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Apparent weight measures support rather than the gravitational force itself. In an ideal scale model, it is the magnitude $N$ of the normal force exerted on the body. A scale calibrated to display mass divides that support force by its calibration value of $g$.

Take upward as positive in a constant downward gravitational acceleration of magnitude $g$. If a body stays in contact with a horizontal scale and accelerates upward at $a$, Newton's second law gives $N-mg=ma$, hence $N=m(g+a)$. At rest, $N=mg$. In common free fall, $a=-g$, so $N=0$ despite the nonzero gravitational weight $mg$. If the formula would require $N<0$, ordinary supporting contact cannot realize it: the body loses contact instead.

A laboratory gravity gradiometry experiment used two vertically separated clouds of chemical atoms in vacuum, launched into free fall. Shared laser beams applied three pulses ($\pi /2,\pi ,\pi /2$), splitting and recombining matter waves. Balanced photodiodes measured probe-light absorption to infer internal-state populations; scanning the last pulse's wave phase produced an interference fringe.^[ref]

The difference of fringe wave phases obeys $\delta \varphi ={\mathbf{k}}_{\mathrm{eff}}\cdot \Delta \mathbf{g}{\tau }^2$, with effective wave vector ${\mathbf{k}}_{\mathrm{eff}}$ and pulse spacing $\tau$: shared reference motion cancels, leaving differential gravitational acceleration. Eight lead bricks, each approximately $12.5\mathrm{kg}$, were placed symmetrically around the lower chamber, about $0.2\mathrm{m}$ from the fountain apex. The measured signal was $(8.1\pm 2.1)\times 10^{-9}g$, versus a calculated $8.2\times 10^{-9}g$, where $g$ denotes terrestrial gravitational acceleration.^[ref]

Controls deliberately shook the shared reference reflector and compared actively isolated and unisolated operation. Tilting tests exposed sensitivity to Coriolis acceleration and centrifugal acceleration. Thus shared translation is rejectable, while rotation and finite measurement uncertainty remain limitations. The brick signal detects spatially varying gravity; it does not establish a precision value of the gravitational constant.^[ref]

The equivalence principle states that gravity couples universally to energy-momentum and that a freely falling local frame removes uniform gravitational acceleration at a point. Tidal curvature remains measurable over a finite region.

Imagine releasing a ball inside a small falling elevator. Relative to the cabin it initially floats, because cabin and ball share the same free fall. The weak equivalence principle says this trajectory is independent of the test particle's composition for identical initial conditions. The Einstein equivalence principle further identifies local nongravitational experiments with those of special relativity. It is a statement about experiments in a local inertial frame, not just equality of two masses.

The word local is essential. Coordinates can make the metric tensor equal to the Minkowski metric and its first derivatives vanish at one event. They generally cannot erase the Riemann curvature tensor, which involves second derivatives. Two separated balls can therefore approach or recede even inside the falling cabin. A sufficiently extended experiment detects these tidal forces; a long observation can accumulate them even when the cabin is small. The equivalence principle does not identify an entire planetary gravitational field with acceleration in flat spacetime. See Tong's discussion of normal coordinates and local equivalence.

Universal metric coupling and local special-relativistic nongravitational physics do not select a unique dynamical metric action. Additional assumptions are needed to obtain the Einstein field equations. A local freely falling coordinate choice can remove connection coefficients at a point but cannot remove curvature or tidal effects over a finite region. Tests of redshift, free fall and self-gravitating dynamics therefore address related but distinct claims.

For a circular orbit, gravity supplies centripetal acceleration, giving $v_c^2/r=\mu /r^2$. ThusA body in a circular orbit is bound despite its continuous free fall. At the same $r$, escape needs twice its kinetic energy per unit mass. An instantaneous forward tangential velocity increment from the circular orbit requires $\Delta v=(\sqrt{2}-1)v_c$, rather than the full escape speed from rest. In the synthetic example this is exactly $2(\sqrt{2}-1)m/s\approx 0.828427m/s$.^[ref]

Escape speed is a coasting threshold, not a required continuously maintained speed or a measure of rocket propellant. A powered vehicle can start at rest and acquire energy gradually. The rocket equation relates an ideal velocity increment to rocket exhaust speed and rocket mass ratio; rocket thrust, drag, and the timing and direction of propulsion affect a real ascent. Rotation also changes the relation between surface-relative velocity and the inertial frame used in the calculation.

Reaching another body is a finite-destination spacecraft trajectory problem. In the synthetic radial example, a launch at $2m/s$ reaches $r=4\mathrm{m}$ while remaining bound. A spacecraft can similarly encounter the Moon on a bound geocentric transfer; actual planning must include the Moon's gravity, target motion, arrival velocity, and any propulsion needed for capture. Escaping Earth alone does not imply escaping the Sun, because the relevant reference frame and attracting mass have changed. The isolated formula describes one gravitational well; a full mission requires more than its scalar threshold.

One motivating laboratory demonstration compares two balls initially at the same height. A switch releases one vertically when the other leaves a horizontal track; arranging their paths to meet gives a collision as the readout of equal downward displacement over equal time. Equal initial height and synchronized release control the comparison. This tests the local independence of horizontal motion and free fall, under nearly constant gravitational acceleration; it does not measure an inverse-square law.^[ref]

A complementary astronomical comparison uses the Moon's distance and orbital period to infer inward centripetal acceleration, $a=4{\pi }^{2}r/T^2$, and compares it with surface gravitational acceleration scaled by $(R/r{)}^2$. The approximately sixtyfold distance ratio gives the approximately $1/3600$ prediction. The inference links terrestrial free fall to a lunar orbit. Distance calibration matters; the circular orbit approximation and the Sun's tidal force limit this elementary comparison.^[ref]

Free fall is motion under gravity alone: there is no supporting normal force, engine thrust or appreciable drag force. In the Newtonian mechanics model, a test particle of mass $m$ obeys $m\ddot{\mathbf{r}}=m\mathbf{g}(\mathbf{r})$, hence $\ddot{\mathbf{r}}=\mathbf{g}(\mathbf{r})$. The mass cancels. Bodies released at the same position with the same velocity follow the same path in this model, even if their gravitational weights differ. ^[ref]

A dropped stone is one example; a coasting satellite is another. Free fall does not require motion straight down or decreasing distance to the attracting body. An ideal circular orbit has constant radius while its velocity continually turns under centripetal acceleration. The circular orbit as continuous free fall connects the familiar drop with orbital motion.

Near a surface, treating the gravitational acceleration as constant gives a useful first prediction. Over larger heights, the inverse-square law changes the acceleration during the fall. Radial free fall from rest uses conservation of mechanical energy to account for that change exactly, within the specified Newtonian gravitational field. The contrast matters: the simplest formula is an approximation to a physical model, rather than a law with the same numerical $g$ everywhere.

History also makes the idealization tangible. Galileo's Two New Sciences (1638), Third Day, Theorem II, derives distance proportional to the square of time for uniformly accelerated motion from rest; its first corollary gives successive distances in the ratios $1,3,5,7,\dots$ over equal time intervals. This is a kinematic statement, predating the gravitational inverse-square law. ^[ref]

On Apollo 15's final lunar surface excursion, David Scott released a hammer and a feather together. The near-vacuum removed the large difference in drag force familiar on Earth. NASA's report describes their arrival together within the accuracy of the release; this demonstration illustrates the free fall prediction without establishing infinite experimental precision. ^[ref]

A test particle released from rest at $r_0$ in a fixed spherical mass density $\rho (r)={\rho }_0(r_0/r{)}^2$ obeys $\ddot{r}=-4\pi G{\rho }_0{r}_0^2/r$. Integrating once gives ${\dot{r}}^2=8\pi G{\rho }_0{r}_0^2\ln (r_0/r)$. The substitution $u=\sqrt{\ln (r_0/r)}$ then gives $t(r)=(8G{\rho }_0{)}^{-1/2}\mathrm{erf}(u)$, and the arrival time at the centre is $(8G{\rho }_0{)}^{-1/2}$. This is free fall in a frozen singular isothermal sphere, not the self-consistent pressureless collapse of the background. An initially resting material shell with the same initial enclosed mass instead arrives after $\sqrt{\pi /(32G{\rho }_0)}$, because its enclosed mass stays fixed while the cloud contracts.

Gravitational acceleration is the acceleration of a test particle in free fall in a Newtonian gravitational field. It is a vector pointing along that field; $g$ commonly denotes its positive magnitude. Near Earth's surface, the magnitude is approximately constant over shallow vertical scales. For a fixed spherical source it instead varies as $g(r)=\mu /r^2$, where $\mu$ is the standard gravitational parameter.

This is distinct from gravitational weight, the force $m\mathbf{g}$, and from apparent weight, the support force a scale measures. An orbiting body can have substantial gravitational acceleration while its apparent weight is zero.

For a fixed source point mass, the magnitude $g=\mu /R^2$ of the gravitational acceleration measures the local pull; the radial eigenvalue ${\lambda }_r=2\mu /R^3$ of the tidal tensor measures how that pull changes across space. Doubling $R$ divides $g$ by four but ${\lambda }_r$ by eight. For a fixed small radial separation $L$, the linear relative acceleration is $\Delta a\simeq {\lambda }_{r}L$, soA small body can therefore experience strong common gravitational acceleration with a comparatively small tidal force. A measurement of $g$ alone does not determine the tidal tensor: sources at different distances can have the same $g$ but different spatial variation.

The local stretching time scale is ${\lambda }_r^{-1/2}=\sqrt{R^3/(2\mu )}$: it marks when an initially resting separation could change by order itself if the tidal tensor stayed fixed. During free fall, $R$ changes, so this scale is not a complete trajectory or a universal disruption time. Tidal disruption additionally compares the tidal force with binding by self-gravity and material forces; $g$ by itself supplies no such comparison.

The inner radius at which a stellar magnetic dipole field disrupts a surrounding accretion disk. In a free fall pressure-balance estimate with fixed aspect ratio $ϵ=H/r$, $r_m\sim ({ϵ}^2{\mu }^4/(GM{\dot{M}}^2){)}^{1/7}$. Geometry and inflow structure change its numerical coefficient.

A satellite experiment testing the universality of free fall. Electrostatic control keeps test masses made from titanium and platinum alloys on matched orbital motion; the required differential forces constrain composition dependence. A reference sensor uses masses of the same composition. The experiment tests the weak equivalence principle, rather than the full strong equivalence principle.^[ref]

A massive main-sequence donor loses material in a stellar wind. Gravitational capture by the neutron star can produce a nearly radial, supersonic infall when the captured material has little specific angular momentum. A magnetic field may channel the final infall toward polar accretion columns, where a standing shock decelerates the flow. Large angular momentum instead favours an accretion disk; strictly radial accretion is an idealization.

For mean particle mass $\mu$, the ideal gas law is $p_{\mathrm{d}}={\rho }_{\mathrm{d}}{k}_{\mathrm{B}}{T}_{\mathrm{d}}/\mu$. Combining it with b givesFor Newtonian free fall from a distance large compared with the stellar radius $R_{\ast }$, $v_{\mathrm{u}}^2\simeq 2G{M}_{\ast }/R_{\ast }$. ThereforeTODO: a unique numerical temperature and characteristic photon band require source data not supplied here. The held question gives $M_{\ast }=1.4M_{\odot }$ but gives neither $R_{\ast }$ nor a composition fixing $\mu$. The mass alone does not fix the free-fall speed or the temperature. For example, doubling $R_{\ast }$ at the same mass and composition halves both quantities in the last box.

A normalization, rather than an additional claimed datum, makes the scale explicit:Here $k_{\mathrm{B}}T$ is the characteristic particle energy; the mean translational energy of a nonrelativistic thermal particle is $3k_{\mathrm{B}}T/2$. If one additionally chose $R_{\ast }=10\mathrm{km}$ and $\mu =m_p$, this ideal one-temperature estimate would correspond to gamma-ray-scale thermal photon energies. Those are conditional choices, not information recovered from the source. Real accreting neutron stars commonly emit X-rays: electron-ion disequilibrium, radiative cooling, radiation pressure, reprocessing and relativistic infall invalidate a direct identification of this ideal ion-heating estimate with the escaping photon spectrum.

There are two different physical interpretations. The supplied logarithmic integral signals free fall in a fixed singular isothermal sphere, with a test particle moving through a frozen background. Write $C={\rho }_0{r}_0^2$. By the shell theorem, the enclosed mass is $M(r)=4\pi Cr$ and the radial equation of motion isThe second equality follows by multiplying by $\dot{r}$ and using the initial rest condition. The free fall trajectory is given implicitly byThus the intended fixed-profile answer isA collapsing material shell gives a different answer. If the entire cloud undergoes pressureless gravitational collapse from rest, the edge follows the fixed enclosed mass $M_0=4\pi {\rho }_0{r}_0^3$, not the instantaneous fixed-profile value $4\pi Cr$. With no shell crossing, $\ddot{r}=-G{M}_0/r^2$, soThis is a concrete counterexample to treating the fixed-profile result as the exact collapse time of the whole nebula. The initial mean density inside $r_0$ is $3{\rho }_0$, consistent with the usual free-fall time of a uniform sphere expressed using that mean density. The logarithmic integral is valid for the explicitly frozen background approximation.

The strong equivalence principle extends universality of free fall to bodies with appreciable gravitational self-energy, and states that the outcome of any local experiment, including a gravitational experiment, in a freely falling laboratory is independent of its location and velocity. In a sufficiently small freely falling laboratory the external gravitational influence can be removed to first order by choosing a local inertial frame.

This does not remove tidal forces or the Riemann curvature tensor. The laboratory must be small compared with the length scale over which the external gravitational field varies. These qualifications distinguish a local equivalence statement from the incorrect assertion that gravity can be transformed away throughout an extended curved region.

Use metric signature $(-,+,+,+)$ and $x^0=ct$. At an event in the ship's instantaneous rest local inertial frame, $u^{\alpha }=(c,0,0,0)$. Since its squared norm is a scalar,in every coordinate system. Equivalently it follows directly from $d{s}^2=-c^{2}d{\tau }^2$ and $u^{\alpha }=d{x}^{\alpha }/d\tau$.

Metric compatibility lets us differentiate this identity covariantly:In the instantaneous rest local inertial frame, the last equation gives $a^0=0$. The spatial components are the acceleration recorded by an onboard accelerometer, ${\mathbf{a}}_{\mathrm{rest}}$, whose magnitude is $g$. ThereforeThis proper acceleration is zero in ideal free fall, even when the coordinate position has a nonzero second derivative. A coordinate acceleration and an accelerometer reading are different quantities.

Put $F(r)=1-r_s/r$ and $r_s=2GM/c^2$. For radial motion $u^{\theta }=u^{\varphi }=0$. The given Christoffel symbols implyandThe four-velocity normalization of part c isIts bracketed combination is therefore exactly $c^2$, independently of the instantaneous velocity or the thrust history. ConsequentlyThe derivative here is with respect to proper time. This expression does not say that the ship measures a gravitational acceleration $GM/r^2$ during free fall; setting $a^r=0$ gives its radial proper-time equation of motion in Schwarzschild spacetime.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

General relativity models spacetime by a smooth four-dimensional Lorentzian manifold whose metric tensor governs clock readings, causal cones and free-particle motion. The Einstein equivalence principle motivates local special relativity and universal geodesic free fall for test bodies; it does not by itself select a unique gravitational field equation. We assume a Levi-Civita connection, a metric tensor as the gravitational field with no additional dynamical gravitational variables, and matter with a covariantly conserved stress-energy tensor. Units have $c=1$.

The existence of local inertial frames is a local statement. At any regular event choose an orthonormal frame and Riemann normal coordinates, so that $g_{ab}={\eta }_{ab}$ and ${\Gamma }^a{}_{bc}=0$ at the event, with metric signature $(+---)$. Metric compatibility then gives ${\partial }_c{g}_{ab}=0$ there. The geodesic equation at that point is the special-relativistic inertial equation. Along a timelike geodesic, a parallel-transported orthonormal frame gives Fermi normal coordinates, eliminating the connection along the central worldline. These constructions do not in general eliminate second derivatives of the metric tensor in a neighborhood. The Riemann curvature tensor is built from the derivatives of the connection coefficients as well as their quadratic terms; nonzero Riemann curvature tensor components can remain when all connection coefficients vanish at a chosen point. Conversely, Polar coordinates in Minkowski spacetime can have nonzero connection coefficients but zero Riemann curvature tensor. Connection coefficients alone are therefore not an invariant measure of gravity.

For the tensorial content, consider a two-parameter family of geodesics parametrized by an affine parameter, with tangent $U$ and connecting field $J$. Their coordinate construction gives $[U,J]=0$, and torsion-freeness gives ${\nabla }_{U}J={\nabla }_{J}U$. Since ${\nabla }_{U}U=0$, commuting derivatives in the convention of Question 1 yields geodesic deviation:The left side is covariant relative acceleration, rather than the second coordinate derivative of an arbitrary separation. In an orthonormal frame in free fall with $U=(1,0,0,0)$, its spatial components are $D^2{J}^i/d{\tau }^2=-R^i{}_{00j}{J}^j$. This is a measurable tidal force, and a coordinate transformation cannot make a nonzero Riemann curvature tensor disappear. It is the distinction between gravity at one event and the behavior of an extended freely falling laboratory.

In a static Newtonian limit, $g_{00}=1+2\Phi$ and ${\Gamma }^i{}_{00}={\partial }_i\Phi$, so the test-body acceleration is $-\nabla \Phi$. The curvature-sign convention in Killing derivative identities gives $R^i{}_{00j}\simeq {\partial }_i{\partial }_j\Phi$. Hence relative acceleration is the negative Hessian of the gravitational potential. For the Newtonian potential of a point mass $\Phi =-GM/r$, the Hessian has radial eigenvalue $-2GM/r^3$ and two tangential eigenvalues $GM/r^3$: radial separations stretch and tangential ones compress. Outside the source the Ricci tensor can vanish while the Riemann curvature tensor remains nonzero. Its vacuum tidal part is encoded by the Weyl tensor; vacuum does not mean flatness. Gravitational waves likewise carry Riemann curvature tensor disturbances without a local material source.

To obtain the field equations, make the additional dynamical assumptions explicit: locality and diffeomorphism invariance of general relativity; a symmetric metric tensor field equation using at most second derivatives of the metric tensor; a geometrical left-hand side linear in the Riemann curvature tensor with constant coefficients; and compatibility with local matter conservation. These restrictions exclude higher-curvature or extra-field alternatives. Under them the available rank-two curvature contractions give the ansatzThe contracted Bianchi identity and metric compatibility giveFor this to vanish identically for arbitrary metric tensors, $b=-a/2$. Assuming $a\ne 0$ and absorbing its normalization into the coupling givesThe permitted constant term is the cosmological constant. Its value is not fixed by the equivalence principle. Conservation of $T$ follows for generally covariant matter on its equations of motion: varying the matter action under a compactly supported infinitesimal diffeomorphism, then integrating its term $T^{ab}{\nabla }_a{\xi }_b$ by parts, gives ${\nabla }_a{T}^{ab}=0$. The divergence-free Einstein tensor is consistent with this identity.

The coupling sign must be fixed in the actual curvature-sign convention in Killing derivative identities, with $R_{ab}=R^c{}_{acb}$. For the static weak-field approximation, $R_{00}\simeq -\Delta \Phi$. Set $\Lambda =0$ for the local asymptotically flat Newtonian comparison. In four dimensions the trace equation gives $R=-\kappa T$, so $R_{00}=\kappa (T_{00}-\frac{1}{2}{g}_{00}T)\simeq \kappa \rho /2$ for slowly moving matter. Matching $\Delta \Phi =4\pi G\rho$ forces $\kappa =-8\pi G$. Thus, with this convention,Under the opposite curvature-sign convention in Killing derivative identities the coupling sign reverses, as in Einstein-equation coupling under reversed curvature. This sign statement is compatible with the supplied linearized handout. Local inertial frames remove the connection at an event; tidal Riemann curvature tensor effects remain tensorial, and the field equations require dynamical assumptions beyond the equivalence principle.

Use an Earth-centred rotating reference frame, with Earth's constant angular velocity vector $\mathbf{\omega }$. The vectors $\mathbf{r}$, $\mathbf{v}$, and $\mathbf{a}$ are the particle's position, velocity, and acceleration measured in that rotating frame. The vector $\mathbf{g}$ is the gravitational acceleration due to the actual gravitational force, expressed in the same axes; it excludes the separate centrifugal acceleration. The remaining term $-2\mathbf{\omega }\times \mathbf{v}$ is the Coriolis acceleration. A constant rotation axis and an unaccelerated Earth-centred origin account for the absence of an Euler term or a translational-origin term.

For the size estimates, Earth's angular speed isWith drop height $H=20\mathrm{m}$, unperturbed free fall gives the fall timeand final speed $gT=20{\mathrm{m}\mathrm{s}}^{-1}$. Thus the gravitational term has size $10{\mathrm{m}\mathrm{s}}^{-2}$, the Coriolis acceleration is at most about $2\omega gT\approx 2.8\times 10^{-3}{\mathrm{m}\mathrm{s}}^{-2}$, and the centrifugal acceleration is at most ${\omega }^2{R}_0\approx 2.9\times 10^{-2}{\mathrm{m}\mathrm{s}}^{-2}$. These are characteristic sizes $10^1$, $10^{-3}$, and $10^{-2}$ in acceleration units; geometric factors can reduce them, and the Coriolis term is zero at release.

For a suitable approximation to the rotational deflection of a falling body, take $\mathbf{g}$ constant over the short fall, start at rest relative to Earth, and use the zeroth-order velocity ${\mathbf{v}}^{(0)}(t)=\mathbf{g}t$ inside the small Coriolis term. In the centrifugal term replace $\mathbf{r}$ by ${\mathbf{R}}_0$, because $H/R_0\ll 1$. Integrating the resulting acceleration twice, with initial position ${\mathbf{r}}_{\mathrm{i}}$, givesSince ${\mathbf{r}}_{\mathrm{i}}+\mathbf{g}{T}^2/2={\mathbf{R}}_0$, evaluating at the unperturbed fall time yields the printed first-iterate estimate

The approximation keeps the leading Coriolis contribution and the leading centrifugal contribution enhanced by Earth's large radius. Feeding the Coriolis velocity correction back into the equation produces terms of size ${\omega }^{2}g{T}^4$, smaller than the retained centrifugal displacement by a factor of order $H/R_0$. Replacing $\mathbf{r}$ by ${\mathbf{R}}_0$ in that term has the same relative error. Here $\omega T\approx 1.4\times 10^{-4}$ and $H/R_0\approx 3.3\times 10^{-6}$, so both approximations are well controlled. Strictly, the displayed vector is the position at time $T$: its small upward component is removed by the slightly longer actual fall time. That arrival-time correction changes the retained horizontal displacements only at higher order. Thus the horizontal components of the printed estimate give the landing point on the locally flat surface; its vertical component should not be interpreted as a landing above the ground.

Let ${\hat{\mathbf{e}}}_E,{\hat{\mathbf{e}}}_N,{\hat{\mathbf{e}}}_U$ point east, north, and vertically upwards. They form a right-handed basis, with ${\hat{\mathbf{e}}}_N\times {\hat{\mathbf{e}}}_U={\hat{\mathbf{e}}}_E$. At latitude $\lambda$, in the spherical-Earth approximation,The required cross products areAt $\lambda =45^{\circ }$, the northerly and easterly displacements are thereforeThe negative northerly component means a southward displacement. With the supplied values, these are approximately $2.9\mathrm{cm}$ south and $1.3\mathrm{mm}$ east. The eastward effect comes from Coriolis acceleration, while the leading southward effect here comes from centrifugal acceleration.