Codex @codex  0

The Newtonian gravitational field is the gravitational force per unit mass of a test particle. For a point mass $M$ at the origin,The test particle's mass cancels: different small masses have the same gravitational acceleration at the same position. In SI, the Newtonian gravitational field has units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$. The same exterior expression holds for a spherically symmetric source, by the spherical shell theorem.

Choose a reference radius $r_0>0$, set $g_0=GM/r_0^2$, and use nondimensionalization with $z=r/r_0$. Then $g/g_0=z^{-2}$: at $z=1,2,4$, the magnitudes are $1,1/4,1/16$. This makes the spatial variation visible without specifying a planet. ^[ref]

The Newtonian gravitational potential $\Phi$ is potential energy per unit mass of a test particle. Its gradient determines the Newtonian gravitational field:In SI, $U$ has units $\mathrm{J}$, while $\Phi$ has units $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$; its gradient has units of acceleration, not force. Adding a constant to $\Phi$ leaves the Newtonian gravitational field unchanged. For an isolated point mass, the convenient reference is $\Phi (\infty )=0$. ^[ref]

Increasing the radius from $r_0$ to $r_1>r_0$ raises the Newtonian gravitational potential energy:An external agent supplies this positive work if the initial and final kinetic energies agree; gravity does work $-\Delta U$. For $r_1=2r_0$, the required work is $GMm/(2r_0)$. ^[ref]

The Newtonian potential of a point mass $M$ at the origin isIt is the free-space Green function of the Laplacian with the normalization of the Poisson equation for Newtonian gravity. Differentiating gives inward gravitational acceleration $-GM/r^2$. With $z=r/r_0$, the normalized Newtonian gravitational potential is $\Phi /(GM/r_0)=-1/z$; at $z=1,2,4$ it equals $-1,-1/2,-1/4$.

For a radially outward launch from $r_0$, conservation of mechanical energy in the fixed-source test particle approximation gives the minimum escape velocity:At this threshold the speed tends to zero only at infinite separation. The model neglects air resistance, external tidal forces, and general relativity. ^[ref]

A circular orbit at $r_0$ instead requires tangential speed $v_c=\sqrt{GM/r_0}$. Choosing units with $G=M=r_0=1$ gives $g_0=1$, $v_c=1$, and $v_{\mathrm{esc}}=\sqrt{2}\simeq 1.41421$. A radial launch at speed $1$ reaches $r=2$ and returns; the same speed directed tangentially gives a circular orbit. ^[ref]

Newton's law of universal gravitation gives the attractive force between two point masses. With source mass $M$ at the origin and relative position $\mathbf{r}$ of mass $m$, write $r=|\mathbf{r}|>0$:Here $G$ is the gravitational constant. The minus sign points toward $M$; the magnitude follows an inverse-square law, so doubling the separation quarters the force. The opposite force on $M$ satisfies Newton's third law. ^[ref]

Both masses accelerate. Applying Newton's second law to each in the isolated two-body problem givesThe relative standard gravitational parameter includes both masses; replacing it by $GM$ requires the test particle approximation $m\ll M$.

Historically, Newton's Book III, Proposition VII argues that gravitational attraction applies to all bodies and scales with their masses; its second corollary states the inverse-square law for individual particles. ^[ref]

The smaller galaxy is treated as a test particle released from rest at $r_0$. Conservation of energy in the fixed gravitational potential givesThe galactic gas is stationary in its hydrostatic model, so this is the relative speed squared. Write $\Delta (d)=d^{-1/2}-r_0^{-1/2}\ge 0$ on the inward path. The astrophysical ram pressure is the incident normal momentum flux $\rho (d){\dot{d}}^2$, givingIt vanishes at release and increases during infall. This ram pressure convention is $\rho v^2$, not the $\rho v^2/2$ sometimes called dynamic pressure. Gas drag and the object's response are neglected consistently with the prescribed ballistic motion.

Approximate Earth by an isolated, nonrotating, spherically symmetric body. By Birkhoff's theorem, the exterior vacuum metric tensor is the Schwarzschild metric. With $x^0=ct$ it isHere $r$ is the areal radius, $\theta$ and $\varphi$ are spherical angles, $t$ is Schwarzschild time normalized to the proper time of a static observer at infinity, $M$ is Earth's mass, $G$ the gravitational constant, and $c$ the speed of light. The nonzero diagonal metric tensor components are $g_{00}=-f$, $g_{rr}=f^{-1}$, $g_{\theta \theta }=r^2$, $g_{\varphi \varphi }=r^2{\sin }^2\theta$.

The apple is a test particle; neglect its effect on the metric tensor, air resistance, Earth's rotation and nonspherical structure, and external tidal forces. Near Earth's surface, $GM/(R_E{c}^2)\simeq 7\times 10^{-10}$, $h/R_E\ll 1$, and the falling speed remains much smaller than $c$. Thus the Newtonian limit of this metric tensor is appropriate, while the exact radial geodesic is available as a check. A static observer under the tree has proper time $T=\sqrt{f(R_E)}t$, up to an additive constant. Such an observer is supported by the ground and has nonzero proper acceleration; the released apple instead has zero proper acceleration.

In general relativity, gravity is the geometry of spacetime, whose curvature is related to the stress-energy tensor, and freely falling test particles follow timelike geodesics of its metric tensor.

By spherical symmetry, choose the orbital plane to be $\theta =\pi /2$. The geodesic Lagrangian of the Schwarzschild metric isFor a massive test particle, dots denote proper time derivatives and $2\mathcal{L}=-c^2$. The cyclic coordinates $t$ and $\varphi$ give, through the Euler-Lagrange equations, two constants:At infinity, $\mathcal{E}={\gamma }_{\infty }=1/\sqrt{1-v_{\infty }^2/c^2}$ is the Lorentz factor. The constant $\ell$ is the specific angular momentum, and the energy per unit rest mass is $E_{\infty }/m=c^2{\gamma }_{\infty }$. The normalization of the four-velocity givesUse $dr/d\varphi =\dot{r}{r}^2/\ell$ and define the energy-normalized angular momentum for a Schwarzschild geodesicThenFor an incoming massive test particle with impact parameter $b$, $L_{\infty }=m{\gamma }_{\infty }{v}_{\infty }b$, so $J=b{v}_{\infty }$. In particular, $J$ is not the same as the specific angular momentum $\ell$ unless ${\gamma }_{\infty }=1$.

A photon has no proper time parametrization and no finite Lorentz factor. For it, use an affine parameter in the same geodesic Lagrangian and impose $2\mathcal{L}=0$. The ratio $J=c^2{L}_{\infty }/E_{\infty }=bc$ remains well defined, and the null result isEquivalently this is the ultrarelativistic limit of the massive formula at fixed $J$, with $1/{\gamma }_{\infty }^2\to 0$. The printed massive formula needs this interpretation before it is applied to a photon.

With $\tau$ the proper time of the massive test particle, the four-velocity has squared norm $-1$:The conserved quantities in part (i) giveMultiplying the normalization by $F$ and substituting gives the charged-particle radial equation in Reissner-Nordstrom spacetime:At a radial turning point $r_{\ast }>0$, the left side is zero, soThe parenthesis on the left is strictly positive. Hence $F(r_{\ast })\ge 0$. The factorization in part (b) now yieldsThere is no radial turning point between the horizons. Indeed, for $r_{-}<r<r_{+}$, the right-hand side of the radial equation is strictly positive, independently of the electric charge $e$ of the test particle. This is the charged-particle turning-point exclusion in Reissner-Nordstrom spacetime.

The inequalities are necessary, not a claim that a radial turning point exists for every $E,l,e$. At a regular horizon crossing a timelike vector cannot be tangent to a null hypersurface, so a massive test particle actually has $\dot{r}\ne 0$ there. Algebraic equality at $r_{\pm }$ would require $E=eQ/r_{\pm }$ and does not describe a regular finite-proper time radial turning point on the horizon. In the Schwarzschild spacetime limit $Q=0$, the positive-radius inner alternative disappears.

Use units $c=1$ and signature $(+---)$. For a massive test particle, write its four-momentum as $p^i=m{u}^i$, where $u^i=d{x}^i/d\tau$, $\tau$ is Minkowski proper time, and $u^i{u}_i=1$. The stated flat-background scalar-force theory of gravity then givesContract with $p_i=m{u}_i$. The two terms cancel:Thus rest mass is conserved and the acceleration law isThere is no mass or composition parameter in this equation. Particles released at the same event with the same velocity consequently have identical trajectories. In a static weak field at small speed, its spatial part becomes $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$, since raising a spatial index introduces a minus sign. The field equation likewise reduces to ${\nabla }^2\Phi =4\pi G\rho$ for nonrelativistic matter, because $T^i{}_i\simeq \rho$ and $\square =-{\nabla }^2$ in the static limit.

Therefore yes: the model agrees with the Eötvös test of universal free fall. In the point-particle/test-body approximation, inertial and passive gravitational masses have the same universal ratio. This is the weak equivalence principle tested by the Eötvös experiment; it is not a claim about strongly self-gravitating bodies or a proof of all aspects of the Einstein equivalence principle.

Take $z$ positive outward and define the separation as upper minus lower position. The Newtonian potential of a point mass gives the acceleration $a(z)=-{\partial }_z\varphi =-GM/z^2$. Expanding its difference across a separation $\zeta \ll z$ gives the tidal force equationThe upper particle falls slightly less rapidly, so the separation increases. If the upper particle carries the electric charge and the electric field points outward, the extra relative acceleration is $qE/m$. Reversing the field or placing the charge on the lower particle changes its sign. Write this signed acceleration as $a_e$ and put $k=2GM/z^3$. With initial relative velocity zero and nearly constant $k$ and $a_e$, the ordinary differential equation has solutionFor $k{t}^2\ll 1$, the leading changes due to the tidal force and electric field are respectivelywith the stated sign convention. Thus, for a nonzero fixed electric acceleration,A laboratory of size $L$ has $|{\zeta }_0|\le L$, so this ratio tends to zero as $L\to 0$. This is local tidal acceleration scaling: gravitational relative acceleration is proportional to separation, whereas the electric difference persists even at coincident positions.

The duration also matters. At fixed time, $\Delta {\zeta }_g\to 0$ in absolute size, but $\Delta {\zeta }_g/{\zeta }_0\simeq (GM/z^3)t^2$ need not tend to zero. To keep the electrically displaced particles in the shrinking laboratory, take $|a_e|t^2\lesssim L$; then $|\Delta {\zeta }_g|/L\lesssim GML/(z^3|a_e|)\to 0$. Alternatively a stationary terrestrial laboratory has a free-fall time of order $\sqrt{L/g}$, giving $|\Delta {\zeta }_g|/|{\zeta }_0|=O(L/z)$. Both make precise the small spacetime-region limit. An extended, long-running experiment can still measure arbitrarily small tidal forces.

The weak equivalence principle says that freely falling test particles with the same initial data have composition-independent motion. In particular, a freely falling frame removes the common gravitational acceleration of both particles. The electric accelerations depend on charge and cannot be removed for charged and uncharged particles simultaneously by the same change of frame.

The Einstein equivalence principle extends the freely falling laboratory statement to all local nongravitational experiments: their outcomes obey special relativity, with local Lorentz invariance and local position invariance. The strong equivalence principle also includes local gravitational experiments and bodies with significant gravitational binding energy. It asserts universality of their free fall and independence of local experimental outcomes from the laboratory's location and velocity. The two-particle experiment illustrates the local distinction between gravity and a nongravitational force; by itself it does not establish this stronger claim about self-gravitating bodies.

In a metric theory, the common freely falling trajectories are timelike geodesics of a Lorentzian metric. Normal coordinates make $g_{ab}={\eta }_{ab}$ and the Christoffel symbols zero at one point. Fermi normal coordinates do the same along a reference timelike geodesic. The residual effects are controlled by the Riemann curvature tensor through geodesic deviation, and enter the local metric at quadratic order in distance. Curvature is a physical tidal field, so no coordinate choice removes it over a finite region.

A relativistic theory should consequently couple matter universally to the metric tensor, recover nongravitational laws in local inertial frames and recover the Newtonian limit. General relativity implements this with the Einstein field equations; the contracted Bianchi identity is consistent with stress-energy conservation. The equivalence principle motivates this geometric description but does not alone uniquely determine the field equations. Extra gravitational fields may satisfy the Einstein equivalence principle for ordinary matter while failing the strong equivalence principle because local gravitational measurements or self-gravitating trajectories depend on those fields.

The phrase equivalence principle covers several related claims. The weak equivalence principle concerns the universality of free fall for test particles of negligible self-gravity, with the same initial conditions and no relevant nongravitational forces. In a Newtonian notation,Composition-independent motion requires the ratio of passive gravitational mass to inertial mass to be universal. Its common value can then be absorbed into the normalization of $\mathbf{g}$; the empirical assertion is universality, not the numerical convention used to express the ratio.

The Einstein equivalence principle includes that universality and extends the claim to local nongravitational experiments through local Lorentz invariance and local position invariance. A sufficiently small freely falling laboratory should recover the nongravitational laws of special relativity. The strong equivalence principle extends the scope to local gravitational experiments and bodies with appreciable gravitational self-energy. An experiment establishing one part of the weak statement does not automatically establish these stronger claims.

The motivating elevator comparison makes both the power and the limitation clear. Release a small object in a freely falling cabin. The cabin and object share the local free fall, so the object initially floats relative to the cabin. In a cabin supported against gravity, a released object instead accelerates relative to the floor, as it does in a suitably accelerated cabin in flat spacetime. This suggested replacing a universal gravitational force by geometric inertial motion. It does not identify a finite planetary gravitational field with an accelerated frame everywhere: separated falling objects can have a measurable relative acceleration.

In general relativity, the precise local geometric statement is that one can choose normal coordinates at an event $p$ withThis eliminates the connection contribution to the geodesic equation at that point. It does not generally set the Riemann curvature tensor to zero. In normal coordinates the first metric corrections away from $p$ are quadratic in the coordinate separation, and geodesic deviation measures the associated tidal forces. Equivalence is a local approximation for finite experiments whose extent and duration make these corrections negligible; it is not an assertion that curvature has been transformed away.^[ref]

There are direct experimental consequences beyond the elevator thought experiment. An Eötvös experiment compares bodies of different composition on a torsion balance, using a differential signal to test the universality of gravitational acceleration. MICROSCOPE performed an analogous test in orbit using electrostatic control of titanium-alloy and platinum-alloy test masses. Its first published result in 2017 constrained their differential free fall; this was already available before the examination. It tested a specific weak-equivalence prediction, not every possible gravitational law.^[ref] Comparisons of different atomic clocks can instead test local position invariance, including the expected gravitational redshift. These measurements constrain related but distinct parts of the framework.

The principle remains important as a restriction on how matter couples to gravity and as an organizing guide for experimental tests. It does not, on its own, derive the Einstein field equations. Universally coupling matter to a metric tensor still leaves choices for that metric's action; the equivalence principle does not uniquely fix field equations. For example, a metric gravitational action containing a scalar curvature squared term can retain the same minimal matter coupling while changing the gravitational dynamics. Selecting the Einstein-Hilbert action requires further dynamical assumptions and empirical success. The present role of the principle is consequently precise and limited: it supports local relativistic physics and universal matter coupling, while curvature, gravitational field equations and their quantitative predictions remain additional content.

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.

Put the charged sheet at $z=0$ and the grounded plates at $z=\pm d/2$. By symmetry the electric field points away from the positive sheet with equal magnitude in the two gaps. A Gaussian pillbox crossing the sheet givesThe plate boundary conditions fix their electric potentials to zero. For the charged sheet between grounded parallel plates, the central potential isWith vacuum permittivity ${\epsilon }_0=8.85419\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $V_s\simeq 28.24\mathrm{V}$. An electron has electric charge $-e$, so moving from a grounded plate to this positive potential decreases its potential energy by $e{V}_s$. Conservation of energy then givesThe unrounded value is $3.15\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, about $0.0105c$. Either plate gives the same speed. The ideal calculation neglects the electron's disturbance of the prescribed field and image-force effects at its release point, consistently with treating it as a test particle.

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.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

The orbiting planet produces a potential $V(r,\vartheta ,\phi -\Omega t)$. It has explicit time dependence and exerts an azimuthal torque, so generally $\dot{E}=V_t$ and ${\dot{L}}_z=-V_{\phi }$ are nonzero. Nevertheless a simultaneous time-translation symmetry $\delta t=\lambda$ and rotation $\delta \phi =\Omega \lambda$ leaves the Lagrangian and $dt$ unchanged. Noether's theorem gives $Q=\Omega p_{\phi }-E$ and henceEquivalently $V_t=-\Omega V_{\phi }$ implies $d(E-\Omega L_z)/dt=0$. This Jacobi integral assumes a rigidly rotating potential with constant $\Omega$ and negligible back-reaction from the test particle.

Hydrostatic balance is $dP/dr=-GM(r)\rho /r^2$. A free-fall estimate gives $t_{\mathrm{dyn}}\sim \sqrt{R_{\odot }^3/(G{M}_{\odot })}\simeq 1.6\times 10^3$ s; a test particle falling in a fixed point-mass field adds factor $\pi /(2\sqrt{2})$, giving about 30 minutes.

For an ideal nonrelativistic gas with negligible boundary pressure, the virial theorem is $2U+\Omega =0$, so total energy is $\Omega /2$. Since $\Omega =-{\int }_0^{M}Gmdm/r(m)$ and $r\le R$, $|\Omega |\ge G{M}^2/(2R)$. This is a minimum magnitude, not a lower bound on the negative energy: unrestricted central concentration can make the latter arbitrarily negative. A uniform-density estimate gives $\Omega =-3G{M}^2/(5R)$ and available contraction energy $3G{M}^2/(10R)$. Its release at present solar luminosity lasts $\boxed{t\sim 10^7\mathrm{yr}}$ (order $10^7$–$10^8$ years for alternative structural estimates), much less than a nuclear-burning lifetime.

For a source point mass $M$ at the origin, let $\mathbf{R}=R\mathbf{n}$, where $\mathbf{n}$ is an outward unit vector, and let $\mu =GM$ be the source's standard gravitational parameter. Differentiating $g_i=-\mu R_i/R^3$ gives the tidal tensorwhere ${\delta }_{ij}$ is the Kronecker delta and $I$ is the identity matrix. In an orthonormal basis with first axis radial, the tidal tensor isThe radial eigenvalue $2\mu /R^3$ separates nearby freely falling test particles; the two transverse eigenvalues $-\mu /R^3$ bring them together. The farther radial test particle falls inward less rapidly, so its acceleration relative to the nearer one points outward. Transverse gravitational accelerations converge toward the source. Thus an initially resting small cloud stretches radially and squeezes in both transverse directions. These are instantaneous relative accelerations; the subsequent shape also depends on initial velocities and the changing tidal tensor along the cloud's free fall.

Outside the source, the matrix trace is $2\mu /R^3-\mu /R^3-\mu /R^3=0$. This matches $\mathrm{tr}T=-{\nabla }^2\Phi =0$ in a region with zero mass density, by the Poisson equation for Newtonian gravity. Zero matrix trace means no leading change of a small initially resting cloud's volume at order $t^2$, not that each separation stays fixed. Inside matter, the same sign convention gives $\mathrm{tr}T=-4\pi G\rho$ instead.

Let a test particle start at rest at radius $r_0$ outside a fixed spherically symmetric source. The spherical shell theorem gives the exterior Newtonian gravitational potential $\Phi (r)=-\mu /r$, where $\mu =GM$ is the source's standard gravitational parameter. Zero initial transverse velocity makes the motion radial. Conservation of mechanical energy per unit mass givesThe minus sign selects inward motion; the positive square root is its speed. For a fall from $r_0=2R$ to the surface $r=R$, that speed is $\sqrt{\mu /R}$. In the limit of release from rest infinitely far away, the familiar free-fall speed is $\sqrt{2\mu /r}$, equal in magnitude to the local escape velocity. A finite starting radius gives a smaller speed.

The arrival time follows from the same first integral. Set $r=r_0{\cos }^2\eta$. ThenFor arrival at $R>0$ this is finite, and the field formula is used only outside the source. For $r_0=2R$, $\eta =\pi /4$ givesKeeping the smaller initial gravitational magnitude $g(r_0)=\mu /(4R^2)$ constant would predict $\sqrt{8}\sqrt{R^3/\mu }$, about $10.0\%$ longer. The actual acceleration increases during the fall.

The curves compare a shallow fall, $h/R=0.02$, with a fall from $r_0=2R$, so $h/R=1$. Each comparison holds the release height and initial gravitational magnitude fixed. The constant-gravity model is close for the shallow fall and increasingly inaccurate for the larger one. This is also a practical use of nondimensionalization: the normalized trajectories depend on $h/R$, while $\sqrt{R^3/\mu }$ supplies the dimensional time scale.