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Choose $\mathbf{B}=B\hat{\mathbf{z}}$ and a signed tangential velocity $\mathbf{v}=u\hat{\mathbf{\varphi }}$, with $u>0$ counterclockwise viewed from $+z$. The magnetic part of the Lorentz force is $euB\hat{\mathbf{r}}$, while the electric part is $-eV\hat{\mathbf{r}}/(\eta r)$. The radial acceleration for a circular orbit is $-u^2/r$, regardless of the sign of $u$. ConsequentlyOne valid signed-velocity convention therefore has $a=m$, $b=eBr$, $c=-eV/\ln (r_2/r_1)$. If $v$ is required to be a nonnegative speed, the two senses of rotation instead satisfywith the plus sign for counterclockwise motion. For positive $V$, the electric force is inward, so it can supply circular motion in either direction; in one direction the magnetic force opposes it, while in the other it helps. A signed negative root must not be discarded as an unphysical speed; it denotes the opposite orbit direction.

Write $\mu =GM$ and $A=\mu m$. For a circular orbit of radius $r$, centripetal balance gives $v^2=\mu /r$. Thus $E=m{v}^2/2-A/r=-A/(2r)$ and $J^2=m^2{r}^2{v}^2=Amr$. Since $e=0$, $a=r_0=r$, these agree with the general expressions.

For an outward coplanar Hohmann transfer, burn prograde at the inner circle to put its radius at the ellipse's pericentre, then burn prograde at apocentre to circularize at the station's radius. The station must be phased to meet the spacecraft there. With $r_p=R$, $r_a=\alpha R$,Energy at the first burn, with $v_0^2=\mu /R$, givesAngular momentum is conserved along the transfer ellipse, so $v_{2-}=v_1/\alpha$, while the outer circular speed is $v_{2+}=v_0/\sqrt{\alpha }$. For $\alpha =3$,The correct second impulse adds $\Delta v=v_{2+}-v_{2-}$. The reversed circularization burn in a Hohmann transfer, using the same impulse magnitude, instead leavesIts magnitude is below the circular speed at $3R$, so that point is the new apocentre. Define $q=r_a{v}_w^2/\mu =(\sqrt{2}-1{)}^2=3-2\sqrt{2}$. For an ellipse at apocentre, $q=1-e_w$, givingThe lunar radius is $R/3$. Since $r_{p,w}<R/3$, the osculating orbit intersects the Moon: the spacecraft crashes before reaching its formal pericentre. The calculation uses impulsive tangential burns and a spherical central gravitational field; it does not assume that a trajectory continues through the solid body.

The original plot uses $R$ as the length unit and $R/3$ as the lunar radius. The initial station position shown is phased for rendezvous: its initial angular lead is $\pi -\pi [(1+\alpha )/(2\alpha ){]}^{3/2}$, modulo $2\pi$.

The radial force is $F_r=-dU/dr=-3A{r}^2$. A nontrivial circular orbit therefore requires $A>0$, so that the force is inward. Balancing its magnitude against the required centripetal acceleration givesThusThis is the circular-orbit energy ratio in a power-law potential, $2K=r{U}^{\prime }(r)=sU$ for $U=A{r}^s$, with $s=3$. It uses the zero of potential energy specified by $U=A{r}^3$; adding a constant would change the ratio without changing the motion. For $A<0$, the force is outward and no such free circular orbit exists.

For point masses, Newton's force is attractive along their separation with magnitude $G{m}_1{m}_2/r^2$. Treat the narrow rod as uniform, with linear mass density $\mu =M/(2a)$; a mass profile would otherwise be needed beyond total mass and length.

For axial and transverse gravity of a uniform rod, on the axis at $b>a$, integrating the element force givesdirected toward the rod. On the perpendicular bisector, axial components cancel and transverse components add:At $r=b\gg a$,Both approach the point-mass force, but $F_1/F_2=1+3a^2/(2b^2)+O(a^4/b^4)$, so the axial attraction is slightly larger.

For a circular orbit perpendicular to a uniform rod, rotational symmetry about the rod makes a centred circular orbit possible in its perpendicular-bisector plane. With $m\ll M$, centrifugal balance givesFor $r\gg a$, it is Keplerian, $v\sim \sqrt{GM/r}$. For rod thickness $\ll r\ll a$, the force approximates that of an infinite line and $v\to \sqrt{GM/a}$, a flat speed limit rather than a divergence.

For two isolated point masses, introduce total mass $M_T=M+m$, reduced mass $\mu =Mm/M_T$ and relative separation $r$. The relative acceleration is $\ddot{\mathbf{r}}=-G{M}_T\mathbf{r}/r^3$. Conserved specific angular momentum $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ fixes a plane and a constant areal velocity $h/2$. The specific orbital energy isFor $h\ne 0$, the trajectory is a conic:Negative energy gives an ellipse, including the circular limit $e=0$; zero energy gives a parabola; positive energy gives a hyperbola. Bound ellipses obey $\epsilon =-G{M}_T/(2a_o)$ andThus Kepler's laws of planetary motion follow from angular momentum conservation and the inverse-square force. For a circular orbit of radius $r$, $v_c=\sqrt{G{M}_T/r}$, while escape from that radius requires $v_{\mathrm{esc}}=\sqrt{2G{M}_T/r}$. Energy determines whether the orbit is bound; angular momentum also determines its shape. These ideal results neglect extended-body perturbations, drag and relativistic corrections.

For a test mass moving around a fixed central mass $M$, gravity is central and conservative. Specific angular momentum $h=|\mathbf{r}\times \mathbf{v}|$ and specific orbital energy $ϵ=v^2/2-GM/r$ are constant. The motion is planar and sweeps equal areas in equal times, $dA/dt=h/2$.

For a nonradial orbit, the inverse-square force gives a conic section,Negative energy gives an ellipse, zero energy a parabola and positive energy a hyperbola. For an ellipse of semimajor axis $a$,A circular orbit is the $e=0$ case, with $v=\sqrt{GM/r}$. Escape from radius $r$ requires $v\ge \sqrt{2GM/r}$ in the absence of further forces. For two comparable masses, apply the relative two-body equation with $GM$ replaced by $G(M+m)$, rather than treating one body as fixed.

For a circular orbit, $v^2/r=GM/r^2$. The relation for circular orbital speed from surface gravity eliminates the planet's gravitational parameter: $GM=g{R}^2$. The orbital radius is $r=R+h=7.4\times 10^6$ m, soThis is about 7.4 km/s, under the circular-orbit and spherical-Earth approximation. Using surface $g$ unchanged at orbital height would overestimate the centripetal acceleration; its role here is to determine $GM$.

For a test particle outside a spherical central mass, the gravitational force is central, so angular momentum is conserved and the orbit is planar. With specific angular momentum $h=r^2\dot{\theta }$ and specific orbital energy $ϵ={\dot{r}}^2/2+h^2/(2r^2)-GM/r$, the radial effective potential combines attraction with the centrifugal barrier. Its nonradial solutions are conic sections:For nonzero $h$, negative energy gives an ellipse with $0\le e<1$, zero energy a parabolic Kepler orbit and positive energy a hyperbolic Kepler orbit. The circle is the special case $e=0$. Exactly radial motion is a separate zero-angular-momentum case and may hit the central body; a mathematical point-mass trajectory is not automatically a physically noncolliding orbit.

Conservation of $h$ gives constant areal velocity $h/2$, so equal areas are swept in equal times. For a bound ellipse of semimajor axis $a$, $ϵ=-GM/(2a)$ and $T^2=4{\pi }^2{a}^3/(GM)$. A circular orbit has speed $\sqrt{GM/r}$, while the minimum local escape speed is $\sqrt{2GM/r}$, obtained by setting the total energy to zero. These are the test-mass two-body results: a comparable second mass requires the relative-coordinate gravitational parameter $G(M+m)$, and nonspherical structure or drag can perturb the conic and its conserved quantities.

Powered vertical flight. Take upward velocity $v$ as positive, and let $v_e>0$ be the downward exhaust speed relative to the rocket. During $dt$ the rocket's mass changes by $dm<0$, and exhaust mass $-dm$ leaves with Earth-frame velocity $v-v_e$. Compare the momentum of the rocket plus this newly expelled exhaust before and after the interval:Discard the second-order product $dmdv$. The result is $mdv+v_{e}dm=-mgdt$, orThe positive rocket thrust is $-v_e\dot{m}$. Integrating with $v(0)=0$ and $m(0)=m_0$ gives the rocket equation with gravitational loss:Using $d(mv)/dt$ for the rocket alone without the exhaust momentum flux would give the wrong equation for this open system.

Orbital speed under the stated gravity assumption. A circular orbit of radius $r_o=R_E+h_o=7.90\times 10^6\mathrm{m}$ requires centripetal acceleration $v_o^2/r_o$. If the magnitude of gravitational acceleration is the stipulated constant $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ even at the orbit,This treats gravity as centrally directed for the orbital calculation, while retaining its constant magnitude. The subsequent flat-surface approximation treats its direction as locally vertical.

Burnout height. During the unpowered projectile motion, horizontal velocity is constant. At the top of the trajectory the vertical component vanishes, so the horizontal component must equal the orbital speed $v_o$. At burnout, therefore,These follow either by integrating $\ddot{y}=-g$ or by the vertical work-energy theorem. Substituting the preceding constant-$g$ orbital result givesThe vertical component is initially positive, as required to reach the orbit after burnout.

Alternative inverse-square interpretation. A real Earth's exterior Newtonian gravitational field is not constant over a $1500\mathrm{km}$ altitude change. If $9.8\mathrm{m}{\mathrm{s}}^{-2}$ instead means the surface value $g_0$, then Newton's law of universal gravitation gives $GM=g_0{R}_E^2$ andUsing this orbital speed but the paper's constant-$g$ flat projectile motion approximation for the coast would giveThis alternative changes the global constant-gravity assumption; it is not the same model as the $183\mathrm{km}$ result. An exact curved Kepler orbit coast would also abandon the prescribed flat trajectory. The printed assumptions support the constant-magnitude answers above, while the inverse-square values explain the different answer obtained from the usual physical Earth-orbit formula.

Field of discrete masses. Newton's law of universal gravitation and superposition give the gravitational field away from the source points:It is the force per unit test mass, with units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$, and points toward each positive source mass.

Infinite circular cylindrical rod. Let $b$ be the radius of the circular cross-section, so $A=\pi b^2$, and write $\lambda =\rho A$ for the linear mass density. Choose a coaxial cylindrical Gaussian surface with radius $r>b$ and length $L$. Translation and rotation symmetry make $\mathbf{g}=g_r(r)\hat{\mathbf{r}}$. The end caps have zero field flux because their normals are axial; on the curved surface the outward-normal component is the constant negative value $g_r$. Gauss's law for gravity therefore givesThe gravitational field of an infinite uniform cylinder inside the material instead has enclosed mass $\rho \pi r^{2}L$, giving $g_r=-2\pi G\rho r$. The exterior result assumes a circular cylindrical rod; area by itself would not specify the exact near field of a noncircular cross-section.

In the left panel the dashed circle is a transverse section of the coaxial Gaussian surface; the field points inward and the surface normals outward. Its side area is $2\pi rL$.

For an exterior circular orbit, Newton's second law equates the force required for centripetal acceleration to gravity:The orbital speed is independent of radius and of the orbiting object's mass, for orbits outside the infinite rod.

Finite rod. Taking the supplied field as the gravitational field of a finite uniform line segment, the corresponding balance at radius $R$ isSince $M=2a\rho A=2a\lambda$, its two limits areThe first reproduces the infinite-rod result: a nearby observer sees a very long source. The second is the familiar circular orbit speed about a point mass, since a distant observer cannot resolve the rod's length. The dimensionless curve isIt starts at a constant plateau and decreases smoothly, approaching an $R^{-1/2}$ dependence far away. For $R>0$, $dv/dR=-\sqrt{GM}R/[2(R^2+a^2{)}^{5/4}]<0$, consistent with the sketch.

Derivation of the finite-rod field. Put the source along $-a<z<a$ on the $z$ axis. In its central perpendicular plane, an element $dm=\lambda dz$ is at distance $s=\sqrt{r^2+z^2}$. Its field magnitude is $G\lambda dz/s^2$, and its inward radial component has the extra factor $r/s$. The axial components cancel between $z$ and $-z$. ThusReplacing $2a\lambda$ by $M$ givesThe integral treats the source as a line mass. For a finite cylinder of nonzero cross-section, the displayed finite-rod formula is a thin-rod approximation, not an exact field for all $r$. Its near-rod limit should therefore be read as $b\ll R\ll a$, keeping the orbit outside the material; the apparent line-source singularity at $r=0$ does not describe the interior of a real cylinder.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

For a low circular orbit, Newton's law of universal gravitation gives $P=2\pi \sqrt{R^3/(GM)}$. Using $M=4\pi \bar{\rho }{R}^3/3$ yieldsThus the Moon and Mars have roughly comparable mean densities: low circular-orbit period measures mean density. Equal periods would mean equal mean densities in this approximation. This does not require their masses or radii to be similar. The orbit altitudes must be small compared with the body radii, and the bodies are being treated as approximately spherical.

An active galactic nucleus is a compact, exceptionally luminous central source powered chiefly by black-hole accretion rather than ordinary stellar fusion. Typical signatures include a broad spectral-energy distribution extending from radio to X-rays, rapid variability implying a small emitting region, strong optical or ultraviolet emission lines and, in some objects, radio jets and lobes. Broad permitted lines indicate rapidly moving dense gas; narrower forbidden lines can arise farther out. Obscuration and viewing orientation can hide the broad-line region, and many active nuclei are radio quiet, so no one signature is required in every object.

For the Schwarzschild circular-ring energy shift, set $r_g=GM/c^2$ and let $E_0$ be the photon energy in the orbiting gas frame. A local static observer in Schwarzschild spacetime measures $E_{\mathrm{stat}}$. For gas receding along the line-of-sight component of its velocity, special-relativistic frequency transformation givesPropagation from that static frame to a distant stationary observer contributes $E_{\infty }=E_{\mathrm{stat}}\sqrt{1-2r_g/r}$. HenceThe factors describe gravitational redshift, transverse time dilation and the longitudinal Doppler shift. Here the emitter is on the maximally receding azimuth of a circular disc ring. A general azimuth has a smaller projected velocity, and a fully ray-traced image must use the photon direction at emission rather than equating that local direction with the distant inclination. The supplied expression is the direct-projection transfer model, not a complete light-bending calculation.

The innermost stable circular orbit is the marginal stable timelike circular geodesic separating stable circular motion from radial instability. In Schwarzschild spacetime,For example, the circular-orbit specific angular momentum satisfies $j_c^2=GM{r}^2/(r-3r_g)$; its radial stability changes where $\mathrm{d}{j}_c^2/\mathrm{d}r=0$, at $6r_g$. This is outside both the horizon at $2r_g$ and the circular photon orbit at $3r_g$.

The printed local orbital-speed relation gives ${\beta }^2=r_g/(r-2r_g)$. Combining the gravitational and transverse factors,At the ISCO, $\beta =1/2$, soThe Edge-on Schwarzschild ISCO redshift envelope is reached for an edge-on disc on its receding side:The approaching edge has $1+z=\sqrt{2}(1-\frac{1}{2}\sin i)$. For a fixed inclination below $90$ degrees the maximum is smaller; the quoted number is an upper envelope, not the value for every observer. Within the stated ring model the receding factor increases inward across stable orbits, so the inner edge sets that envelope. Emission from a plunging region, occultation and returning rays are outside this ISCO-emitting model.

The Kerr ISCO efficiency limits follow from equatorial circular orbits in the Kerr metric. Prograde motion has a smaller ISCO than the non-spinning case, while retrograde motion has a larger one. A zero-torque thin disc radiating the orbital binding energy before the gas plunges has efficiency $\eta =1-E_{\mathrm{ISCO}}$, where $E_{\mathrm{ISCO}}$ is energy per unit rest-mass energy at infinity. The ideal limiting values areThus more orbital binding energy is available in the prograde inner disc. The extremal values are formal geodesic limits: photon capture, inner stresses, outflows and finite accretion thickness change the efficiency of real flows. An ISCO inside the ergoregion is not a claim that material or radiation escapes from inside the horizon.

For accretion output relative to galactic binding energy, distinguish accreted rest mass from the mass retained by the black hole. At constant radiative efficiency, building a hole from a negligible seed to $M_{\mathrm{BH}}=\alpha M$ emitsA non-negligible seed replaces $\alpha M$ by the actual mass increase. Let the first use of ${\sigma }_v$ mean the total three-dimensional rms random speed. For an approximately virialized galaxy with negligible boundary term, $K=\frac{1}{2}M{\sigma }_v^2$ and $W=-2K=-M{\sigma }_v^2$. The ratio to the magnitude of its gravitational potential energy is thereforeIf “binding energy” means the net energy needed to unbind the virialized system, $|K+W|=M{\sigma }_v^2/2$, the ratio is twice as large. The widely used order-of-magnitude estimate drops the retained-mass correction and writes $\eta \alpha (c/{\sigma }_v{)}^2$; it should not be confused with a precisely fixed structural coefficient.

For the numerical parameters given,No efficiency is specified, so this function is the numerical result. For an explicitly chosen illustration $\eta =0.1$, it is about $111$, or $222$ relative to the net virial binding energy. The non-spinning zero-torque efficiency gives about $60.6$ relative to $|W|$; the ideal retrograde and prograde limits give about $39.2$ and $731$, respectively. Spin affects the ratio through the adopted efficiency and orbital orientation, not through a universal rule that a larger unsigned spin always means a larger efficiency.

Such large energy ratios make even a modest coupling of the output to galactic gas dynamically important. Radiation and winds can heat or eject gas and reduce star formation or later black-hole feeding. For the $\eta =0.1$ illustration a percent-level coupling is already comparable to $|W|$. Energy available is not energy necessarily deposited: cooling, escape directions, gas geometry, photon opacity and the duration of activity determine the actual feedback. It does not follow that the entire stellar galaxy must be disrupted.

For the swept-shell calculation, the problem now explicitly redefines ${\sigma }_v$ as a one-dimensional line-of-sight dispersion. The specified isothermal profile hasNeglecting central-hole gravity and the shell's self-gravity, the inward force due to the remaining interior component isIn a single-scattering momentum-driven model with the shell intercepting the central luminosity, $F_{\mathrm{rad}}=L/c$. At the hydrogen Eddington luminosity this becomes $4\pi G{M}_{\mathrm{BH}}{m}_p/{\sigma }_T$. Sustained outward acceleration against that weight requiresand henceThe radius cancels because the swept isothermal mass is proportional to $R$. This is the isothermal-galaxy momentum-balance shell threshold, with its fourth-power dispersion scaling.

The coefficient identifies the approximations. Self-gravity in a thin spherical shell expansion adds the shell's inward pressure-equivalent force $G{M}_{\mathrm{sh}}^2/(2R^2)$ and replaces $f(1-f)$ by $f(1-f/2)$. A shell intercepting only a fraction of the photon momentum needs a larger luminosity; Thomson-thin material does not automatically receive $L/c$. Conversely, multiple scattering or an energy-driven hot wind changes the force model. A shell already carrying sufficient outward kinetic energy need not satisfy this instantaneous acceleration condition everywhere to escape. The printed inequality is the stated quasistatic, fully coupled momentum-balance estimate, not an assumption-free lower bound for every possible shell trajectory.

geodesic integrals and affine normalization. Put $F(r)=1-2\mu /r$. Choose the affine parameter to be proper time, so $g_{ab}{\dot{x}}^a{\dot{x}}^b=c^2$. Cyclic $t$ and $\varphi$ in the geodesic Lagrangian give $F\dot{t}=k$ and $r^2\dot{\varphi }=h$. Substituting into the four-velocity normalization giveshenceAn arbitrary constant rescaling of affine parameter changes its normalization; the displayed unit constant assumes the proper-time choice.

circular orbit. Define $V(r)=h^2/r^2-2\mu h^2/r^3-2\mu c^2/r$. The radial geodesic equation is $\ddot{r}=-V^{\prime }/2$. A circular geodesic therefore needs $V^{\prime }(R)=0$, not merely $\dot{r}=0$ at an instant. Solving and substituting into the radial first integral givesExistence requires $R>3\mu$. circular orbits below $6\mu$ are radially unstable, although they still define the instantaneous comparison here.

orthogonal local velocities in Schwarzschild spacetime. Use Alice's static orthonormal frame at the meeting event. Her proper clock is $d{\tau }_A=\sqrt{F}dt$ and tangential proper distance is $Rd\varphi$. Bob's local tangential speed isCarol has $h_C=0$ and $k_C=1$, because she starts from rest at infinity. The radial integral gives ${\dot{r}}_C=-c\sqrt{2\mu /R}$. Her static-frame radial speed is $(dr/\sqrt{F})/(\sqrt{F}dt)=dr/(Fdt)$, and $F{\dot{t}}_C=1$, soThis is inward radial motion, perpendicular to Bob's tangential motion in Alice's frame.

Their Lorentz factors relative to Alice are ${\gamma }_{\mathrm{AB}}=\sqrt{(R-2\mu )/(R-3\mu )}$ and ${\gamma }_{\mathrm{AC}}=\sqrt{R/(R-2\mu )}$. The four-velocity contraction therefore gives

Special-relativistic consistency. Let Bob's tangential velocity in Alice's local frame be $u$ and Carol's inward radial speed be $w$. A boost to Bob's frame gives Carol components $v_{\mathrm{tang}}^{\prime }=-u$ and $v_{\mathrm{rad}}^{\prime }=-w/{\gamma }_u$. ConsequentlyThus the three measured relative speeds agree with local Lorentz velocity addition. Ordinary Euclidean subtraction of the two velocity vectors would omit the transverse-velocity transformation.

In Newtonian gravity, the radial equation with conserved specific angular momentum is $\ddot{r}=-GM/r^2+h^2/r^3$. The substitution $u=1/r$ therefore gives the linear differential equationChoosing $\varphi =0$ along perihelion, its elliptic orbit solution isHere $p$ is the semilatus rectum. One can check the last equality directly from $r_{\mathrm{peri}}=p/(1+e)$, $r_{\mathrm{apo}}=p/(1-e)$, and $a_{\mathrm{orb}}=(r_{\mathrm{peri}}+r_{\mathrm{apo}})/2$.

To find Schwarzschild perihelion precession, write $u=u_0+\delta u$ in the relativistic equation and retain first order in $\mu /p$. The linear ordinary differential equation for the correction isThe constant and double-frequency forcing change the orbit's shape without accumulating an angular shift. The resonant $\cos \varphi$ term producessince $(d^2/d{\varphi }^2+1)(\varphi \sin \varphi )=2\cos \varphi$. The oscillating part of the perturbed orbit is consequently the first-order expansion ofSuccessive perihelia are separated by $2\pi /(1-3\mu /p)$, slightly more than a full revolution. Their forward angular advance is thusin radians per orbit. This leading expression applies when $\mu /p\ll 1$. For a perfectly circular orbit there is no uniquely located perihelion; the same expression describes the limiting mismatch of radial and azimuthal frequencies for a small nonzero orbital eccentricity.

Use the $+---$ metric signature and abbreviate $f(r)=1-2\mu /r$, with $\mu =GM/c^2$ a length. Dots initially mean differentiation with respect to an affine parameter $\lambda$. The geodesic Lagrangian in the equatorial plane isThe planarity of geodesics in spherical symmetry permits this plane to be chosen. More explicitly, the full angular Euler-Lagrange equation is $\ddot{\theta }+2\dot{r}\dot{\theta }/r-\sin \theta \cos \theta {\dot{\varphi }}^2=0$, so initial data $\theta =\pi /2$, $\dot{\theta }=0$ preserve it.

Both $t$ and $\varphi$ are cyclic coordinates. Their Euler-Lagrange equations give conserved conjugate momenta $p_t=c^{2}f\dot{t}$ and $p_{\varphi }=-r^2\dot{\varphi }$. Defining constants $k,h$ therefore givesThe corresponding second-order geodesic equations areand variation of $r$ givesThe geodesic Lagrangian has no explicit $\lambda$ dependence. Its conserved Lagrangian energy function is ${\dot{x}}^{\alpha }\partial \mathcal{L}/\partial {\dot{x}}^{\alpha }-\mathcal{L}=\mathcal{L}$, since it is quadratic in the tangent components. Thus twice this function is a constant:This is the constant squared tangent norm of an affinely parametrized geodesic.

For a massive particle choose $\lambda =\tau$, its proper time, and obtain $\kappa =c^2$. For a photon, the tangent is null and $\kappa =0$. A general rescaling of a massive particle's affine parameter rescales a positive $\kappa$; the numerical value $c^2$ requires the proper time choice. With that choice, $h=L_z/m$ is signed specific angular momentum, of units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, and $k=E/(m{c}^2)$ is dimensionless conserved specific Killing energy. The time coordinate is normalized to proper time at infinity, so this $E$ is the particle's conserved energy referred to infinity. A photon's constants scale with the normalization of its null tangent; it has no rest frame or proper time parametrization.

Substitution of $\dot{t}=k/f$ and $\dot{\varphi }=h/r^2$ into the norm gives the general radial first integralFor a massive particle this becomesAdopting the usual half-normalization for the effective potential for timelike Schwarzschild geodesics, defineAll terms have units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$, or energy per unit mass. The first two terms are the Newtonian gravitational and centrifugal contributions; the last is the relativistic correction.