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For a bound orbit let $E<0$ be the specific energy and $L$ the specific angular momentum. Introduce the dimensionless radius suggested in the question,The radial energy equation becomeswhere $x_{-}$ and $x_{+}$ are the two turning points. Comparing coefficients givesThe time from periapsis to apoapsis isThus the radial orbital period isIt depends on $E$ but not on $L$, which is the defining isochrone property.

At periapsis, $\lambda =\varpi$. If the planet has longitude ${\lambda }_0$ there, thenso ${\varphi }_1/p$ is the angular displacement of periapsis from the planet, modulo $2\pi /p$. At an astronomical conjunction, $\lambda ={\lambda }_1={\lambda }_c$, andso ${\varphi }_1/q$ is the conjunction longitude measured from periapsis, with the $q$ possible branches differing by $2\pi /q$.

The planet's mean motion is $n_1=(p+q)n/p$. Hence the synodic period, or mean interval between conjunctions, iswhere $T=2\pi /n$ is the planetesimal's orbital period.

Orbital energy. Treat the star as stationary to leading order in $m/M$, with Newtonian gravitational parameter $\mu =GM$. For velocity components in plane polar coordinates, $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$. The kinetic energy is $m{v}^2/2$. Integrating Newton's law of universal gravitation with zero gravitational potential energy at infinite separation gives $U(r)=-GMm/r$. Thuswhere $J=m{r}^2\dot{\varphi }$ is the satellite's angular momentum. The first term is radial kinetic energy; the second is transverse kinetic energy, also the centrifugal contribution to the radial effective potential; the last is the attractive gravitational potential energy. These are not three independent sources of energy: the centrifugal term is already part of the ordinary kinetic energy.

For an elliptic Kepler orbit of semi-major axis $a$ and orbital eccentricity $e$, the polar equation of a Kepler orbit $r=p/(1+e\cos \varphi )$ has semilatus rectum $p=J^2/(GM{m}^2)=a(1-e^2)$. Substituting $\dot{r}=(Je/(mp))\sin \varphi$ into the preceding expression reduces it toFor a circular Kepler orbit, $a=R$, $v_0^2=GM/R$, $K=GMm/(2R)$, and $U=-GMm/R$. Negative total energy describes a bound orbit.

Rapid, symmetric mass loss. Assume the mass is removed on a timescale short compared with the satellite's orbital period, without an impulsive kick or appreciable wind drag on the satellite. The post-loss gravitational parameter is ${\mu }^{\prime }=G(1-f)M$ with $f=0.01$. At the change, $r=R$, $\dot{r}=0$, and $v=v_0=\sqrt{GM/R}$ are unchanged. Spherical symmetry keeps the force central, so the specific angular momentum $j=J/m=\sqrt{GMR}$ is conserved. Orbital energy is not conserved across the changing gravitational field:The new semi-major axis is consequentlyThe original speed now exceeds the new circular speed at $R$, and the outward radial acceleration is $v_0^2/R-{\mu }^{\prime }/R^2=fGM/R^2>0$. Thus the release point is the new pericentre, not the apocentre. The impulsive orbital expansion under isotropic mass loss givesFor the stated one-percent loss,An independent angular momentum check is $j^2={\mu }^{\prime }{a}^{\prime }(1-e^{\prime 2})=GMR$. The impulsive orbit stays bound for $f<1/2$; loss of half the central mass is the zero-orbital energy limit.

What the wind timescale changes. The source specifies a spherically symmetric wind but does not give its duration. The preceding nearest and furthest distances require the rapid-loss assumption. In the opposite, slow limit, adiabatic orbital expansion under isotropic mass loss preserves an initially circular orbit to leading adiabatic order. Conserved specific angular momentum then gives $j^2=GM(t)r(t)$, hence $M(t)r(t)=MR$. After the same loss,For a finite intermediate timescale, the orbit depends on the mass-loss history; the percentage alone does not select a unique pair of turning distances. Neglecting wind recoil and wind drag is common to both idealizations.

Neglecting the planets' masses compared with the Sun's mass, Newton's law of universal gravitation supplies the centripetal acceleration of each circular orbit:Thus Kepler's third law, calibrated by the Earth's orbital period of one year and orbital radius of one astronomical unit, givesBoth orbits are taken to have the same direction. Their relative angular speed is $2\pi (1/T_J-1/T_S)$, so the Sun-centred interval between successive conjunctions is the synodic periodUsing the supplied rounded orbital radii explains the small difference from the commonly quoted approximately twenty-year interval.

From the Earth, the two lines of sight are displaced differently because Jupiter and Saturn are at different distances. The Earth's annual motion along its orbit therefore shifts the dates of apparent conjunctions by amounts of order months. The broad recurrence remains about twenty years, but successive apparent conjunctions need not be equally spaced: apparent retrograde motion can produce a group of three conjunctions separated by months during one encounter. The Earth's daily rotation adds a much smaller change of viewpoint. This is a geometrical comparison in the stipulated coplanar model; daylight and the horizon can prevent a particular conjunction from being visible at a chosen surface location.

For a satellite of mass $m$ moving in the Newtonian gravitational field of a much larger, effectively fixed mass $M$, put $\mu =GM$. Its conserved mechanical energy, with gravitational potential energy zero at infinity, isHere $r$ is the instantaneous centre-to-centre distance, $v$ the satellite's speed, $G$ the gravitational constant, and $a$ the semi-major axis of its bound elliptic orbit. The kinetic energy and potential energy vary around the orbit, while their sum remains constant. Dividing by $m$ and solving for $v$ gives the vis-viva equation, $v^2=\mu (2/r-1/a)$.

For this Hohmann transfer orbit, the periapsis and apoapsis distances are $r_1=1\mathrm{AU}$ and $r_2=1.52\mathrm{AU}$, soThe vis-viva equation gives the endpoint speedsAs a check, conservation of angular momentum at the two endpoints gives $r_1{v}_{\mathrm{p}}=r_2{v}_{\mathrm{a}}$ because each velocity is perpendicular to its radius. Using $1\mathrm{AU}=1.49598\times 10^{11}\mathrm{m}$ and $\mu =1.32712\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ gives $v_1=29.78\mathrm{km}{\mathrm{s}}^{-1}$ andThe departure impulse must increase the speed in the direction of the initial circular orbit:These are Sun-centred speeds; leaving the Earth's own gravitational field would require an additional calculation beyond this idealized solar orbit transfer.

At $r_2$, the desired circular orbit speed is $v_2=\sqrt{\mu /r_2}=24.16\mathrm{km}{\mathrm{s}}^{-1}$. The arrival impulse is thereforeThe required thrust on the spacecraft is forward, tangent to its orbit, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's velocity; this distinguishes the direction of thrust from the direction in which the exhaust is fired.

The Hohmann transfer time is half the orbital period of the transfer ellipse. From Kepler's third law, relative to the Earth's one-year orbital period,Meanwhile Mars's orbital period is $T_M=1.52^{3/2}\mathrm{yr}=1.8740\mathrm{yr}$. The spacecraft arrives $\pi$ radians from its departure position. For a rendezvous, Mars must reach that same position after advancing through $2\pi t_H/T_M$. Thus its initial lead angle over the departing spacecraft isMars must be ahead by this angle in the common direction of motion around their orbits. At departure the spacecraft and the Earth share the inner orbital position; arrival occurs on the opposite side of the Sun.

In the ideal semidiurnal tide model, successive high tides occur after the Earth–Moon relative angle advances by $\pi$. Since the motions have the same sense, subtract their angular speeds:The lunar orbital period is sidereal, so the consistent terrestrial rotation period is $T_E=23.9345\mathrm{h}$. With $T_M=27.3\times 24\mathrm{h}$, this gives about 12 h 25 min. The sidereal terrestrial period is distinct from the 24-hour solar day.^[ref]

If the Earth's rotation is approximated as one revolution in 24 hours, the same calculation instead gives 12 h 27 min. That small difference is the day convention, not a change of sign in the relative angular speed. Actual coastal tide times also depend on ocean dynamics and the solar tide.

The Hohmann transfer orbit has periapsis $r$ and apoapsis $2r$, so its semi-major axis is $3r/2$. Kepler's third law gives a period $(3/2{)}^{3/2}$ times the original orbital period. Only half the ellipse is traversed:The relevant radii are measured from Earth's centre, not its surface.

For the attractive inverse-square force, choose zero potential energy at infinity, so $U=-A/r$. At either radial turning point $r_p,r_a$, the kinetic energy is purely tangential, with conserved angular momentum $J$. ThusIts two roots are the periapsis and apoapsis distances. Their sum is $r_p+r_a=-A/E$, whereas the ellipse has $r_p+r_a=2a$. Therefore $a=-A/(2E)$, with $E<0$ for a nondegenerate bound ellipse. The circular case follows by the double-root limit.

For the two-body problem, define $\mathbf{r}={\mathbf{r}}_B-{\mathbf{r}}_A$, total mass $M=m_A+m_B$, reduced mass $\mu =m_A{m}_B/M$, and center of mass $\mathbf{R}$. The relative equation is $\mu \ddot{\mathbf{r}}=\mathbf{F}$, while the isolated center of mass has $\ddot{\mathbf{R}}=0$. Once the relative orbit is known,The relative energy uses $\mu$ in its kinetic energy and $A=G{m}_A{m}_B$ in the gravitational potential energy.

In the equal-mass case the initial relative speed is $u$, $\mu =m/2$, and the center-of-mass velocity is half B's initial velocity. ThusNegative relative energy requires $u^2<4Gm/a_1$. The escape threshold is $u_E=2\sqrt{Gm/a_1}$; equality is marginal parabolic escape, so a strictly bound pair has $u<u_E$.

For $u=u_E/2=\sqrt{Gm/a_1}$, $E_{\mathrm{rel}}=-3G{m}^2/(4a_1)$ and the relative semimajor axis is $a=2a_1/3$. The initial transverse speed is below the relative circular value $\sqrt{2Gm/a_1}$, so the initial turning point is apoapsis. HenceChoose the initial separation along $+x$ from A to B, with B's initial velocity $+u\hat{\mathbf{y}}$. Conservation of angular momentum gives relative periapsis speed $3u$. At the opposite end of the ellipse the direction is $-\hat{\mathbf{y}}$, soThus B has the same speed as initially but the opposite velocity. A has $2u\hat{\mathbf{y}}$ at that instant. A value $3u/2$ for B would be its center-of-mass-frame speed, not the requested initial laboratory frame result.

The relative orbital period follows from Kepler's third law with total gravitating mass $2m$:In the center-of-mass frame each star follows an ellipse of semimajor axis $a_1/3$ and eccentricity $1/2$, with the center of mass at a focus; they remain opposite on their simultaneous position vectors. In the laboratory frame add the uniform drift $ut/2$ along $y$. The laboratory paths of an equal-mass binary are therefore not closed ellipses, even though their relative motion repeats.

For two bodies interacting through a potential energy that depends only on their separation, introduce the centre of mass coordinate and the two-particle relative coordinate:Their kinetic energy separates exactly:The centre of mass moves uniformly if there is no external force, while the internal two-body problem reduces to one effective particle of reduced mass $\mu$:The original trajectories follow from ${\mathbf{r}}_1=\mathbf{R}+(m_2/M)\mathbf{r}$ and ${\mathbf{r}}_2=\mathbf{R}-(m_1/M)\mathbf{r}$.

For Newtonian gravity, $U=-G{m}_1{m}_2/r$, so the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3$. Conserved relative angular momentum and energy then give the usual conic sections. For a bound orbit of relative semimajor axis $a$,The two bodies share this orbital period and orbit their common centre of mass with semimajor axes in inverse proportion to their masses. If $m_2\gg m_1$, then $\mu \simeq m_1$, recovering the fixed-heavy-centre approximation. Using $M$ as the internal inertial mass would confuse centre-of-mass translation with relative motion.

For two circular Kepler orbits about the same body, Kepler's third law gives $T^2/r^3=\text{constant}$. The reference radius is measured from Earth's centre:A geostationary orbit is prograde and equatorial, with one sidereal day as its orbital period. Using $T_g\simeq 1436\min$ and the supplied $T_1=109\min$,Using $24\mathrm{h}$ instead changes this estimate by less than one per cent. This is the orbital radius, not the altitude; the corresponding altitude is about $3.00\times 10^4\mathrm{km}$.

There is a numerical inconsistency in the source data. The quoted reference orbit implies $G{M}_E=4{\pi }^2{r}_1^3/T_1^2\simeq 2.56\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, appreciably below the measured terrestrial value, about $3.99\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$. Using the latter instead gives the familiar geostationary orbit radius $4.22\times 10^4\mathrm{km}$. The boxed estimate consistently uses the period actually printed in the question.

For the two-body approximation with a dominant central mass, Kepler's laws of planetary motion are:

The first law includes a circular Kepler orbit as a special case. The third-law constant is the same for bodies orbiting the same dominant mass.

Use a Hohmann transfer orbit. At the initial radius $r_0$, make a short prograde burn parallel to the existing velocity. The increased speed changes the circular Kepler orbit into an elliptic orbit whose periapsis is the burn point and whose apoapsis lies on the desired larger circle. Coast for half the transfer orbital period to apoapsis.

At that point the transfer velocity is again tangent to the target circle, but its magnitude is smaller than the circular speed there. A second short prograde burn raises it to the required circular speed. Thus both burns increase the instantaneous speed and preserve its direction at the burn, even though the final circular speed is lower than the original one. During the coast the spacecraft loses kinetic energy while gaining gravitational potential energy.

The figure illustrates $3r_0$ for the subsequent parts, but the same construction works for any larger target radius. The impulsive approximation neglects the distance travelled during each burn and keeps both circles and the transfer ellipse coplanar.

The transfer is a Hohmann transfer orbit. Write $r_0$ for the original circular Kepler orbit radius and $a_t$ for the transfer ellipse's semi-major axis. The diagram measures both radii from Earth's centre, giving $r_1=1.2r_0$, and henceThe Hohmann transfer time is half the transfer orbital period, taking the satellite to the opposite end of the elliptic orbit. By Kepler's third law, relative to the original geostationary orbit's orbital period $T_0$,Using one sidereal day, $T_0=86164\mathrm{s}$, gives $t=4.9703\times 10^4\mathrm{s}=13.806\mathrm{h}$; the usual $24\mathrm{h}$ approximation gives $13.844\mathrm{h}$. The arrival impulse changes the velocity into that of the new circular Kepler orbit without adding coast time in the impulsive model.

There is a real discrepancy between the wording and the diagram: an increase in altitude above Earth's surface is different from the radius increase drawn. If altitude is interpreted literally, let $R_E$ be Earth's radius. Thenusing $R_E\simeq 6371\mathrm{km}$ and $r_0\simeq 42164\mathrm{km}$. Thus $13.8\mathrm{h}$ answers the explicitly labelled geometry; $13.5\mathrm{h}$ answers the literal altitude wording.

In a fixed point-mass Newtonian gravitational field, a test mass $m$ experiencesBecause this is a central force, its torque about the attracting centre vanishes and angular momentum is conserved. The orbit lies in a fixed plane and sweeps out equal areas in equal times. Since the force is also conservative, the specific orbital energy is constant:The last two terms form an effective potential. They show how the angular-momentum barrier competes with attraction and determines the turning radii. For nonradial motion the orbit is a conic section:Negative energy gives an ellipse, with $e<1$; zero energy gives a parabola; positive energy gives a hyperbola. The circular limit has $e=0$, $v_c=\sqrt{GM/r}$ and $\epsilon =-GM/(2r)$. For a bound ellipse of semimajor axis $a$, $\epsilon =-GM/(2a)$ and the orbital period is $T=2\pi \sqrt{a^3/(GM)}$. The local escape speed is $\sqrt{2GM/r}$. These are the point-mass, test-particle limits; extended mass distributions or noncentral perturbations change the orbit laws.

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.

Rearrange the supplied form of Kepler's third law to make the gravitational constant the subject:Convert the semi-major axis and orbital period to SI units before substitution:The central value is thereforeFor monomial uncertainty propagation, logarithmic differentiation givesIf the quoted errors are independent standard uncertainties, the variances of these contributions add:Thus ${\sigma }_G=1.61335\times 10^{-12}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, and the measured result under that statistical convention isThe semi-major axis measurement supplies the largest individual contribution to the relative uncertainty.

The paper does not specify whether its errors are statistical uncertainties or bounds. If they are maximum input errors, the first-order worst-case error propagation estimate adds magnitudes:Correlation between input errors would instead require the covariance terms in the general propagation of uncertainty formula; they are not specified.

mass from motion. A resolved Keplerian orbit gives $M=4{\pi }^2{a}^3/(G{P}_{\mathrm{orb}}^2)$ when the orbiting object has negligible mass. At the Galactic centre, repeated infrared astrometry and spectroscopy measure sky motion, acceleration and line-of-sight velocity of stars around Sagittarius A*. Resolving a compact enclosed mass and fitting complete orbits avoids reliance on a single projected velocity. Source confusion, reference-frame stability, distance, inclination and possible extended mass must be included in the fit.^[ref]

For an unresolved stellar-mass binary in a circular orbit, the donor's velocity semiamplitude $K_2$ and orbital period give the binary mass functionIt provides a lower mass bound; donor mass and orbital inclination are needed for an actual mass estimate. For an eccentric orbit, multiply the observational combination $P_{\mathrm{orb}}{K}_2^3/(2\pi G)$ by $(1-e^2{)}^{3/2}$ when evaluating the binary mass function. Resolved gas or maser rotation and stellar-dynamical models also measure masses, provided pressure support, noncircular motion and line-of-sight projection are modeled. Resolving the black-hole sphere of influence improves separation of central mass from the host galaxy.

For active nuclei, reverberation mapping measures a lag $\tau$ between variable continuum and broad-line response. With a line velocity scale $\Delta v$,The factor $f_{\mathrm{geom}}$ accounts for geometry and kinematics and is not known simply from the lag. Cadence, line responsivity, inclination and winds require control. Rapid variability demonstrates compactness but is not by itself a unique mass measurement.

black-hole spin from strong-field emission. Define $a_{\ast }=cJ/(G{M}^2)$. Large-radius Keplerian motion is principally a mass probe and usually insensitive to black-hole spin. In the Kerr model, the radius and binding energy of the innermost stable circular orbit vary with $a_{\ast }$ and orbital orientation, making inner-disk radiation useful.

black-hole continuum fitting infers an inner emitting radius from the disk's thermal spectrum and flux normalization, then identifies it with the innermost stable circular orbit in a relativistic thin-disk model. Independent mass, distance and inclination, together with atmospheric spectral hardening, absorption and contamination, are needed. Spectra should be selected in a disk-dominated regime where truncation and strong advection are limited. A 2011 Cygnus X-1 analysis combined such spectra with independently constrained binary parameters and reported a near-extreme black-hole spin conditional on black-hole spin–orbit alignment and its disk model.^[ref]

black-hole reflection spectroscopy instead models a relativistically broadened iron-line and reflection continuum: orbital Doppler shifts and beaming broaden the profile, while gravitational redshift contributes to the extended red wing. Its innermost emitting radius can constrain black-hole spin if reflection extends to the stable inner edge. Ionization, abundance, absorption, coronal illumination, disk thickness and possible truncation are competing parameters; a broad line is not a direct black-hole spin reading independent of these assumptions. Timing or precession models can add information but introduce their own dynamical identification.

mass is principally inferred from gravitating dynamics; black-hole spin is inferred from strong-field effects with a specified spacetime and accretion model. Agreement between independent methods is especially valuable because disk inner radius, inclination and radiation-transfer assumptions can otherwise mimic a change of black-hole spin.

Kepler's third law for two point masses in a Newtonian gravitational field isHere $a$ is the semimajor axis of their relative orbit, $P$ the orbital period and $G$ the gravitational constant. Equivalently $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, to the accuracy with which these astronomical units normalize the solar mass.

The intact source figure gives orbital period $P=6.0\mathrm{yr}$, stellar radial velocity semiamplitude $K=30.0\mathrm{m}{\mathrm{s}}^{-1}$, and orbital eccentricity $e=0$. A planetary exoplanet transit implies a nearly edge-on orbit, so $\sin i\simeq 1$. Adopt a solar mass for the Solar-type star and initially neglect the companion's mass relative to it.

For a circular solution of the two-body problem, the star's distance from the centre of mass is $a_{\ast }=a{M}_p/(M_{\ast }+M_p)$, and the measured semiamplitude is $K=2\pi a_{\ast }\sin i/P$. Kepler's third law givesHenceThe inferred binary mass ratio is only $0.00183$, validating the approximation. Since $e=0$, the star–planet separation is constant and equals the orbital semimajor axis. Without the exoplanet transit, Doppler spectroscopy would generally give only a minimum companion mass through $M_p\sin i$.

Transit photometry. A planet crossing the stellar disc produces a periodic reduction in flux, approximately $\delta =(R_p/R_{\ast }{)}^2$ for a small, dark planet away from the limb. The orbital period and stellar radius allow an inferred planetary radius, but not its mass by this method alone. The geometric geometric transit probability is roughly $(R_{\ast }+R_p)/a$ for a circular orbit, so the method strongly favours close orbits and large planets. Small signals compete with instrumental noise and stellar activity; eclipsing stellar binaries and blended sources can mimic a planetary exoplanet transit, and long orbital periods require long observing baselines.

Doppler spectroscopy. Periodic shifts of stellar spectral lines reveal the stellar radial velocity caused by motion around the centre of mass. For a light companion, $K\propto M_p\sin i{P}^{-1/3}{M}_{\ast }^{-2/3}/\sqrt{1-e^2}$, favouring massive, short-period companions and low-mass host stars. Unknown orbital inclination gives a minimum mass; stellar activity, line broadening, and instrumental calibration limit precision. Long orbital periods require years of measurements.

Direct imaging. Spatially resolved reflected light or thermal emission separates a planet from its host. The large star–planet brightness contrast and small angular separation demand starlight suppression and precise correction of optical aberrations. The method favours wide orbits around nearby stars, and young, hot, massive planets with stronger thermal emission. Translating brightness to mass depends on planetary cooling models and age; it is difficult to image an Earth analogue close to a bright star.