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About a specified axis, the moment of inertia is $I=\int r_{\perp }^{2}dm$, giving rotational kinetic energy $I{\omega }^2/2$ for fixed-axis rotation. With square coordinates centred on the plate,The in-plane inertia isotropy of a square plate follows using the perpendicular axis theorem, which gives $I_z=I_x+I_y=M{l}^2/6$. For any in-plane central axis at angle $\varphi$, the perpendicular in-plane axis has the same inertia by the square's $90^{\circ }$ rotational symmetry. Their sum is $I_z$, so every such in-plane axis has $I=M{l}^2/12$.

The parallel axis theorem gives hinge inertia $I_H=M{l}^2/3$. During the short impulse gravity is negligible, and the hinge impulse has no moment about the hinge. ThereforeThe hinge impulse of an edge-pivoted square plate follows from linear momentum: $J_H=Mv-K$. Taking its positive direction to be that of the applied impulse,Generally $K$ alone does not supply the centre-of-mass momentum compatible with rotation about the fixed hinge, so the hinge supplies the difference. Its impulse vanishes at the centre of percussion,

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

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

Measure displacement from each arrangement's gravitational equilibrium. The series and parallel spring stiffness follows by adding forces or extensions. In the parallel arrangement the two spring forces change by $kx$ each, giving equivalent stiffness $2k$. In series the common force $F$ extends each spring by $F/k$, so total extension is $2F/k$ and the equivalent stiffness is $k/2$. HenceThe corresponding angular frequencies are $2\pi f$. gravity shifts the equilibrium extensions but does not alter the small-oscillation stiffness of these ideal linear, massless springs.

Newton's first law identifies inertial frames in which a force-free body maintains constant velocity. The second is $\dot{\mathbf{P}}={\mathbf{F}}_{\mathrm{net}}$ for a closed system; for fixed mass it becomes $m\mathbf{a}=\mathbf{F}$. The third gives equal and opposite pair forces. Summing the second law over a closed collection cancels internal force pairs, so total momentum is conserved when the resultant external force vanishes. A rocket alone is an open variable-mass subsystem, so one must include the momentum carried by exhaust before applying this reasoning.

Choose downward velocity as positive and write $M(t)=M_0-\alpha t$. During $dt$, exhaust mass $dm=\alpha dt$ leaves downward with speed $v+v_E$ in the surface frame. To first order, momentum balance for rocket plus this exhaust givesIntegration, with $v(0)=v_0$, yieldsSetting $g=0$ gives the gravity-free result. This assumes a constant downward relative exhaust speed, constant mass-loss rate and $M_0-\alpha t>0$ throughout the burn; the negative logarithm supplies the upward thrust effect.

Height satisfies $\dot{h}=-v$. Put $u(t)=1-\alpha t/M_0$. Integrating the logarithm and choosing $h(0)=h_0$ givesEquivalently, the last term is $v_E[t+(M_0-\alpha t)\log u/\alpha ]$. Differentiating either form recovers $-v$ and checks the sign.

For this constant-rate retro-rocket landing, at touchdown $u_L=3/4$ and $v_L=0$, henceThe fuel condition gives $\alpha =M_0/(4t_L)=200.30$ kg/s. Requiring $h(t_L)=0$ then givesIn this burn the downward speed decreases monotonically to zero: the initial thrust acceleration already exceeds gravity and increases as fuel is consumed. The result is the idealized vertical landing, without atmospheric drag, changing gravity or a separate terminal burn phase.

An inverse-square point-source field has constant flux through enclosing spheres because their area grows as $r^2$. In electrostatics, Gauss's law is ${\oint }_S\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{\epsilon }_0$, or $\nabla \cdot \mathbf{E}={\rho }_e/{\epsilon }_0$. For a spherical charge distribution, rotational symmetry makes the exterior field radial and constant in magnitude on a concentric sphere, giving $\mathbf{E}=Q\hat{r}/(4\pi {\epsilon }_0{r}^2)$. Symmetry is needed to take the field outside the surface integral; Gauss's law alone does not make an arbitrary charge distribution spherical.

For gravity the corresponding relations are ${\oint }_S\mathbf{g}\cdot d\mathbf{S}=-4\pi G{M}_{\mathrm{enc}}$ and $\nabla \cdot \mathbf{g}=-4\pi G{\rho }_m$. A spherical mass has exterior field $\mathbf{g}=-GM\hat{r}/r^2$. Inside a uniform sphere, $M_{\mathrm{enc}}(r)=4\pi {\rho }_m{r}^3/3$, so $\boxed{\mathbf{g}=-(4\pi G{\rho }_m/3)r\hat{r}}$. Inside an empty spherical shell the enclosed mass is zero and symmetry gives zero field. These examples distinguish an inverse-square exterior field from a field inside distributed matter.

The law also treats other symmetries, but their field-distance dependence differs: an infinitely long line charge has cylindrical rather than spherical spreading and a $1/r$ field. Flux conservation and geometry must both be used, rather than assuming every Gaussian surface leads to an inverse-square answer.

A time-independent conservative force defines a conservative vector field on a domain when the work between two points is independent of path, equivalently when every closed-path work integral vanishes. Define a potential energy $U$ by $\boxed{\mathbf{F}=-\nabla U}$ and $U(B)-U(A)=-{\int }_A^B\mathbf{F}\cdot d\mathbf{r}$. The additive constant is arbitrary; only differences affect work. A vanishing curl is necessary, and is sufficient on a simply connected domain under the usual smoothness assumptions. Local zero curl by itself need not guarantee a single-valued global potential on a domain with holes.

For a test mass outside a point or spherical mass, $U=-GMm/r$ with zero at infinity reproduces the inward gravitational force. Near a surface with approximately uniform downward gravity, $U=mgz$ when $z$ increases upward. A linear spring has $U=k{x}^2/2$. In electrostatics the electric potential is energy per unit charge, so $U=qV$ and $\mathbf{E}=-\nabla V$; potential and potential energy have different units and roles.

For motion under a time-independent conservative force, $d(K+U)/dt=0$. nonconservative forces change the mechanical energy by their work, and an explicitly time-dependent potential can exchange energy with the external agency changing it. A time-varying magnetic field can give a circulating electric field, so not every electric field is electrostatic or conservative. Conversely, zero magnetic work on a charge does not turn a velocity-dependent magnetic force into an ordinary position-potential force.

Measure the mass's displacement $x$ from its static equilibrium height. This removes the constant spring extension supporting its weight. If the spring's top moves with the box, its incremental extension is $y_2-x$, so its incremental upward force is $k(y_2-x)$. The specified damping force is against the mass's velocity in the static frame. Newton's second law therefore givesIn absolute height coordinates an additional constant on the right expresses the spring's rest length, its fixed attachment height and gravity; translating to static equilibrium removes it. Replacing the given damping by $-b(\dot{x}-{\dot{y}}_2)$ would describe a different base-excited oscillator.

After the decaying transient, write the steady-state response as $x=A(\omega )\cos (\omega t-\Phi )$. Using a complex sinusoid givessoFor positive frequency, the phase lag lies between zero and $\pi$, passing through $\pi /2$ at ${\omega }_0=\sqrt{k/m}$. For small $b$, the frequency response begins at $a_s$, develops a narrow resonance near ${\omega }_0$, and falls asThe precise amplitude-peak frequency is $\sqrt{k/m-b^2/(2m^2)}$ when that expression is positive.

The pen records the relative displacement $z=x-y_2$. Its complex transfer function isAt high frequency, $x$ is small while $z\simeq -y_2$. The inertial mass then stays nearly stationary and the recorded relative motion reproduces the ground displacement, with reversed sign. At low frequency the mass follows the box and their relative motion is small; near resonance the response is strongly frequency-dependent. Here “high” means compared with the suspension's natural frequency, rather than an absolute seismic-frequency threshold.

Equilibrium and undamped motion. Resolve forces along the incline, taking positive displacement down the slope. At equilibrium, Hooke's law balances the downhill component of weight:If $x$ is measured from that equilibrium position, Newton's second law becomesThus the motion is a harmonic oscillator withGravity shifts the equilibrium extension but drops out of this oscillation frequency because the downhill force component is constant.

Viscous damping. For a damped harmonic oscillator, put $\beta =b/(2m)$. Substitution of $z=A{e}^{-pt}$ in the supplied linear ordinary differential equation givesIn the underdamped regime $\beta <{\omega }_0$, define ${\omega }_d=\sqrt{{\omega }_0^2-{\beta }^2}$. The two characteristic roots are $p=\beta \pm i{\omega }_d$, and the general real displacement isHere $C,D$, or equivalently $A_{\mathrm{env}}$ and $\varphi$, are fixed by the initial conditions. For release from rest at $x_0$, the condition $\dot{x}(0)=0$ requires $D=\beta x_0/{\omega }_d$, not $D=0$:The curve oscillates about zero between exponentially shrinking envelopes $\pm A_{\mathrm{env}}{e}^{-\beta t}$, where $A_{\mathrm{env}}=|x_0|\sqrt{1+(\beta /{\omega }_d{)}^2}$ for this release. The upper panel illustrates that shape; its damping strength is chosen to make the envelope visible.

Frequency change. Viscous damping reduces the angular frequency and lengthens the oscillation period:For light damping, ${\omega }_d\simeq {\omega }_0[1-b^2/(8mk)]$, so the frequency shift is second order in $b$, even though the envelope decay rate is first order in $b$.

Will the block reach the stop? The given parameters giveFor $x_0=-0.260\mathrm{m}$, differentiating the release-from-rest solution givesIt is positive throughout the first half-cycle. The first positive turning point therefore occurs at $t_{+}=\pi /{\omega }_d=0.993460\mathrm{s}$. The turning points of an underdamped release from rest give its unobstructed height:Since $x_{+}>0.250\mathrm{m}$ and the displacement increases continuously up to that point,Solving $x(t)=0.250\mathrm{m}$ gives first contact at approximately $0.91117\mathrm{s}$. The plotted dashed continuation is the hypothetical free motion after that time; the post-impact motion is not determined by the supplied damping model.

For an elastic eccentric impact of a thin disc, take upward velocity as positive and let $J$ be the short upward impulse. The disc's moment of inertia about a horizontal diameter is $I=M{a}^2/4$. With impact offset $d=a/2$,The contact point has upward velocity $v_C+d\omega$. Unit restitution against the fixed post requires that velocity to be $+u$, soSince $I=M{d}^2$, $J=Mu$ and the centre's instantaneous speed is zero. The disc rotates with $\omega =2u/a$; its rotational kinetic energy $I{\omega }^2/2=M{u}^2/2$ equals the entire initial energy. Gravity's impulse during the impact is neglected. Using the polar-axis moment of inertia would describe the wrong rotation axis.

For a normal from medium 1 to 2, the magnetic boundary conditions areWithout a free surface current, tangential $H$ is continuous. To obtain the magnetization of a permeable sphere of radius $R$ and permeability ${\mu }_0(1+\chi )$, choose magnetic scalar potentialsTangential continuity gives $H_i=H_0-C/R^3$, and normal $B$ continuity gives $(1+\chi )H_i=H_0+2C/R^3$. HenceThe fields are uniform inside. For negative susceptibility, magnetization is antiparallel to the applied field; the scalar formula retains its negative sign.

For a small weakly magnetic body, the induced magnetic energy is $U_m=-\chi V{B}^2/(2{\mu }_0)$. The factor one-half accounts for building the induced moment. The vertical force is $F_z=\chi VB{B}^{\prime }/{\mu }_0$. Balancing gravity therefore givesA diamagnet needs a field decreasing in magnitude with height for an upward force. For diamagnetic levitation in an inverse-height field,The force falls as $z^{-3}$, giving a restoring vertical derivative at equilibrium. Full lateral stability would require the transverse field geometry, which is not specified.

For the single-line greenhouse deficit, use hydrostatic balance, fixed surface pressure $p_s$, constant gravity $g$, and the dry adiabatThe gas constant and specific heat refer to the background gas. Let ${\mu }_d=\cos \tilde{\theta }>0$ be the supplied effective ray cosine. The vertical optical depth and slant transmittance from level $p$ to space areThis permits pressure-dependent line widths. With local thermodynamic equilibrium, negligible infrared scattering, a black thermal surface, and no incident thermal radiation from space, the formal radiative transfer solution isThe weights add to one. Subtract the transparent-surface radiance and put $B(p)=B_{{\nu }_0}(T(p))$, using the narrow-line approximation only in the line deficit. Define its pressure-dependent equivalent widthApproximating each hemispheric flux by $\pi$ times the effective-ray radiance preserves the black-surface flux. The outgoing longwave flux is thereforeIntegration by parts gives the particularly transparent formBoth boundary terms vanish: $E(0,q)=0$ and $B(p_s)-B(p_s)=0$. The dry adiabat and monotonic Planck function give $dB/dp>0$. Atmospheric emission replaces part of the warm surface radiation by emission from cooler levels, reducing emission to space for a specified $T_s$. The inequality is strict when opacity occurs over a nonzero range with temperature contrast; an isothermal absorbing layer at the surface temperature would not produce such a contrast.

For a spherical planet the mean absorbed instellation is $F_{\ast }=(1-\alpha )L_{⊛}/4$. The required global energy balance isConsequently every equilibrium of this model has $T_s\ge T_{\mathrm{bare}}=[(1-\alpha )L_{⊛}/(4{\sigma }_{\mathrm{SB}}){]}^{1/4}$, with strict warming when $D>0$. At fixed temperature profile, increasing $q$ increases $E$ and hence $D$. This argument does not assume that the greenhouse gas changes the stellar-band albedo, surface pressure, or background thermodynamics; those are held fixed. For sufficiently broad or overlapping saturated wings the full frequency-dependent Planck function must replace the narrow-line approximation.

Hawaii: short-wavelength volcanic loading and long-wavelength mantle support must be distinguished. A dense volcanic edifice replacing seawater adds mass and gives a positive free-air gravity anomaly. Its load bends the lithosphere down, producing a surrounding moat and an outer flexural arch. The associated displaced density interfaces contribute negative and positive gravity signals at neighbouring distances. Observed gravity near the Hawaiian ridge includes strong positive anomalies over the ridge, negative flanks, and a broader positive belt. At much longer wavelengths, the Hawaiian swell also correlates with a positive gravity anomaly, with a smaller gravity-to-topography ratio that requires compensation; the precise source of support is not determined by that correlation alone.^[ref]

The mechanical distinction can be expressed with a simple elastic-plate model. For load height $h$, downward deflection $w$, plate rigidity $D$, loading density contrast $\Delta {\rho }_l$ relative to water, and restoring density contrast $\Delta {\rho }_b$ at an interface,For a Fourier component of horizontal wavenumber $q$, this gives $w_q=\Delta {\rho }_{l}g{h}_q/(D{q}^4+\Delta {\rho }_{b}g)$. The load's direct gravity contribution competes with the negative contribution of a depressed interface at reference depth $t$. In a linear thin-interface approximation,At short wavelengths rigidity suppresses compensation, leaving relatively high positive gravity admittance. At long wavelengths the plate can deflect and compensate more completely, reducing admittance. Fitting the wavelength dependence constrains effective elastic thickness through $D=E_Y{T}_e^3/[12(1-{\nu }^2)]$, subject to density structure, load distribution and subsurface loading assumptions. This explains why local Airy compensation alone misses the spatially extended flexural response.

The broad swell needs a deeper explanation than the narrow volcanic pile: hot, buoyant mantle and mantle flow can support regional topography, while deep density anomalies and surface elevation jointly determine its gravity. A free-air anomaly keeps the attracting mass of the topography; a Bouguer correction removes an assumed direct topographic attraction and can reveal its compensating mass deficit. Gravity plus topography constrain support as a function of wavelength; neither alone uniquely fixes the mantle density distribution.

For the OR alternative, triple junctions are points where three plate boundaries meet. Each boundary can be a spreading ridge, subduction trench, or transform fault. The three plate velocities satisfyThis velocity triangle is necessary but not enough: a single junction velocity ${\mathbf{u}}_J$ must also keep the meeting point on all three moving boundaries. For symmetric spreading between A and B, the ridge constraint is $[{\mathbf{u}}_J-({\mathbf{v}}_A+{\mathbf{v}}_B)/2]\cdot \mathbf{n}=0$. For a transform, ${\mathbf{v}}_A\cdot \mathbf{n}={\mathbf{v}}_B\cdot \mathbf{n}$, and the junction must have that same normal component. For a trench, its normal migration must match the adopted subduction/rollback model; a trench fixed relative to the overriding plate shares that plate's normal velocity.

In velocity space each boundary supplies a line of possible junction velocities. A common intersection means the junction can migrate while preserving its local configuration; incompatible constraints force boundary reorganization. This is kinematic stability, not the claim that the junction is stationary. A symmetric ridge–ridge–ridge example has three arms meeting at $120^{\circ }$ and a common junction velocity; unequal spreading rates do not require those equal angles. Transform and trench junctions depend on the allowed relative motions and boundary geometry. Stable and unstable evolution were formulated through plate-velocity geometry in the original triple-junction analysis.^[ref]

These original diagrams show an illustrative flexural response and a symmetric ridge-junction construction; they are not maps or measured Hawaiian profiles.

Assume the original ice sheet had time to reach local isostatic compensation, melted rapidly relative to subsequent rebound, and left one exponentially relaxing depression mode. Neglect elastic layering and assign the whole present gravity deficit to that unrelaxed depression. Load balance givesThe supplied sheet-gravity approximation gives the residual amplitudeFrom $h(t)=h_0{e}^{-t/\tau }$,For an order-of-magnitude single-mode estimate, choose the ice-sheet diameter as wavelength, $\lambda \simeq 2R=6000\mathrm{km}$. Taking $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the more accurate specified relation then givesA circular load has a spectrum, not one uniquely specified wavelength. Choosing $\lambda \sim R$ instead gives about $1.3\times 10^{21}\mathrm{Pa}\mathrm{s}$; another wavelength convention rescales the answer linearly. The robust inference is viscosity of order $10^{21}\mathrm{Pa}\mathrm{s}$ within this simple model. A realistic estimate would also model gradual melting, other gravity anomalies, density interfaces, lithospheric flexure and depth-dependent mantle viscosity.

Ice accumulation adds a load that produces an immediate elastic response and slower viscous displacement of the mantle away from the loaded region. The depressed interior may be accompanied by a peripheral forebulge. A long-lived load approaches an approximately compensated state. Melting removes the load, giving rapid elastic recovery followed by mantle return flow and gradual uplift. The forebulge can subside while the formerly loaded centre rises.

Postglacial rebound can continue long after the ice disappears, because mantle relaxation is slow. Raised shorelines, changing relative sea level, vertical geodetic motion and gravity changes record glacial isostatic adjustment. Present motion depends on the whole loading and melting history, mantle rheology and lithospheric stiffness, not only the current absence of ice.

gravity and surface topography. gravity responds to all density structure, while surface topography records both load and support. First specify the reference field and measurement height. latitude and planetary rotation affect normal gravity. At elevation $h$, the reference field decreases approximately by $2gh/R$; a free-air gravity anomaly restores this elevation effect but retains topographic mass. A Bouguer gravity anomaly additionally subtracts an approximate topographic slab contribution $2\pi G{\rho }_{c}h$. terrain correction replaces the unrealistic slab near valleys and ridges, and spherical geometry matters over large areas. A density choice is required for these reductions; a numerical crustal density is not provided here.

In Airy isostasy, surface topography of height $h$ and crustal density ${\rho }_c$ is compensated by a root of thickness $b$ replacing mantle of density ${\rho }_m>{\rho }_c$. Equal column masses give ${\rho }_{c}h=({\rho }_m-{\rho }_c)b$. The positive gravity of the relief is partly cancelled by the negative density contrast of the root. At long wavelengths a compensated mountain can therefore have a modest free-air anomaly and a negative Bouguer anomaly. Pratt isostasy instead varies column density at a roughly common compensation depth. Neither model says that surface topography itself exerts no gravity.

Potential-field depth filtering makes wavelength crucial. In a planar Fourier approximation, a density interface at depth $z$ is attenuated by $e^{-kz}$ at horizontal wavenumber $k$. For a small Airy-compensated height perturbation and a compensation interface at depth $D$, the gravity-topography admittance is approximatelyLong wavelengths allow strong cancellation between load and root. At short wavelengths the root's gravity is filtered out, so the observed near-surface mass can dominate even if the column is compensated. In a flexural isostasy model, an elastic lithosphere supports and spreads the load; for downward deflection $w$ and imposed load $q$, $D_f{\nabla }^{4}w+\Delta \rho gw=q$, with $D_f=E{T}_e^3/[12(1-{\nu }^2)]$. Thus $w_k=q_k/(D_f{k}^4+\Delta \rho g)$, and the effective elastic thickness controls the support scale.

Estimate admittance from cross-spectra of gravity and surface topography, and compare its wavelength dependence and coherence with local-compensation, elastic-flexure and subsurface-load models. A primary Atlantic study sampled ship-track free-air gravity and bathymetry and compared their Fourier-domain relationship with compensation models, illustrating this observational route rather than inferring a root from the sign of one gravity value.^[ref] Check navigation, instrument drift, terrain/density reductions, sediment and water loads, and independent seismic crustal constraints. Density-depth trade-offs and mantle dynamic support make the inverse problem nonunique; surface topography and gravity together constrain support, but do not uniquely map every density anomaly.

mantle viscosity. A second, independent route concerns deformation rates rather than static compensation. glacial isostatic adjustment follows loading and unloading of a deformable Earth: an ice sheet depresses the surface, and subsequent removal leaves time-dependent uplift, horizontal displacement, sea-level and gravity changes. Fit dated relative sea-level indicators and raised shorelines together with geodetic displacement rates and a reconstructed ice/water load history to a viscoelastic Earth model. Relative sea level includes solid-surface motion, changing geoid and redistribution of ocean water; treating it only as local vertical rock motion biases the inference.

For scale, a sinusoidal surface perturbation of wavenumber $k$ on a uniform incompressible viscous half-space relaxes under gravity with ${\tau }_k\sim 2\eta k/(\rho g)$. A Maxwell material separately has a stress-relaxation scale ${\tau }_M=\eta /\mu$. These are different experiments and generally different times: rebound modes also involve gravity, spherical geometry, a layered mantle and elastic lithosphere. Fit a family of relaxation modes or a full load-history convolution, rather than equating every observed rebound time to $\eta /\mu$.

Use independent ice-margin, exposure-age and sea-level constraints to reduce the trade-off between ice history and viscosity; compare horizontal as well as vertical GNSS motions, and separate present ice loss, groundwater changes and tectonic signals from the older loading response. A primary global analysis jointly tested ice-history and mantle-response predictions against space-geodetic observations, rather than treating a local uplift rate as a unique viscosity measurement.^[ref]

postseismic relaxation probes shallower or faster relaxation after a stress perturbation, but must be distinguished from afterslip and poroelastic deformation. Long-wavelength geoid and mantle-flow models provide complementary constraints when density structure is estimated independently from seismic structure and mineral physics; unknown density-to-velocity conversion and boundary tractions remain trade-offs. Laboratory creep experiments constrain constitutive laws as functions of temperature, stress and grain size, but require a large extrapolation to geological conditions. Combine spatially and temporally distinct observations; there is no single directly measured mantle viscosity independent of depth, rheology and loading assumptions.

I would not assign the lunar mare an automatic terrestrial 100–125 km compensation level. For present-day mare basins, a local equal-column-mass model may have no physically valid compensation depth; first test whether lithospheric flexure supports the load. For ancient thermally driven subsidence that did relax, choose a common datum beneath the cooling layer in deformable lunar mantle appropriate to that epoch.

The Moon is smaller than Earth and has lost a larger fraction of its original heat. Its cold, mechanically strong outer shell can be much thicker relative to its radius than the terrestrial oceanic lithosphere. Lunar mare basalts filled pre-existing impact basins: their loading, basin excavation, mantle uplift, cooling and the evolving strength of the lithosphere all matter. A thick elastic lithosphere can support lateral stress differences, so equal column mass need not hold even after a long time. The strong gravity anomalies of lunar mascons, measured by GRAIL, are evidence that the present mare-basin system is not everywhere locally compensated.^[ref]

For an explicitly assumed ancient relaxed thermal model, estimate its thermal penetration depth from thermal diffusivity and the elapsed cooling time, infer the contemporaneous mechanical lid from gravity-topography admittance or basin deformation, and place $z_c$ below that layer where lateral pressure anomalies relax. For example, a hypothetical several-hundred-kilometre-thick lid requires a deeper datum than that lid, not the shallow crust-mantle boundary. The paper supplies neither an epoch-specific thermal model nor a load wavelength, so a unique kilometre value is not determined. If the chosen datum is a substantial fraction of the lunar radius, spherical geometry and radial variation in gravity must replace the flat, constant-gravity column approximation.

Does the Moon have an asthenosphere? Not a demonstrated shallow, globally connected Earth-like layer underlying mobile plates; but “no weak or partially molten mantle anywhere, ever” is too strong. Apollo lunar seismology indicates a cold outer interior and more attenuating deep regions. Reanalyses support a deep low-velocity or partially molten region near the lunar core, while lunar laser ranging and tidal Love numbers constrain deep deformability and tidal dissipation. Such a deep weak region is sometimes called a lunar asthenosphere; it is not automatically the correct local compensation surface for a mare load.^[ref]

The inferred melt fraction and distribution remain model dependent: Dissipation at tidal and seismic frequencies in a melt-free Moon by Francis Nimmo and others demonstrates that much of the dissipation can also be explained by hot solid-state anelasticity, although its model does not reproduce every frequency-dependent constraint. The Moon was hotter during mare emplacement and could relax loads differently then. Choose the datum from the relevant thermal and mechanical history, or use a flexural model when local isostasy fails.

The momentum of a material volume $V(t)$ is ${\int }_{V(t)}\rho \mathbf{u}dV$. Newton's second law equates its time derivative to the pressure force and gravity:Apply the given Reynolds transport theorem componentwise to the left side, and the divergence theorem to the pressure term. ThenSince this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive density gives the Euler equations for an inviscid fluidAt a fixed physical time $t_0$, a streamline is a curve tangent to $\mathbf{u}(\cdot ,t_0)$, which may be parametrized by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t_0)$. The chain rule gives $df(\mathbf{x}(s),t_0)/ds=\mathbf{u}\cdot \nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.

The coordinate $\psi$ is cyclic, so $p_{\psi }=C(\dot{\psi }+\dot{\varphi }\cos \theta )$ is conserved. Initially the spin is zero, hence $\boxed{\dot{\psi }+\dot{\varphi }\cos \theta =0}$ throughout. The coordinate $\varphi$ is also cyclic. After imposing this zero spin, its conserved momentum becomes $(A+m{h}^2)\dot{\varphi }{\sin }^2\theta$. Initially $\theta =\pi /2$ and $\dot{\varphi }=\omega$, so $\boxed{\dot{\varphi }{\sin }^2\theta =\omega }$.

The equation for $z$ isThe quantity in parentheses is the vertical velocity of the centre of mass. Initially that velocity and its height are both zero: the initial rotation is horizontal and the vertex starts at rest at the reference height. Twice integrating givesIndependently, the frictionless vertical wire exerts only horizontal force. Gravity is therefore the only vertical external force on the body, and the centre of mass undergoes vertical free fall. This gives the same equation without the angular equations.

Use the largest Newtonian tidal-tensor eigenvalue, the radial gravity gradient $2GM/r^3$. Imposing the supplied gradient threshold givesThis is about $1590\mathrm{km}$ from the centre, or $1580\mathrm{km}$ above the stated surface. The tangential eigenvalues have half the radial magnitude. Since $GM/(r_{\min }{c}^2)\simeq 1.4\times 10^{-3}$, the Newtonian tidal estimate is adequate; this calculation uses the given threshold without inferring any additional material-response model.

inertial mass sets the response to force, $\mathbf{F}=m_i\mathbf{a}$. Passive gravitational mass sets the response to a given gravitational potential, ${\mathbf{F}}_g=-m_g\nabla \Phi$; active gravitational mass describes a body's gravitational sourcing. Universality of the passive-to-inertial mass ratio makes freely falling trajectories independent of composition. torsion balance comparisons and differential free-fall experiments test this weak equivalence principle by searching for composition-dependent acceleration, with environmental forces and instrumental asymmetries controlled.

The Einstein equivalence principle further says that local nongravitational experiments in a freely falling laboratory obey special relativity, independent of the laboratory's location and velocity. A local inertial coordinate system can remove Christoffel symbols at one event, but cannot remove a nonzero Riemann tensor throughout a finite region. Relative acceleration of separated freely falling particles reveals tidal gravity. Thus a uniformly accelerating laboratory in flat spacetime can mimic a local gravitational field, while extended tidal measurements distinguish genuine curvature. Equality of the measured mass responses does not make every gravitational field globally equivalent to acceleration.