)=6.03\times 10^6\mathrm{m};r_a=a(1+e)=1.2\times 10^7\times 1.497=1.797\times 10^7\mathrm{m}.$

WHAT: Kepler's second law kehta hai ki radius vector equal time mein equal areas sweep karta hai, rate $\frac{dA}{dt}=\frac{h}{2}$ par. Dekho Kepler's Second Law. $\frac{dA}{dt}=\frac{h}{2}=\frac{6.0\times 10^{10}}{2}=3.0\times 10^{10}{\mathrm{m}}^2/s.$ Total area: $b=a\sqrt{1-e^2}=1.2\times 10^7\times \sqrt{0.7526}=1.2\times 10^7\times 0.8675=1.041\times 10^7\mathrm{m}.$ $A=\pi ab=\pi \times 1.2\times 10^7\times 1.041\times 10^7=3.925\times 10^{14}{\mathrm{m}}^2.$ Quarter area: $A/4=9.812\times 10^{13}{\mathrm{m}}^2.$ $t=\frac{A/4}{dA/dt}=\frac{9.812\times 10^{13}}{3.0\times 10^{10}}\approx 3.27\times 10^3\mathrm{s}(\approx 54.5\text{min}).$

Circular speed gravity aur centripetal requirement ko balance karne se aati hai: $\frac{GM}{r^2}=\frac{v^2}{r}\Rightarrow v_{\text{circ}}=\sqrt{GM/r}$. Tab $v_{\text{circ}}r=\sqrt{GM/r}\cdot r=\sqrt{GMr}=h$. Same cheez hai. ✓ Numbers: $v_{\text{circ}}=\sqrt{3.986\times 10^{14}/7.0\times 10^6}=\sqrt{5.694\times 10^7}=7546m/s.$ $h=v_{\text{circ}}r=7546\times 7.0\times 10^6=5.28\times 10^{10}{\mathrm{m}}^2/s$, jo $\sqrt{GMr}$ se match karta hai.

Step 1: periapsis par $\vec{v}\perp \vec{r}$, isliye $h=r_p{v}_p=1.5\times 10^{11}\times 5.0\times 10^4=7.5\times 10^{15}{\mathrm{m}}^2/s.$ Step 2: $p=\frac{h^2}{G{M}_{\odot }}=\frac{(7.5\times 10^{15}{)}^2}{1.327\times 10^{20}}=\frac{5.625\times 10^{31}}{1.327\times 10^{20}}\approx 4.239\times 10^{11}\mathrm{m}.$ Step 3: $r_p=\frac{p}{1+e}$ se, $e=\frac{p}{r_p}-1=\frac{4.239\times 10^{11}}{1.5\times 10^{11}}-1=2.826-1=\mathrm{1.826.}$ Kyunki $e>1$, orbit indeed hyperbolic hai — comet bound nahi hai aur escape kar jaayega.

Same $h$ ⟹ same $p=\frac{h^2}{GM}=\frac{(5.0\times 10^{10}{)}^2}{3.986\times 10^{14}}=\frac{2.5\times 10^{21}}{3.986\times 10^{14}}=6.272\times 10^6\mathrm{m}.$ Circle ($e=0$): $a_1=p=6.272\times 10^6\mathrm{m}.$ Ellipse ($e=0.6$): $a_2=\frac{p}{1-e^2}=\frac{6.272\times 10^6}{1-0.36}=\frac{6.272\times 10^6}{0.64}=9.800\times 10^6\mathrm{m}.$ Ellipse ka $a$ bada hai. Key insight: equal $h$ half-width $p$ fix karta hai, length $a$ nahi. Eccentricity add karne se (fixed $p$ ke saath) orbit major axis ke along bahar ki taraf stretch ho jaati hai, isliye $a$, $1/(1-e^2)$ ki tarah badhta hai.

Periapsis par $h=r_{p}v$ (kyunki $\vec{v}\perp \vec{r}$). $r_p$ fixed hone par, $h$ directly $v$ ke proportional hai, isliye $v_2>v_1\Rightarrow h_2>h_1$. Tab $p=\frac{h^2}{GM}$, $h$ ke saath badhta hai, isliye $p_2>p_1$. Kyunki $h$ aur $p$ saath badhte hain (yahi exactly $h=\sqrt{GMp}$ hai), same closest point par zyada tez dhakka ek wider, higher-energy orbit produce karta hai. Yahi reason hai ki periapsis par prograde burn orbit ke far side ko raise karta hai — yeh $h$ aur isliye $p$ ko pump up karta hai. Links to Vis-viva equation aur Specific orbital energy.