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

False. It means the third body's mass is negligible — it feels the two primaries but exerts no gravity back on them. A restricted third body could be a whole moon in size as long as its mass is dynamically ignorable.

True — that is the "circular" in CR3BP. If their mutual orbit were an ellipse we would have the elliptic restricted problem, and the potential would no longer be time-independent even in the rotating frame.

False. Coriolis is $2\mathbf{\omega }\times \dot{\mathbf{r}}$, always perpendicular to the velocity, so $\mathbf{F}\cdot \dot{\mathbf{r}}=0$. It bends the path but never changes speed — which is exactly why the Jacobi constant stays conserved.

False. $C_J=2U-({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)$ is an energy-like integral of the rotating frame; it is not the inertial energy, which is not conserved because the primaries' gravity field rotates in inertial space.

False on the geometry, true on the outcome. $L_4,L_5$ are actually maxima of $U$; they are stable (for small $\mu$) only because the Coriolis force curls the runaway motion into a bounded orbit.

True. Equilibrium requires $\dot{}=\ddot{}=0$, which reduces the equations of motion to $\nabla U=0$. All five points satisfy this; they differ only in their stability.

False. Their instability comes from $U_{xx}{U}_{yy}-U_{xy}^2<0$ (with $U_{xy}=0$ and $U_{xx}>0>U_{yy}$), a sign condition independent of $\mu$. They are saddle-type and always unstable.

True. Real negative $\Lambda$ gives $\lambda =\pm i\omega$, pure oscillation ($e^{i\omega t}$). Any $\Lambda >0$ produces a real positive $\lambda$ and exponential blow-up.

True. ${\mu }_{\text{Jup}}\approx 0.00095\ll 0.0385={\mu }_{\text{crit}}$, so Jupiter's $L_4,L_5$ are linearly stable and can trap the Trojan asteroids.

Error. The Coriolis terms are exactly what produce the "$4$" in ${\lambda }^4+(4-U_{xx}-U_{yy}){\lambda }^2+\dots$. Dropping them gives the non-rotating (wrong) stability test.

Error. The hilltop analogy ignores rotation. The Coriolis term steers the departing motion sideways into a closed loop, stabilising the maximum whenever $\mu <0.0385$.

Error. It is degree four in $\lambda$: ${\lambda }^4+(4-U_{xx}-U_{yy}){\lambda }^2+(U_{xx}{U}_{yy}-U_{xy}^2)=0$. Only after substituting $\Lambda ={\lambda }^2$ does it become quadratic in $\Lambda$.

Error. Rotation is about $\hat{z}$, so the centrifugal push acts only in the plane perpendicular to the axis: the term is $\frac{1}{2}(x^2+y^2)$, with no $z^2$. That is why out-of-plane motion has no centrifugal support.

Error. Product of roots equals $c$; if $c<0$ the two roots have opposite signs, forcing one positive $\Lambda$ — hence instability. Negative $c$ is a red flag, not a safety check.

Error. The bigger primary $m_1=1-\mu$ sits at $x=-\mu$ and the smaller $m_2=\mu$ at $x=1-\mu$. The heavy body hugs the origin; the light body swings out to distance 1.

In the inertial frame the two primaries whirl around, making the gravitational potential time-dependent and the problem non-autonomous. Rotating with them freezes the primaries, kills the time dependence, and hands us a conserved Jacobi constant plus fixed equilibria.

The general three-body problem has no closed-form (integrable) solution. Taking $m_3\to 0$ with circular primaries reduces it to a solvable, well-behaved model with only one free parameter $\mu$.

It comes from the $2\dot{\eta }$ and $2\dot{\xi }$ Coriolis terms in the linearised equations; when the determinant is formed, the two factors of $2$ multiply to give $2\times 2=4$ as the coefficient of ${\lambda }^2$.

Those collinear points are always unstable (saddle-type), so any residual error grows like $e^{+\lambda t}$. Thrusters must periodically cancel the exponential drift, as covered in Halo orbits & station-keeping.

We only care whether a small nudge grows or oscillates. Near equilibrium the leading behaviour is set by the first non-zero (linear-in-displacement) force terms, whose exponents are the eigenvalues $\lambda$ — see Eigenvalues & linear stability analysis.

At $L_4,L_5$ the second derivatives take fixed values ($U_{xx}=\frac{3}{4}$, $U_{yy}=\frac{9}{4}$, $U_{xy}=\pm \frac{3\sqrt{3}}{4}(1-2\mu )$), so the discriminant collapses to $1-27\mu (1-\mu )$ — a single-variable inequality in the mass ratio.

The $z$-equation $\ddot{z}=\partial U/\partial z$ has no Coriolis coupling (rotation is about $\hat{z}$), and at $z=0$ the cross second-derivatives $U_{xz},U_{yz}$ vanish by symmetry, so vertical oscillation separates cleanly from planar motion.

Then $\mu =\frac{1}{2}$, its maximum value. The primaries sit symmetrically at $x=\pm \frac{1}{2}$, and since $\frac{1}{2}>0.0385$, the triangular points $L_4,L_5$ are unstable — equal masses cannot host Trojans.

The discriminant $1-27\mu (1-\mu )$ is exactly zero, so the two $\Lambda$ roots coincide (repeated root). This marginal case is degenerate; the linear theory gives neutral/algebraically-growing behaviour rather than clean oscillation.

With $\mu \to 0$ the small primary vanishes and the problem collapses toward the two-body case; $L_4,L_5$ become deeply stable ($27\mu (1-\mu )\to 0\ll 1$), matching a lone Keplerian primary.

No. "At rest" means zero velocity in the rotating frame. In the inertial frame the point itself is orbiting the barycentre, so the co-located spacecraft carries that orbital velocity too.

The zero-velocity surfaces $2U=C_J$. Where $2U<C_J$ would require imaginary speed, so those regions are forbidden — the surfaces act as walls the spacecraft cannot cross without changing $C_J$.

It runs away exponentially. $L_3$ is a collinear point, hence a saddle with a real positive $\lambda$; the displacement grows along the unstable eigendirection until nonlinear terms or an escape take over.

Not in the CR3BP sense. A saddle of $U$ gives $U_{xx}{U}_{yy}-U_{xy}^2<0$, forcing a positive $\Lambda$ and exponential instability — exactly the situation at $L_1,L_2,L_3$.