Question bank — Three-body problem — restricted (CR3BP), characteristic equation

False. Iska matlab hai ki third body ki mass negligible hai — woh do primaries ko feel karti hai lekin unpe apna gravity wapas nahi dalti. Ek restricted third body poore moon ke size ki bhi ho sakti hai jab tak uski mass dynamically ignorable ho.

True — yahi CR3BP mein "circular" ka matlab hai. Agar unka mutual orbit ellipse hota toh hum elliptic restricted problem mein hote, aur potential rotating frame mein bhi time-independent nahi rehta.

False. Coriolis $2\mathbf{\omega }\times \dot{\mathbf{r}}$ hai, hamesha velocity ke perpendicular, isliye $\mathbf{F}\cdot \dot{\mathbf{r}}=0$. Yeh path ko bend karti hai lekin speed kabhi nahi badlati — yahi wajah hai ki Jacobi constant conserved rehta hai.

False. $C_J=2U-({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)$ rotating frame ka ek energy-jaisa integral hai; yeh inertial energy nahi hai, jo conserved nahi hoti kyunki primaries ka gravity field inertial space mein rotate karta hai.

Geometry par False, outcome par True. $L_4,L_5$ actually $U$ ke maxima hain; woh (chhote $\mu$ ke liye) sirf isliye stable hain kyunki Coriolis force runaway motion ko curl karke ek bounded orbit mein daal deti hai.

True. Equilibrium ke liye $\dot{}=\ddot{}=0$ chahiye, jo equations of motion ko $\nabla U=0$ tak reduce kar deta hai. Paanch saare points yeh satisfy karte hain; woh sirf apni stability mein differ karte hain.

False. Unki instability $U_{xx}{U}_{yy}-U_{xy}^2<0$ se aati hai (jahan $U_{xy}=0$ aur $U_{xx}>0>U_{yy}$), jo ek sign condition hai jo $\mu$ se independent hai. Woh saddle-type hain aur hamesha unstable rehte hain.

True. Real negative $\Lambda$ se $\lambda =\pm i\omega$ milta hai, pure oscillation ($e^{i\omega t}$). Koi bhi $\Lambda >0$ ek real positive $\lambda$ aur exponential blow-up produce karta hai.

True. ${\mu }_{\text{Jup}}\approx 0.00095\ll 0.0385={\mu }_{\text{crit}}$, isliye Jupiter ke $L_4,L_5$ linearly stable hain aur Trojan asteroids ko trap kar sakte hain.

Error. Coriolis terms hi woh hain jo ${\lambda }^4+(4-U_{xx}-U_{yy}){\lambda }^2+\dots$ mein "$4$" produce karte hain. Unhe drop karne se non-rotating (galat) stability test milta hai.

Error. Hilltop analogy rotation ko ignore karta hai. Coriolis term departing motion ko sideways steer karke ek closed loop mein daal deta hai, jo maximum ko tab stabilise karta hai jab $\mu <0.0385$.

Error. Yeh $\lambda$ mein degree four ka hai: ${\lambda }^4+(4-U_{xx}-U_{yy}){\lambda }^2+(U_{xx}{U}_{yy}-U_{xy}^2)=0$. $\Lambda ={\lambda }^2$ substitute karne ke baad hi yeh $\Lambda$ mein quadratic banta hai.

Error. Rotation $\hat{z}$ ke around hai, isliye centrifugal push sirf rotation axis ke perpendicular plane mein act karta hai: term $\frac{1}{2}(x^2+y^2)$ hai, koi $z^2$ nahi. Isliye out-of-plane motion ko koi centrifugal support nahi milta.

Error. Roots ka product $c$ ke barabar hota hai; agar $c<0$ toh do roots ke opposite signs hain, jo ek positive $\Lambda$ ko force karta hai — hence instability. Negative $c$ ek red flag hai, safety check nahi.

Error. Bada primary $m_1=1-\mu$ $x=-\mu$ par baitha hai aur chhota $m_2=\mu$ $x=1-\mu$ par. Heavy body origin ke paas rehti hai; light body distance 1 tak jhulti hai.

Inertial frame mein do primaries ghoomte rehte hain, gravitational potential ko time-dependent bana dete hain aur problem ko non-autonomous kar dete hain. Unke saath rotate karne se primaries freeze ho jaate hain, time dependence khatam ho jaati hai, aur hume ek conserved Jacobi constant plus fixed equilibria milte hain.

General three-body problem ka koi closed-form (integrable) solution nahi hai. $m_3\to 0$ lena aur circular primaries ke saath use ek solvable, well-behaved model mein reduce kar deta hai jisme sirf ek free parameter $\mu$ hota hai.

Yeh linearised equations mein $2\dot{\eta }$ aur $2\dot{\xi }$ Coriolis terms se aata hai; jab determinant form hota hai, to $2$ ke do factors multiply hokar $2\times 2=4$ dete hain ${\lambda }^2$ ke coefficient ke roop mein.

Woh collinear points hamesha unstable (saddle-type) hote hain, isliye koi bhi residual error $e^{+\lambda t}$ ki tarah grow karta hai. Thrusters ko periodically exponential drift cancel karni padti hai, jaisa Halo orbits & station-keeping mein cover kiya gaya hai.

Hume sirf yeh jaanna hai ki kya ek chhota nudge grow hoga ya oscillate karega. Equilibrium ke paas leading behaviour pehle non-zero (linear-in-displacement) force terms se set hoti hai, jinke exponents eigenvalues $\lambda$ hain — dekho Eigenvalues & linear stability analysis.

$L_4,L_5$ par second derivatives fixed values lete hain ($U_{xx}=\frac{3}{4}$, $U_{yy}=\frac{9}{4}$, $U_{xy}=\pm \frac{3\sqrt{3}}{4}(1-2\mu )$), isliye discriminant $1-27\mu (1-\mu )$ mein collapse ho jaata hai — mass ratio mein ek single-variable inequality.

$z$-equation $\ddot{z}=\partial U/\partial z$ mein koi Coriolis coupling nahi hai (rotation $\hat{z}$ ke around hai), aur $z=0$ par cross second-derivatives $U_{xz},U_{yz}$ symmetry se vanish ho jaate hain, isliye vertical oscillation planar motion se cleanly separate ho jaati hai.

Tab $\mu =\frac{1}{2}$, uski maximum value. Primaries symmetrically $x=\pm \frac{1}{2}$ par baithte hain, aur kyunki $\frac{1}{2}>0.0385$, triangular points $L_4,L_5$ unstable hain — equal masses Trojans host nahi kar sakte.

Discriminant $1-27\mu (1-\mu )$ exactly zero hai, isliye do $\Lambda$ roots coincide karte hain (repeated root). Yeh marginal case degenerate hai; linear theory clean oscillation ki jagah neutral/algebraically-growing behaviour deta hai.

$\mu \to 0$ ke saath chhota primary disappear ho jaata hai aur problem two-body case ki taraf collapse ho jaati hai; $L_4,L_5$ deeply stable ho jaate hain ($27\mu (1-\mu )\to 0\ll 1$), jo ek akele Keplerian primary se match karta hai.

Nahi. "Rest mein" ka matlab hai rotating frame mein zero velocity. Inertial frame mein woh point khud barycentre ke around orbit kar raha hai, isliye co-located spacecraft bhi woh orbital velocity carry karta hai.

Zero-velocity surfaces $2U=C_J$. Jahan $2U<C_J$ hoga waahan imaginary speed ki zaroorat hogi, isliye woh regions forbidden hain — surfaces walls ki tarah act karti hain jinhe spacecraft $C_J$ change kiye bina cross nahi kar sakta.

Woh exponentially bhaagg jaayegi. $L_3$ ek collinear point hai, hence ek real positive $\lambda$ wala saddle; displacement unstable eigendirection ke along grow karta rahega jab tak nonlinear terms ya escape na aa jaaye.

CR3BP sense mein nahi. $U$ ka ek saddle $U_{xx}{U}_{yy}-U_{xy}^2<0$ deta hai, jo ek positive $\Lambda$ aur exponential instability force karta hai — exactly $L_1,L_2,L_3$ par wali situation.