Question bank — Specific orbital energy ε = −GM - 2a

False. $\epsilon$ energy per unit mass hai, isliye $m$ cancel ho jaata hai; ek feather aur ek boulder jo identical orbit mein hain, dono ek hi value $-GM/2a$ share karte hain.

False. Speed akela kuch nahi bataata — yeh usi orbit ke perigee par sabse zyada aur apogee par sabse kam hoti hai. Sirf sum $\frac{v^2}{2}-\frac{GM}{r}$ (jo ki $\epsilon$ hai) orbits ko energy ke hisaab se rank karta hai.

False. Negative sign ek convention hai: PE ko infinity par zero define kiya gaya hai, isliye koi bhi finite $r$ us reference se neeche hota hai. Kuch khoya nahi — iska matlab sirf yeh hai ki infinity tak climb karne ke liye kaam karna padega.

True. Ek circle mein $r_p=r_a=r$ hota hai, isliye $a=\frac{1}{2}(r_p+r_a)=r$. Yeh ek maatra case hai jahan "$a$ is the radius" sahi hai.

False. $v=\sqrt{GM/r}$ se, bada $r$ chhota $v$ deta hai. Uski total energy zyada hoti hai lekin woh use height (PE) ke roop mein store karti hai, speed (KE) kam rehti hai.

False. $\epsilon =0$ ek parabola hai ($a\to \infty$, ek open escape trajectory), na ki ek closed circle. Body infinity tak coast karti hai aur zero speed ke saath pahunchti hai.

True. $\epsilon >0$ ke liye, $\epsilon =-GM/2a$ solve karne par $a=-GM/2\epsilon <0$ milta hai. Negative $a$ ek unbound path ka formal fingerprint hai (dekho Conic sections in orbits).

False. Dono continuously change karte rehte hain (woh ek doosre ke saath trade karte hain); sirf unka sum $\epsilon$ fixed rehta hai.

True. $|\epsilon |=GM/2a$, $a$ ke inversely proportional hai, isliye $a\to 2a$ karne par $|\epsilon |$ aadha ho jaata hai (yaani $\epsilon$ zero ke paas aa jaata hai, kam tightly bound hota hai).

Error yeh hai ki $a$ (poore ellipse ki ek fixed geometric property) aur $r$ (moving position) ko confuse kiya ja raha hai. $a$ free flight ke dauran kabhi nahi badalta, isliye $\epsilon$ ek single constant hai, position se independent.

Escape speed zero leftover speed ke saath infinity tak pahunchti hai, kisi orbit tak nahi. Yeh $\epsilon =0$ set karne se milti hai: $v_{\text{esc}}=\sqrt{2GM/r}$ (dekho Escape velocity). Ek stable circular orbit ke liye kam speed chahiye, $\sqrt{GM/r}$.

Circle ke liye $r=a$ set hota hai (na ki $0$), jisse $v^2=GM(\frac{2}{a}-\frac{1}{a})=GM/a$ milta hai. $r=0$ set karne ka matlab hoga centre par baithna — physically impossible hai aur infinity deta hai.

Gravity conservative hai, isliye total energy conserve hoti hai — yeh ghatti nahi hai. Inward girne par PE, KE mein convert hoti hai, lekin sum $\epsilon$ unchanged rehta hai.

Sirf ek prograde burn (velocity ke saath) perigee par apogee raise karta hai. Ek radial ya retrograde burn speed momentarily badha sakta hai phir bhi orbit ko lower ya reshape kar sakta hai. Direction, na sirf speed change, naya $a$ determine karta hai.

Kinetic energy per mass $\frac{1}{2}{v}^2$ hamesha $\ge 0$ hoti hai. Poora $\epsilon$ negative hai kyunki negative PE term $-GM/r$ positive KE term se zyada hai; koi bhi piece akele galat tarike se negative nahi hota.

Derivation dikhata hai ki $\epsilon =-GM/(r_p+r_a)=-GM/2a$: sirf extreme distances ka sum enter karta hai, aur woh sum $2a$ ke barabar hota hai chahe ellipse kitna bhi squashed ya round ho. Shape control karti hai ki energy KE aur PE ke beech kaise split hogi, total nahi.

Taaki satellite ki apni mass cancel ho jaaye, aur ek number orbit ko describe kar sake — isi wajah se sab cheezein ek hi rate se girti hain. Isse $\epsilon$ trajectory ki property ban jaati hai, payload ki nahi.

Humne $h=r_p{v}_p=r_a{v}_a$ (from Conservation of angular momentum) sirf tool ke roop mein use kiya taaki unknown speeds eliminate ho sakein. Jab speeds chali jaati hain, geometry bachti hai aur result purely $r_p+r_a=2a$ par depend karta hai.

Jaise-jaise woh inward girti hai, PE ($-GM/r$) drop hoti hai (aur zyada negative ho jaati hai), aur kyunki $\epsilon$ fixed hai, KE badhni chahiye, isliye $v$ increase hota hai. Apogee tak climb karne par ulta hota hai — ek continuous energy trade.

Kyunki $\epsilon$ constant hai, $\frac{v^2}{2}-\frac{GM}{r}=-\frac{GM}{2a}$ equate karne par $v$ sirf $r$ aur $a$ se link hota hai. Angular position already $r$ mein baked in hai, isliye koi alag angle required nahi.

$\epsilon =0$ ka matlab hai ki total energy barely infinity par zero reference tak pahunchti hai — body $r\to \infty$ par $v\to 0$ ke saath pahunchti hai. Koi bhi kam energy ($\epsilon <0$) use trap karti hai; koi bhi zyada ($\epsilon >0$) speed to spare chhod deti hai (dekho Escape velocity).

$\epsilon =-GM/2a\to 0$. Yeh parabolic escape boundary hai: orbit ab closed nahi hai aur body zero speed ke saath infinity tak pahunchti hai.

Yeh ek hyperbolic (unbound) flyby signal karta hai jisme $\epsilon >0$ ho. "$a$" ab hyperbola ki geometry ka distance parameter hai, koi loop size nahi, aur koi apogee exist nahi karta.

$v^2=GM(\frac{2}{r}-\frac{1}{a})\to \infty$, isliye speed blow up hoti hai. Physically orbit ko centre se guzarna padega — ek real orbit ke liye impossible hai jo finite body ke around ho, isliye perigee surface ke upar rehta hai.

Haan. Koi bhi ellipse (chahe kitna bhi eccentric ho) jo same semi-major axis $a$ ki ho, identical $\epsilon =-GM/2a$ rakhti hai — ek nearly circular orbit aur ek highly elongated orbit energy-twins ho sakti hain. Isi wajah se Kepler's Third Law (period sirf $a$ par depend karta hai) kaam karta hai.

Kahin nahi — $\epsilon$ us single ellipse ke har point par same hota hai. Sirf KE aur PE individually vary karte hain; transfer ki total specific energy ek fixed number hai jo uske apne $a=(r_p+r_a)/2$ se set hoti hai.

Zaroor nahi. $v$ double karne par KE chaar guna ho jaati hai; agar naya $\epsilon =\frac{v^2}{2}-\frac{GM}{r}$, $\ge 0$ ho jaaye, toh orbit parabolic ya hyperbolic ban jaati hai aur body escape kar jaati hai.