Exercises — Kepler's second law — equal areas in equal times, from angular momentum conservation

(a) Areal velocity $\frac{dA}{dt}$ — radius vector dwara unit time mein sweep ki gayi area — constant hoti hai. (b) Gravity ek central force hai ($\vec{F}\parallel \vec{r}$), isliye yeh Sun ke baare mein zero torque exert karti hai; zero torque ka matlab hai angular momentum $L$ conserved hai, aur $\frac{dA}{dt}=\frac{L}{2m}$ constant hai kyunki $L$ aur $m$ dono constant hain.

Perihelion par faster. Kyunki $\frac{1}{2}{r}^2\dot{\theta }$ fixed hai, chhota $r$ ek bada angular rate $\dot{\theta }$ force karta hai, isliye speed bhi badi hoti hai. Swept-area rate DONO points par SAME hai — yahi toh is law ka poora content hai. "Faster" (speed, jo differ karti hai) aur "area per second" (jo har jagah equal hai) ko confuse mat karo.

Dono points apsides hain, isliye $\vec{v}\perp \vec{r}$ aur har ek par $L=mvr$ hai. $L$ ka conservation deta hai $v_p{r}_p=v_a{r}_a\Rightarrow v_a=v_p\frac{r_p}{r_a}.$ Yeh step kyun? Sirf apsides par hum $\sin \varphi$ factor drop karke directly $L=mvr$ likh sakte hain. $v_a=58.98\times \frac{4.60\times 10^{10}}{6.98\times 10^{10}}=58.98\times 0.6591\approx 38.87\text{km/s}.$

Areal velocity $\frac{dA}{dt}=\frac{L}{2m}=\frac{8.4\times 10^{14}}{2}=4.2\times 10^{14}{\text{m}}^2/\text{s}$ hai. Ek ghanta $=3600\text{s}$ hai, aur kyunki $dA/dt$ constant hai, area $=$ rate $\times$ time: $A=4.2\times 10^{14}\times 3600=1.512\times 10^{18}{\text{m}}^2.$ Integral kyun nahi? Constant rate ka matlab hai total sirf rate times elapsed time hai — same reason jaise distance $=$ speed $\times$ time jab speed constant ho.

Cross product ka magnitude $|\vec{r}\times \vec{v}|=rv\sin \varphi$ hai. Toh $\frac{dA}{dt}=\frac{1}{2}rv\sin \varphi =\frac{1}{2}(2.0\times 10^{11})(2.5\times 10^4)\sin 50°.$ $\sin 50°=0.766$ ke saath: $\frac{dA}{dt}=\frac{1}{2}(2.0\times 10^{11})(2.5\times 10^4)(0.766)\approx 1.915\times 10^{15}{\text{m}}^2/\text{s}.$ $\cos \varphi$ kyun nahi, $\sin \varphi$ kyun? Cross product do vectors ka perpendicular piece measure karta hai — $\vec{v}$ ka woh part jo actually radius vector ko sideways swing karta hai aur area paint karta hai. Parallel part sirf $\vec{r}$ ko lengthen/shorten karta hai aur kuch nahi paint karta.

Speeds: apsides par $L=m{v}_p{r}_p=m{v}_a{r}_a$, toh $v_p{r}_p=v_a{r}_a$, jo deta hai $\frac{v_p}{v_a}=\frac{r_a}{r_p}$. Kinetic energies: $KE=\frac{1}{2}m{v}^2$, toh $\frac{K{E}_p}{K{E}_a}=\frac{\frac{1}{2}m{v}_p^2}{\frac{1}{2}m{v}_a^2}={\left(\frac{v_p}{v_a}\right)}^2={\left(\frac{r_a}{r_p}\right)}^2.$ Square kyun? Kinetic energy speed ke square ke saath scale karti hai, toh koi bhi speed ratio energy mein squared ratio ban jaata hai.

(a) Pehle $b=a\sqrt{1-e^2}=1.496\times 10^{11}\sqrt{1-0.0167^2}=1.496\times 10^{11}\times 0.99986=1.49579\times 10^{11}\text{m}$. $A=\pi ab=\pi (1.496\times 10^{11})(1.49579\times 10^{11})\approx 7.030\times 10^{22}{\text{m}}^2.$ (b) Ek poore period mein radius vector puri ellipse ek baar sweep karta hai, toh $A=\frac{L}{2m}T$. $L/m$ ke liye solve karo: $\frac{L}{m}=\frac{2A}{T}=\frac{2(7.030\times 10^{22})}{3.156\times 10^7}\approx 4.455\times 10^{15}{\text{m}}^2/\text{s}.$ Yeh kyun kaam karta hai: $dA/dt$ constant hai, toh ise ek orbit par integrate karna sirf $T$ se multiply karna hai — swept area poori ellipse ke barabar hai kyunki planet exactly ek period mein start par wapas aata hai.

Angular momentum har jagah conserved hai, toh perihelion par $L$ = naye point par $L$: $m{v}_p{r}_p=mrv\sin \varphi \Rightarrow v=\frac{v_p{r}_p}{r\sin \varphi }.$ Kyun: perihelion par $\varphi =90°$ toh $L=m{v}_p{r}_p$; general point par $L=mrv\sin \varphi$. Equal set karo (mass cancel ho jaata hai). $v=\frac{(9.0\times 10^3)(7.0\times 10^6)}{(1.0\times 10^7)(\sin 60°)}=\frac{6.30\times 10^{10}}{(1.0\times 10^7)(0.8660)}=\frac{6.30\times 10^{10}}{8.660\times 10^6}\approx 7.27\times 10^3\text{m/s}.$ Toh $v\approx 7.27\text{km/s}$ hai.

(a) Position 2 aphelion hai, ek apsis hai, toh ${\vec{v}}_2\perp {\vec{r}}_2$ aur $\frac{dA}{dt}=\frac{1}{2}{r}_2{v}_2=\frac{1}{2}(5.0\times 10^{11})(4.0\times 10^3)=1.0\times 10^{15}{\text{m}}^2/\text{s}.$ (b) Areal velocity constant hai, aur generally $\frac{dA}{dt}=\frac{1}{2}r{v}_{\perp }$ (sirf perpendicular part area paint karta hai). Position 1 par: $v_{\perp ,1}=\frac{2(dA/dt)}{r_1}=\frac{2(1.0\times 10^{15})}{8.0\times 10^{10}}=2.5\times 10^4\text{m/s}=25\text{km/s}.$ $v_{\perp }$ kyun, $v$ kyun nahi? Area sirf radius vector ke sideways swing se sweep hoti hai; woh exactly perpendicular velocity component hai. (c) Perpendicular component $v_{\perp ,1}=v_1\sin {\varphi }_1$ hai, toh $\sin {\varphi }_1=\frac{v_{\perp ,1}}{v_1}=\frac{25}{30}=0.8333\Rightarrow {\varphi }_1=\arcsin (0.8333)\approx 56.4°.$ Sanity check: ${\varphi }_1<90°$ ka matlab hai comet mein abhi bhi ek radial velocity component hai — woh andar ya bahar move kar raha hai, apsis par nahi, jo $r_1$ ke perihelion aur aphelion ke beech hone ke saath consistent hai.

Central force ka matlab hai $\vec{F}=f(r)\hat{r}$ kisi function $f$ ke liye — yeh hamesha $\pm \hat{r}$ ke along point karta hai. Tab centre ke baare mein torque hai $\vec{\tau }=\vec{r}\times \vec{F}=\vec{r}\times f(r)\hat{r}=f(r)(\vec{r}\times \hat{r})=\vec{0},$ kyunki $\vec{r}$ aur $\hat{r}$ parallel hain ($\vec{r}=r\hat{r}$, aur koi bhi vector khud ke saath cross zero hota hai). Zero torque deta hai $\frac{d\vec{L}}{dt}=\vec{\tau }=\vec{0}$, toh $L=\text{const}$, isliye $\frac{dA}{dt}=\frac{L}{2m}=\text{const}$. $■$

Shape kyun fix nahi hoti: Second law ne sirf yeh use kiya ki $f(r)$ radially point karta hai — usne $f(r)$ ki form kabhi use nahi ki. Alag radial laws ($1/r^2$ gravity, $-kr$ spring, etc.) alag orbit shapes dete hain; Sun-at-focus wali ellipse specifically inverse-square law require karti hai (woh Kepler's First Law (Ellipse) hai). Equal areas universal hai; ellipse special hai.