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For a Lagrangian $L(q,\dot{q},t)$, its energy function satisfies $d\mathcal{E}/dt=-\partial L/\partial t$ along the Euler-Lagrange equations. This follows by differentiating and cancelling the ${\ddot{q}}_i\partial L/\partial {\dot{q}}_i$ terms. It is conserved when $L$ has no explicit time dependence. For imposed moving constraints it need not coincide with inertial mechanical energy; a Jacobi integral is an important example.

Two point masses released from rest at separation $a$ have negative mechanical energy $-G{m}_1{m}_2/a$ but collide in time $\pi \sqrt{a^3/[8G(m_1+m_2)]}$. Their separation approaches zero along a regular trajectory before collision. The energy sign alone therefore does not supply a positive lower separation bound.

For the damped harmonic oscillator, the mechanical energy is the sum of kinetic energy and spring potential energy:Thus $b{\dot{x}}^2$ is the power transferred out of the oscillator's mechanical energy. For $b>0$, the rate vanishes instantaneously at a turning point, despite nonzero stored potential energy; it is not a constant loss per unit time. In eddy-current damping the loss becomes Joule heating. ^[ref]

For an underdamped response, advancing one oscillation period multiplies both displacement and velocity by $e^{-\zeta {\omega }_0{T}_{\mathrm{d}}}$, so $E(t+T_{\mathrm{d}})=e^{-2\zeta {\omega }_0{T}_{\mathrm{d}}}E(t)$. This exact cycle relation does not imply that instantaneous energy is a pure exponential function throughout each cycle.

For a smooth plane at inclination $\theta$ rotating about a vertical axis at constant angular velocity $\omega$, choose rotating coordinates $\xi$ horizontally and $\eta$ uphill, with their cross product the upward normal vector. A sliding particle under gravity satisfiesThe normal reaction is $N=m(g\cos \theta -2\omega \dot{\xi }\sin \theta +{\omega }^2\eta \sin \theta \cos \theta )$ while contact is maintained. The Coriolis force does no work in the rotating frame, giving the conserved specific energyThe last term is the centrifugal potential; this quantity need not equal the inertial mechanical energy.

The weak-damping mechanical energy is proportional to the squared oscillation amplitude, so it decays as $e^{-\gamma t}$. Over one period $P=2\pi /{\omega }_d$, the fractional loss is $1-e^{-\gamma P}\simeq \gamma P$. HenceFor the release-from-rest solution, comparison at successive corresponding turning points gives the more precise ratio $E/\Delta E=(1-e^{-\gamma P}{)}^{-1}$. The quality factor is $Q\simeq 2\pi E/\Delta E={\omega }_0/\gamma$; the requested energy-to-loss ratio itself has the extra factor $1/(2\pi )$.

The initial conditions in the underdamped solution give $A=x_0$ and $B=\gamma x_0/\Omega$. ConsequentlyThe exact mechanical energy is thereforeFor weak damping, $\gamma \ll {\omega }_0$, the bracket differs from one by a bounded relative correction of order $\gamma /{\omega }_0$. To leading order the energy and its half-life areThe exact identity $\dot{E}=-2m\gamma {\dot{x}}^2\le 0$ confirms dissipation. The small rapid variation of the exponential-normalized energy does not mean that total energy increases; the half-life formula is a weak-damping approximation, not an exact equality for arbitrary $\gamma$.

The characteristic equation of the damped harmonic oscillator is $r^2+\gamma r+{\omega }_0^2=0$. In the underdamped case its roots are $-\gamma /2\pm i{\omega }_d$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2/4}$. Thus the exact real solution is $y=A{e}^{-\gamma t/2}\cos ({\omega }_{d}t+\alpha )$. For $\gamma \ll {\omega }_0$, ${\omega }_d={\omega }_0+O({\gamma }^2/{\omega }_0)$, giving the requested leading approximationFor mass $m$ and spring constant $m{\omega }_0^2$, the mechanical energy is $E=\frac{1}{2}m({\dot{y}}^2+{\omega }_0^2{y}^2)$. Differentiating and using the differential equation gives $\dot{E}=-m\gamma {\dot{y}}^2$. In particular, energy decay of a viscously damped harmonic oscillator is not an exact pointwise exponential decay: $\dot{E}=0$ at a turning point. To leading order in $\gamma /{\omega }_0$, the period average isThe quality factor is $Q=2\pi \overline{E}/(\text{energy lost in one cycle})$. Using the period $T\simeq 2\pi /{\omega }_0$ and the small fractional loss $\gamma T$ gives $\boxed{Q\simeq {\omega }_0/\gamma }$. Equivalently, the energy-decay definition $Q={\omega }_0\overline{E}/{\overline{P}}_{\mathrm{loss}}$ yields the same leading result. Replacing ${\omega }_d$ by ${\omega }_0$ in the oscillatory phase is not uniform over arbitrarily long times; its phase error is $O({\gamma }^{2}t/{\omega }_0)$.

In the steady periodic response, the cycle average of the mechanical energy derivative vanishes. The applied average power therefore equals the power dissipated by the viscous force:Using solution givesThe same result follows from averaging the instantaneous input power $F(t)\dot{z}(t)$, using the response phase relative to the drive.

For positive $m,k,\gamma$ and $\omega >0$, rewrite this asThe last factor is at most one, with equality precisely when $k-m{\omega }^2=0$. Thus the power peak occurs at ${\omega }_0=\sqrt{k/m}$ andThis power-resonance angular frequency is exactly ${\omega }_0$; it need not equal the angular frequency maximizing displacement amplitude. Positive damping is essential: at $\gamma =0$ and exact resonance there is no bounded steady periodic response to nonzero forcing, so the divergent expression is not a finite-power solution.

For the constant small-oscillation matrices, the total mechanical energy satisfiesThe specified conservative pendulum has Lagrangian $L=K-U$, whose Euler-Lagrange equations giveor $8{\ddot{\theta }}_1+2{\ddot{\theta }}_2+(4g/l){\theta }_1=0$ and $2{\ddot{\theta }}_1+{\ddot{\theta }}_2+(g/l){\theta }_2=0$. These equations make $\dot{E}=0$.

The scalar energy law alone is insufficient to infer two unique equations: energy conservation does not uniquely determine equations of motion. In particular, adding $\mathbf{G}\dot{\mathbf{\theta }}$ on the right with antisymmetric $\mathbf{G}$ changes the dynamics but contributes zero power. The conservative-force structure of the given pendulum excludes that extra gyroscopic force and supplies the information needed for the displayed matrix equations.

The quadratic total mechanical energy isFor this frictionless conservative double pendulum, the Euler-Lagrange equations for $L=K-U$ giveor $8{\ddot{\theta }}_1+2{\ddot{\theta }}_2+(4g/l){\theta }_1=0$ and $2{\ddot{\theta }}_1+{\ddot{\theta }}_2+(g/l){\theta }_2=0$. Substitution immediately verifies energy conservation.

Logical qualification. Energy conservation does not uniquely determine equations of motion: the scalar equation $dE/dt=0$ alone does not determine both equations. For example, replacing their right side by $\mathbf{G}\dot{\mathbf{\theta }}$ with any nonzero antisymmetric matrix $\mathbf{G}$ preserves the same energy because ${\dot{\mathbf{\theta }}}^{\mathsf{T}}\mathbf{G}\dot{\mathbf{\theta }}=0$, while changing the dynamics. Such a gyroscopic force is absent from the specified pendulum. The conservative-force structure, supplied here by the Lagrangian, is the additional information needed to justify the displayed equations rather than infer two independent equations from one scalar identity.

The Friedmann acceleration equation gives the gravitational deceleration from a uniform pressureless matter density, plus the acceleration from a cosmological constant. Here $\rho$ is the matter mass density, $G$ is the gravitational constant, $H=\dot{a}/a$ is the Hubble parameter, and $k$ is the constant spatial-curvature coefficient in the chosen normalization of $a$. The Friedmann equation is the first integral relating expansion to density, curvature and vacuum energy. With $c=1$, the matter, curvature and cosmological-constant contributions to $H^2$ are $8\pi G\rho /3$, $-k/a^2$ and $\Lambda /3$. The cosmological constant has effective density $\Lambda /(8\pi G)$ and equation-of-state parameter $w=-1$; its negative pressure explains its positive acceleration contribution when $\Lambda >0$.

For the Newtonian derivation with $\Lambda =0$, follow a test mass on the edge of a uniform sphere of physical radius $R=a\chi$. Conservation of the enclosed matter gives $M=4\pi \rho R^3/3$ constant. The shell theorem removes the force of external spherical shells and givesIntegrating the force equation once gives the conserved specific mechanical energyDivide by $R^2/2=a^2{\chi }^2/2$ to obtainHomologous motion makes $\epsilon /{\chi }^2$ constant between shells. Newtonian mechanics supplies this energy integration constant; its interpretation as intrinsic spatial curvature belongs to the relativistic FRW metric.

An outward acceleration $\Lambda \mathbf{r}/3$ comes from the Newtonian potential per unit masssince $-\nabla {\Phi }_{\Lambda }=\Lambda \mathbf{r}/3$. With conventional dimensional $\Lambda$, restore $c^2$ in this potential and the corresponding acceleration and Friedmann terms.

Introduce the slow time $T=ϵt$ and expand $y=y_0(t,T)+ϵy_1(t,T)+\cdots$. At leading order writeThe first-order equation isIts resonant sine and cosine components must vanish by the solvability condition in the method of multiple scales. Let angle brackets denote a full-period average. The cosine projection of the nonlinear term is zero, since $⟨{\sin }^n\psi \cos \psi ⟩=0$. The sine projection givesThis is the averaged oscillator damping by a power of velocity.

If $n$ is odd, defineThen $R_T=-C_n{R}^n$, givingFor every fixed $n$, these leading multiple-scale expansions are uniform on bounded slow-time intervals $0\le ϵt\le T_0$. The unforced phase remains $t$ to this order. A bounded first-order correction supplies the small adjustment to the initial derivative and initial displacement; it does not change this leading result.

If $n$ is even, the average of the odd power ${\sin }^{n+1}\psi$ vanishes. Thus $R_T=0$ andThe first-order forcing has only a constant and even harmonics, so it produces no fundamental resonance. Its bounded particular solution can be supplemented by homogeneous sine and cosine terms to satisfy the initial conditions. There can be an $O({ϵ}^2)$ frequency correction, but its phase accumulation is only $O(ϵ)$ on the present time scale. In contrast to the odd case, the even-power force is invariant under velocity reversal and the equation is reversible; calling it positive damping for both signs of velocity would be wrong. The mechanical energy identity makes the distinction clear:This is nonpositive for odd $n$ and changes sign for even $n$; its cycle average reproduces the amplitude equation above.

For large odd $n$, the central binomial coefficient givesThis form reveals several time regimes. For $L_nϵt\ll 1$, $R=1-C_nϵt+\cdots$; at $t=O(1/(ϵ\sqrt{n}))$, the amplitude change is only $O(1/n)$. For $L_nϵt\gg 1$ but $\log (L_nϵt)\ll n$, the leading change is $R\simeq 1-\log (L_nϵt)/n$. In particular, at $t=O(1/ϵ)$ it is only $O(\log n/n)$, despite the long elapsed time. High powers damp mainly near velocity maxima, and even a small amplitude reduction strongly suppresses subsequent damping.

An order-one amplitude reduction in the averaged law requires $\log (L_nϵt)=O(n)$. Formally, $t=e^{(n-1)\sigma }/(ϵL_n)$ gives $R\sim e^{-\sigma }$, and for fixed odd $n>1$ its late envelope is an algebraic power $t^{-1/(n-1)}$. These exponential-in-$n$ time scales are predictions of extending the averaged law; the fixed-$n$, bounded-$ϵt$ multiple-scale error estimate alone does not establish uniform accuracy that far. A joint large-$n$, small-$ϵ$ approximation also needs control of the increasingly sharp nonlinear peaks, rather than assuming fixed-$n$ error constants are uniform. Taking $n$ fixed first and then examining the large-$n$ envelope avoids that unsupported interchange of limits.

Dot Newton's second law with $\dot{\mathbf{r}}$. The magnetic cross-product term is perpendicular to this velocity, soThusorThe two halves are the kinetic energy and the isotropic spring's potential energy. The expression in the question is twice their sum. The Lorentz force changes the direction of motion but supplies no power, because it is perpendicular to velocity; this explains why the magnetic field does not appear in the conserved mechanical energy. The sign corresponds to charge $-e$.

Let $r=|{\mathbf{r}}_1-{\mathbf{r}}_2|$ and $\hat{\mathbf{r}}=({\mathbf{r}}_1-{\mathbf{r}}_2)/r$. A differentiable central potential energy gives the internal forcesThere are no external forces. Differentiating the total kinetic energy and using Newton's second law therefore givesSince $\dot{r}=\hat{\mathbf{r}}\cdot ({\mathbf{v}}_1-{\mathbf{v}}_2)$, the last expression is $-U^{\prime }(r)\dot{r}=-dU/dt$. ThusThis conserves total mechanical energy, including the kinetic energy of centre-of-mass motion, not merely the relative kinetic energy.

The potential factors as $V=x^2(3-2x)$, so its zeros are $0$ and $3/2$, with a double zero at the origin. Its derivatives are $V^{\prime }=6x(1-x)$ and $V^{\prime \prime }=6-12x$. Thus there is a local minimum $(0,0)$, a local maximum $(1,1)$ and an inflection at $(1/2,1/2)$. The potential tends to $+\infty$ on the far left and $-\infty$ on the far right. These features determine the left-hand sketch.

Put $v=\dot{x}$. Newton's second law gives the phase plane system $\dot{x}=v$, $\dot{v}=-V^{\prime }=6x(x-1)$. Its conserved mechanical energy and trajectories areThe linearization at $(0,0)$ has eigenvalues $\pm i\sqrt{6}$, and the surrounding energy curves are closed, so it is a center equilibrium. At $(1,0)$ the eigenvalues are $\pm \sqrt{6}$, giving a saddle equilibrium. The figure shows the cubic potential barrier phase portrait; arrows point right when $v>0$ and left when $v<0$.

For $0<E<1$, there are three turning-point roots of $V(x)=E$. The two leftmost enclose a closed orbit around the origin: the particle oscillates between them. There is also a separate open component to the right of the third root, where an incoming particle turns and escapes to $+\infty$. The closed orbits run clockwise in the $(x,v)$ plane.

For $E=1$, the factorization $1-V(x)=(x-1{)}^2(2x+1)$ gives the separatrixThe branch with $-1/2\le x<1$ is a homoclinic orbit: it leaves the saddle asymptotically, turns at $x=-1/2$ and returns asymptotically. On $x>1$, one branch approaches the saddle and the other departs towards infinity. Approaching the saddle takes infinite time. The saddle itself is also an equilibrium trajectory.

For $E>1$, the particle can cross the barrier. There is one left turning point below $-1/2$; an incoming particle passes through the well, turns on the left and escapes to the right. For $E<0$, only a right-hand open component exists, with its turning point beyond $3/2$. At $E=0$, the stationary centre coexists with a right-hand open trajectory turning at $3/2$. Thus negative energy does not imply trapping here: the cubic potential is unbounded below on the right.

For variations $q_i+\epsilon {\eta }_i$ with ${\eta }_i(t_1)={\eta }_i(t_2)=0$, differentiation of the action and integration by parts giveThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equations $d{L}_{{\dot{q}}_i}/dt=L_{q_i}$. An ignorable coordinate has $L_{q_j}=0$, so its conjugate momentum $p_j=L_{{\dot{q}}_j}$ is constant.

For the sphere, $\phi$ is ignorable andThe Hamiltonian is the mechanical energy. Along solutions, $\dot{H}=-\partial L/\partial t=0$, because the Lagrangian has no explicit time dependence.

Choose the gravitational potential energy to vanish at infinity. The total mechanical energy isPut $\gamma =GM/h^2$ and $\phi =\theta -{\theta }_0$. From part (i), $u=\gamma (1+e\cos \phi )$ and $u^{\prime }=-\gamma e\sin \phi$. Consequentlyand the angular terms cancel against the gravitational potential energy:For a nonradial Kepler orbit, negative, zero and positive mechanical energy correspond respectively to $e<1$, $e=1$ and $e>1$.

At infinity, the meteorite's incoming asymptote has speed $V$ and perpendicular distance $b$ from the Earth’s centre. The magnitude of its specific angular momentum is the magnitude of $\mathbf{r}\times \mathbf{v}$. On the asymptote, this is $bV$, so conservation of angular momentum gives $\boxed{h=bV}$. Its mechanical energy at infinity is $E=m{V}^2/2$. Comparing with part (ii) givesFor $b>0$ this Kepler orbit is a hyperbola. The smallest $r=1/u$ occurs at periapsis, where $\cos (\theta -{\theta }_0)=1$. ThusThe middle equality follows by substituting $e^2-1=b^2{V}^4/(G^2{M}^2)$. A trajectory entering the Earth satisfies $r_{\min }<R$, equivalentlyTherefore the sufficient collision condition isThis is gravitational focusing: Newtonian gravity increases the collision radius measured at infinity. Equality is the ideal grazing trajectory, which touches the surface; a strict inequality penetrates it. For $b=0$ the incoming motion is radial and collides, with $r_{\min }=0$ as the limiting value of the above expression.

At infinity the potential energy tends to zero, so the conserved mechanical energy is $E=\frac{1}{2}{v}_{\infty }^2$. The impact parameter $b$ is the perpendicular distance from the origin to the incoming asymptotic line. Conservation of angular momentum gives $|h|=b{v}_{\infty }$.

For $b>0$, a turning point of the radial motion satisfiesOnly the positive root is physical, givingFor $Q=0$ this reduces to the straight-line closest distance $b$.

For nonzero $Q$, put $\eta =b{v}_{\infty }^2/|Q|\ll 1$. If $Q<0$, the numerator is $|Q|(\sqrt{1+{\eta }^2}+1)$, soIf $Q>0$, rationalize the numerator instead:Thus the requested leading distances areRepulsion gives a finite barrier even for zero impact parameter; attraction allows much closer passage. At $b=0$, the attractive formula has the limiting closest distance zero, corresponding to collision with the singular centre rather than a positive turning point.

For the inverse-square potential $V(r)=-Q/r$, the central force is $-V^{\prime }(r)\hat{r}=-Q\hat{r}/r^2$. The transverse component of acceleration in polar coordinates is therefore zero:Thus $\boxed{h=r^2\dot{\theta }\text{ is constant}}$; for unit mass it is the signed angular momentum perpendicular to the orbital plane. The radial component givesConsequentlyThis is the mechanical energy, since the full kinetic energy is $\frac{1}{2}({\dot{r}}^2+r^2{\dot{\theta }}^2)$ and $r^2{\dot{\theta }}^2=h^2/r^2$.

For $h\ne 0$ and $Q>0$, the effective potential goes to $+\infty$ as $r\downarrow 0$ and to zero from below as $r\to \infty$. Its derivative is $U^{\prime }(r)=(Qr-h^2)/r^3$, so it decreases up to $r_0=h^2/Q$ and then increases. Its zero and minimum areFor $Q<0$, $U=h^2/(2r^2)+|Q|/r$ is positive and strictly decreasing, with limits $+\infty$ at zero and $0^{+}$ at infinity. It has no minimum at a finite positive radius.

If $h=0$, the centrifugal term disappears. For $Q>0$ the graph is instead $-Q/r$, increasing from $-\infty$ to $0^{-}$ without a minimum; for $Q<0$ it remains positive and decreasing. The bounded-orbit calculation below uses the stipulated $h>0$, which excludes this radial collision case.