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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.

At closest approach the velocity is tangential. The conserved specific orbital energy and specific angular momentum areThe negative energy gives an elliptic orbit. Its semi-major axis isSince $r_p+r_a=2a$,This treats the star as fixed, appropriate to $m\ll M$. Equivalently, equate the energy at the two turning points using $v_a=\ell /r_a$. The planet mass cancels; a circular-orbit relation would be inappropriate here.

The three classic tests probe orbital motion, the propagation of light, and clock rates in a gravitational field.

Schwarzschild perihelion precession. For an orbit with semi-major axis $a$ and orbital eccentricity $e$, the leading advance per revolution isThe relevant observation for Mercury is the residual advance after the much larger ordinary perturbations have been included, approximately $43$ arcseconds per century.^[ref] An orbital fit must account for the other planets and reference-frame effects before interpreting that residual as a relativistic contribution.

Schwarzschild light deflection. A ray with large impact parameter $b$ passing a nearly spherical mass is deflected throughAt the Sun's limb this is about $1.75$ arcseconds. The Eddington experiment used the 1919 solar eclipse to photograph background stars near the obscured Sun, then compared their positions with reference photographs. The measured effect is an apparent outward displacement of the background stars, not a direct photograph of a curved light path. Telescope focus, plate scale and atmospheric effects must be controlled when extracting such small shifts.^[ref]

Gravitational redshift. For a photon emitted lower in a weak static field and received a height $H$ above it,The Pound-Rebka experiment used the narrow recoil-free nuclear resonance supplied by the Mössbauer effect. A controlled source motion supplies a small Doppler shift that compensates the gravitational shift, allowing a change in resonant absorption to reveal it across a laboratory tower.^[ref] The leading weak-field redshift follows from the equivalence principle and therefore does not, by itself, distinguish every competing metric theory of gravity.

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.

A circular orbit requires both $\dot{r}=0$ and $V_{\mathrm{eff}}^{\prime }(R)=0$. Differentiating at fixed $h$ givesThe radial first integral also givesFor the future-directed circular orbit at $R=4\mu$, these constants are $|h|=4\mu c$, $k=1$, and $f=1/2$. ThereforeSince Schwarzschild time agrees with the clock time of a static observer at infinity, the intended infinity-referred tangential orbital speed isThis uses the circumferential areal radius $R$ with the clock at infinity; it is not a local measurement made at the orbit. A local static observer's clock instead has $d{\tau }_{\mathrm{stat}}=\sqrt{f}dt$ and measures $v_{\mathrm{local}}=v_{\infty }/\sqrt{f}=\boxed{c/\sqrt{2}}$. These two operationally different speeds explain the significance of the phrase “at infinity”.

The effective-potential stability of a circular orbit follows fromConsequently the $R=4\mu$ orbit exists but is unstable to radial perturbations. The original plot below shows its position at a maximum, for $h=4\mu c$.

For a weak-field elliptic orbit, the relativistic correction makes the direction of closest approach advance after each radial cycle. The derivations under i and ii below giveper orbit to first order, where $a_{\mathrm{orb}}$ is the semi-major axis and $e$ the orbital eccentricity. For Mercury, the additional Schwarzschild perihelion precession is approximately 43 arcseconds per century, over and above the much larger Newtonian precession produced primarily by other planets. It is the unexplained residual after those effects have been included that provides the classic quantitative test of general relativity; one should not identify the whole observed precession with this relativistic correction.^[ref]

The labels i and ii in the original paper introduce optional orbit-equation aids, rather than separate mandatory subquestions. They are retained here for derivations of both aids and of the perihelion result.

A Sun-synchronous orbit maintains approximately constant orientation relative to the mean Sun, so corresponding passes over a location occur at nearly the same local solar time. It is usually a nearly polar, slightly retrograde low Earth orbit. Earth rotates beneath the successive tracks, allowing broad coverage; the satellite is not stationary over a location and does not necessarily revisit that location every day.

It is attained by selecting altitude, orbital eccentricity and orbital inclination so that nodal precession caused by Earth's equatorial bulge matches the mean apparent annual motion of the Sun, about $360^{\circ }$ per year eastward. To leading order,Here $a$ is the semi-major axis, $e$ the orbital eccentricity, $i$ the orbital inclination, $n=\sqrt{G{M}_E/a^3}$ the mean motion, and $J_2$ Earth's dimensionless quadrupole coefficient. A retrograde inclination has $\cos i<0$, giving the required positive precession; typical imaging altitudes yield inclinations around $97^{\circ }$–$99^{\circ }$. The launch establishes the required orbital plane, and subsequent orbit maintenance limits drift.

This geometry suits repeated optical and multispectral imaging of land, vegetation and oceans, because comparable local time reduces illumination and shadow differences. Landsat and Sentinel-2 are examples. Radar missions such as Sentinel-1 also use this orbit for regular global sampling even though their illumination is supplied by the radar. Constant local time helps comparison but does not remove seasonal changes in solar elevation.^[ref]

For a Kepler orbit with pericentre distance $q$, apocentre distance $Q$, and semi-major axis $a=(q+Q)/2$, a uniform mean anomaly givesThe two radial passages per orbital period are included, and ${\int }_q^Q{p}_r(r)dr=1$.

A circumplanetary orbit whose orbital period equals the central planet's spin period. Its semi-major axis follows from Kepler's third law. Matching periods alone does not fix the satellite above one ground location; that additionally requires a planet-stationary orbit.

The orbit-averaged Poynting–Robertson drag equations preserve $C=a(1-e^2)e^{-4/5}$ for $0<e<1$. This relates shrinking semi-major axis to decreasing orbital eccentricity; a circular orbit is treated by its separate limiting solution.

For a common semi-major axis $a$ and proper eccentricity $e_p$, randomly distributed proper longitudes of periapsis give an annulus of radii $a(1-e_p)$ and $a(1+e_p)$ about the point displaced from the star by minus $a$ times the forced eccentricity vector, to first order in orbital eccentricity.

A grain with radiation-pressure coefficient $\beta <1$ feels effective central gravitational parameter $G{M}_{\star }(1-\beta )$. Combining its mean motion with the planet's mean motion at an exterior-type $(j+1):j$ frequency commensurability gives the displayed semi-major axis. The commensurability lies geometrically exterior only for $j<\sqrt{1-\beta }/(1-\sqrt{1-\beta })$. Radiation changes the grain's attraction, so the same frequency ratio need not imply the usual radial ordering.

The nominal semi-major axis of an exterior $(p+q):p$ mean-motion resonance follows from $(p+q)n_r=p{n}_p$ and Kepler's third law. Apsidal precession displaces the actual libration center because stationarity of the resonant argument requires $(p+q)n-p{n}_p-q\dot{\varpi }=0$.

Half the peak-to-peak line-of-sight orbital velocity range of a star in a single-companion Kepler orbit. It equals $K=2\pi a_{\star }\sin i/[P\sqrt{1-e^2}]$, where $a_{\star }$ is the stellar barycentric semi-major axis, $P$ the orbital period, $i$ the sky-plane orbital inclination, and $e$ the orbital eccentricity. The Doppler spectroscopy formula follows from Kepler's third law for the relative orbit, $a^3=G(M_{\star }+M_p)P^2/(4{\pi }^2)$.

The motion of a star around the astronomical barycenter caused by an orbiting companion. In an isolated two-body problem with relative semi-major axis $a$, the stellar semi-major axis is $a_{\star }=M_{p}a/(M_{\star }+M_p)$. Doppler spectroscopy measures its line-of-sight velocity and astrometry measures its projected position on the sky.

For an instantaneous in-plane change of velocity of magnitude $\gamma v_K$ on a circular orbit, with angle $\theta$ measured from the forward tangential direction, the new specific orbital energy gives $a^{\prime }/a=(1-2\gamma \cos \theta -{\gamma }^2{)}^{-1}$. The specific angular momentum gives $e^{\prime 2}=1-(1-2\gamma \cos \theta -{\gamma }^2)(1+\gamma \cos \theta {)}^2$. The signed semi-major axis becomes negative for an unbound hyperbolic Kepler orbit.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.

For a Kepler orbit with semi-major axis $a$, the vis-viva equation isFor the convention that a hyperbolic Kepler orbit has positive semi-major-axis magnitude $a$, its final term instead has a plus sign.

Anisotropic thermal radiation produces recoil that changes the semi-major axis of a rotating solid body. Finite thermal inertia provides the temperature lag. Large bodies have small acceleration per unit mass, while very small bodies become nearly isothermal. The Yarkovsky effect can supply bodies to dynamical escape regions.