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For the normalized damped pendulum, $E={\dot{\theta }}^2/2+1-\cos \theta$ obeys $\dot{E}=-c{\dot{\theta }}^2\le 0$. Its mechanical energy is constant only along an equilibrium trajectory, and strictly decreases over every positive-length interval on a nonstationary trajectory. At an individual turning point its derivative vanishes. This excludes nonconstant periodic trajectories despite oscillatory approach to a stable focus.

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 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.

Differentiate the mechanical energy and use the damped pendulum equation:The energy dissipation of a damped pendulum makes $E$ nonincreasing, rather than strictly decreasing at every instant: its derivative vanishes at turning points. It is constant along an equilibrium trajectory. Along every nonstationary trajectory it strictly decreases over any nonzero time interval, because a vanishing integral of $c{\dot{\theta }}^2$ over an interval would force an equilibrium there and hence everywhere by uniqueness.

For $c=1$, the even equilibria are stable foci with eigenvalues $(-1\pm i\sqrt{3})/2$, and the odd equilibria are saddle equilibria with eigenvalues $r_{\pm }=(-1\pm \sqrt{5})/2$. The local saddle directions are $y=r_{\pm }[x-(2j+1)\pi ]$.

The phase portrait shows clockwise spiraling into each even equilibrium: on its right-hand horizontal axis, the flow initially points down. The stable separatrices of the saddles divide attraction basins on the unwrapped angle axis, while the unstable branches flow into neighboring sinks. Initial conditions with enough mechanical energy can cross one or more potential crests before being captured. The curves repeat under $x\mapsto x+2\pi$. They are trajectories, not conservative energy contours: energy dissipation of a damped pendulum rules out nonconstant periodic orbits.

Integrating the standard thin-disk dissipation flux over annular area, with both faces already included in $\mathcal{H}$, givesThus the disk luminosity for $r_{\mathrm{out}}\gg r_{\ast }$ isThe Newtonian gravitational potential decreases by approximately $G{M}_{\ast }/r_{\ast }$ per unit mass from a distant outer edge to the surface, giving a potential-energy release rate $G{M}_{\ast }\dot{M}/r_{\ast }$. The standard thin-disk luminosity is half of this.

The missing half remains as kinetic energy of nearly circular orbital motion: at the inner edge $v_K^2=G{M}_{\ast }/r_{\ast }$, so the specific orbital kinetic energy is $G{M}_{\ast }/(2r_{\ast })$. Equivalently, circular motion has total specific mechanical energy $-G{M}_{\ast }/(2r_{\ast })$. Matter joining a slowly rotating star must shed this orbital motion in an accretion-disk boundary layer, producing approximately another $G{M}_{\ast }\dot{M}/(2r_{\ast })$ of luminosity. A rotating star can retain some energy in spin, so equal disk and accretion-disk boundary layer luminosities assume slow stellar rotation. For a central black hole, there is no material surface, and energy can instead be carried inward.

For unit mass, the inverse-square force has potential energy $-k/r$, with zero potential at infinity. Conservation of angular momentum gives $r^2\dot{\theta }=h$. Differentiate the polar equation of the Kepler orbit:Consequently the total mechanical energy iswhere $k=h^2/\ell$. This also explains the sign distinction between an elliptic orbit, a parabolic Kepler orbit, and a hyperbolic Kepler orbit.

At the original periapsis, the parabolic Kepler orbit has $e=1$, $r=d=\ell /2$, and purely tangential speed $V$. ThusThe outward radial impulse has zero moment about the Sun, so it preserves angular momentum and hence $\ell$. Its additional velocity is perpendicular to the original velocity, so the new mechanical energy is $E^{\prime }={\alpha }^2{V}^2/2$. Applying the energy formula givesFor $\alpha >0$, the new orbit is a hyperbolic Kepler orbit; $\alpha =0$ leaves the parabola unchanged.

To fix the orientation of the sketch, put the Sun at the origin, the impulse point at $(d,0)$, and take the original motion there upwards. The new velocity is $(\alpha V,V)$. The eccentricity vector isIt points towards the new periapsis, which is rotated clockwise through $\beta =\arctan (2\alpha )$ from the original periapsis direction. Thus, measured using the original polar angle $\theta$, the new orbit isIn particular, the impulse point is already on the outgoing part of the new orbit and is not its new periapsis. The old parabola satisfies $x=d-y^2/(4d)$. Both orbits pass through $(d,0)$ and have the same focus, but the hyperbola's new periapsis axis is tilted.

The equation of motion isMultiplication by $\dot{x}$ givesso the mechanical energy $E=\frac{1}{2}m{\dot{x}}^2+V(x)$ is conserved. On a segment where the direction of motion is fixed,