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A gravity anomaly is measured gravitational acceleration, reduced to a stated reference surface and compared with a reference gravity field. A local geological interpretation further removes or models the regional background. An absolute negative Bouguer gravity anomaly is therefore not itself a local negative geophysical density contrast.

Measure relative gravity with a calibrated gravimeter at stations across the profile, tie it to a base station, repeat base readings to estimate instrumental drift, and measure station coordinates and accurate elevations, for example by GNSS and levelling. Record observation times for tide corrections. Subtract the latitude-dependent reference gravity, correct instrument drift and Earth tides, add the free-air correction for station height, and subtract the attraction of the reference-density Bouguer slab. A terrain correction accounts for relief missing from or exceeding that slab. Use a consistent density such as $2670\mathrm{kg}{\mathrm{m}}^{-3}$ for the topographic correction.^[ref]

The question's reference to the anomaly in panel b is a panel-label slip: the upper plot contains gravity, while the lower plot supplies geometry and elevations. A geological slab thickness is the vertical thickness of the anomalous body, not simply its height above sea level.

For a laterally extensive Bouguer slab of thickness $h$ and geophysical density contrast $\Delta \rho$,The greenstone belt reaches from roughly $+1.0$ to $-0.5\mathrm{km}$ near its deepest central part, so use $h\simeq 1500\mathrm{m}$. The gravity maximum is approximately $-100\mathrm{mGal}$; the exposed host-rock background in the south is approximately $-115\mathrm{mGal}$. Taking this as a constant regional reference gives a local excess $\Delta g\simeq 15\mathrm{mGal}=1.5\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-2}$. ThusThis is an approximate density under a constant-background, infinite-slab model. The finite width, variable thickness and neighboring pluton require a fuller gravity inversion for a precise value. In particular, the absolute value $-100\mathrm{mGal}$ must not be inserted as the local excess.

Retain the same regional-background assumption as in part (b). The northern pluton is about $2.0\mathrm{km}$ thick beneath the station near the northern trough, while its gravity anomaly is about $-114\mathrm{mGal}$. Its anomaly relative to the greenstone maximum is approximately $-14\mathrm{mGal}$. Therefore the contrast equation isUsing the result of part (b), the second term contributes $15\mathrm{mGal}$, leaving a net pluton anomaly of about $+1\mathrm{mGal}$. Hence ${\rho }_{\mathrm{gr}}\simeq 2680\mathrm{kg}{\mathrm{m}}^{-3}$, effectively close to the host-rock density at the precision of this diagram.

The apparent trough is strongly negative relative to the greenstone belt, but hardly negative relative to the southern host-rock background. If a northward regional trend is intended instead, it must be stated: in general,with both gravity values in milligals. For example, a northern background of $-109\mathrm{mGal}$ would give ${\rho }_{\mathrm{gr}}\simeq 2610\mathrm{kg}{\mathrm{m}}^{-3}$, but the same reference would also change part (b). No regional curve is explicitly identified in the supplied figure, so the numerical result must retain its reference-field assumption. This is the geological nonuniqueness of geophysical inversion of potential-field interpretation, rather than a reason to force a presumed textbook granite density onto the plot.

Assume the greenstone belt is more magnetically susceptible than the host rocks and granite, its magnetization is mainly induced magnetization, and the body is roughly east–west elongated so that the south–north line crosses it. At southern magnetic latitudes the main field has negative magnetic inclination in a north-and-down coordinate convention. The total-field magnetic anomaly is then generally asymmetric: the principal negative lobe is on the southern side and the positive lobe is displaced northward for the simple induced source illustrated here. The exact lobe sizes and peak positions depend on body geometry and the actual magnetic inclination.^[ref]

An explicit sketching model is an induced magnetic dipole at depth $d$. With northward coordinate $x$, $\mathbf{r}=(x,-d)$ and unit field direction $\hat{\mathbf{F}}=(\cos I,\sin I)$, its normalized total-field magnetic anomaly isThe plot uses $I=-36^{\circ }$ only as an illustrative dipole-field inclination for a latitude near $20^{\circ }$ south, not a measured modern field at the site. A wide greenstone belt would broaden the response and can produce prominent edge anomalies. Remanent magnetization or a magnetic granite could change the sketch substantially; gravity alone cannot identify the magnetic source.

Physical cosmology connects the universe's history to the measurable behaviour of its contents. The Friedmann equations describe cosmic expansion; atomic physics, radiation pressure and gravity determine what that expanding universe looks like. A particularly revealing example is the route from cosmological recombination through the photon-baryon fluid to the Cosmic microwave background acoustic peaks. The universe supplies the initial conditions and the laboratory; the prediction must still pass through a model of the telescope before it can be compared with observations.

The practical reward is that one sky contains several different tests. The Cosmic microwave background power spectrum carries information about the sound horizon, the cosmological baryon density and cosmic geometry. These are related quantities with different observational signatures. Angular sound-horizon degeneracy explains why measuring a characteristic angle alone cannot count the universe's baryons.

A pendulum of length $L$ in vertical gravity $g$ is weakly attracted horizontally by an added long circular cylinder of line density $\lambda$. Outside it, $g_h=2G\lambda /r$ at axis distance $r$. Small-angle equilibrium gives horizontal deflection $d\simeq L{g}_h/g=2GL\lambda /(rg)$. Removing the cylinder's mass reverses the change. It is a before/after differential estimate; finite length, a nonhorizontal radial direction and mechanical motion require corrections.

Potential energy assigns energy to a system's arrangement, so a change of arrangement accounts for the work done by a conservative force. For a time-independent potential energy $V(\mathbf{r})$,The gradient points toward increasing potential energy; the force points toward decreasing potential energy. Only differences matter: adding a constant to $V$ changes neither the force nor its work. This endpoint rule follows by substituting $-\nabla V$ into the line integral for work. ^[ref]

Near Earth's surface, stipulate constant gravitational acceleration $g=9.81m/{\mathrm{s}}^2$ and measure height $h$ upward from a chosen zero. Then gravitational potential energy is $V=mgh$. Raising a $2\mathrm{kg}$ mass by $0.5\mathrm{m}$ increases its gravitational potential energy by $9.81\mathrm{J}$, while gravity does $-9.81\mathrm{J}$ of work. These are computed model values, not measurements of a particular lift. An ideal spring obeying Hooke's law instead has $V=k{x}^2/2$, with displacement $x$ from its relaxed state and spring constant $k$: $k=100N/m$ and $x=0.2\mathrm{m}$ give $V=2\mathrm{J}$ relative to $x=0$. ^[ref]

An object launched into motion whose subsequent trajectory is determined by external forces. The ideal elementary projectile motion model retains uniform gravity and neglects air resistance.

A weak tangential force can move a satellite through a slowly varying sequence of nearly circular Kepler orbits. For constant satellite mass, $E=-\mu m/(2r)$ and $v_c=\sqrt{\mu /r}$ give $\dot{E}=F{v}_c$, hence $F/m=\sqrt{\mu }\dot{r}/(2r^{3/2})$. Integrating the thrust magnitude for a monotone outward transfer proves the displayed formula. The integrated thrust per mass is not the vector difference of the endpoint velocities; gravity acts throughout the transfer. The approximation requires slow orbital evolution and negligible eccentricity.

A particle constrained to a stationary smooth surface of revolution $z=f(r)$ under uniform gravity conserves $h=r^2\dot{\theta }$ and mechanical energy per unit mass. Substitute $\dot{z}=f^{\prime }(r)\dot{r}$ and $\dot{\theta }=h/r^2$ into that energy to obtainThis proves the radial reduction. Simple roots of $E-U_h$ are turning points; a stationary point $U_h^{\prime }(r_0)=0$ supports an exact circular orbit. At a local maximum of $U_h$, another trajectory at the same energy can approach the orbit asymptotically as a separatrix. Existence of roots elsewhere is insufficient for confinement: their dynamically accessible ordering relative to the launch radius matters.

A spatial image after a chosen dark interval is a different readout from continuously recording arrivals at a fixed probe. Here is an original idealised fluorescence imaging model: release a dilute cloud, let it move freely, then apply a short uniform probe pulse. Neglect collisions, residual forces, attenuation and motion during the pulse. Use each image's cloud centre to remove mean drift. Uniform gravity shifts that centre without changing the centred variance.

For each atom let $x(t)=x_0+v_{0}t$ along a horizontal imaging axis. Expanding the square about the expected values gives exactlyNo Gaussian distribution is needed for this variance identity. A jointly Gaussian distribution of initial positions and velocities does make the later density Gaussian. For a separable three-dimensional cloud viewed along $z$, integration along the viewing direction gives the column densityHere $N$ is atom number. Integrating this image along $y$ leaves a one-dimensional Gaussian distribution with standard deviation ${\sigma }_x(t)$. This is a projected density measurement, not a three-dimensional radius or a direct velocity histogram. Under the uniform, optically thin probe assumptions, expected fluorescence is proportional to this column density. A known, independent point spread function with finite variance adds its own variance to the measured width; probe duration and intensity gradients need separate treatment.

For a Maxwell-Boltzmann distribution, $T_x=m{\sigma }_v^2/k_B$, where $m$ is atomic mass and $k_B$ the Boltzmann constant. For a nonthermal distribution this defines a directional second-moment temperature, which need not describe every feature of the distribution. A covariance between position and velocity can initially narrow the cloud even though ${\sigma }_v^2$ is positive. The physical covariance matrix requires $C_{xv}^2\le {\sigma }_{x0}^2{\sigma }_v^2$.

Choose ${\sigma }_{x0}=0.60\mathrm{mm}$, ${\sigma }_v=0.060\mathrm{m}{\mathrm{s}}^{-1}$ and $C_{xv}=-0.018\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-1}$, giving correlation coefficient $-1/2$. With $t$ in milliseconds, the numerical variance in square millimetres is $0.36-0.036t+0.0036t^2$. Its minimum occurs at $5\mathrm{ms}$, with standard deviation $\sqrt{0.27}\mathrm{mm}$. These chosen values satisfy the covariance matrix constraint; early contraction is compatible with free motion.

Generate thirteen synthetic variance observations at equally spaced times from $0$ to $30\mathrm{ms}$, adding independent Gaussian noise of standard deviation $0.012\mathrm{m}{\mathrm{m}}^2$. Linear least squares fits all three coefficients in $a+bt+c{t}^2$. The example yields $a\simeq 0.37026\mathrm{m}{\mathrm{m}}^2$, $b\simeq -0.036413\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-1}$ and $c\simeq 0.0035820\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-2}$. With the chosen approximate sodium mass $m=3.82\times 10^{-26}\mathrm{kg}$, the true $T_x$ is $9.96\mu \mathrm{K}$ and the fitted value is $9.91\mu \mathrm{K}$. This is a fit to generated width data, not historical observations.

If the same data are fitted to $a+c{t}^2$, forcing $C_{xv}=0$, the inferred $T_x$ falls to $6.79\mu \mathrm{K}$. The omitted term is partly absorbed into both fitted coefficients. In noiseless linear least squares with a free intercept, its precise effect on the quadratic coefficient iswhere the subscript denotes equal-weight moments over the chosen observation times, using one consistent unit system. Since the ratio is positive for distinct nonnegative times, negative $C_{xv}$ biases this estimate downwards. An incorrect release-time origin can also generate a linear term, so fitting it does not by itself identify a physical initial covariance. Timing calibration, spatial resolution, residual magnetic fields and probe-induced recoil energy remain separate controls. Neither a cold-looking image nor a single long expansion time identifies all three coefficients.

The ideal rocket equation gives the change in signed velocity produced by ejecting rocket propellant. Assume one-dimensional Newtonian mechanics in an inertial frame, no external force, and a fixed positive rocket exhaust speed $u$ directed rearward relative to the instantaneous rocket. The rocket's instantaneous mass $M$ includes its remaining rocket propellant; burning it gives $dM<0$. For initial and final masses $M_{\mathrm{initial}}\ge M_{\mathrm{final}}>0$,Here $\ln$ is the natural logarithm, and its argument is the rocket mass ratio. This is the constant-rocket exhaust speed, zero-pressure-correction version of the ideal rocket equation. ^[ref]

For a toy rocket with $u=2km/s$, $M_{\mathrm{initial}}=1000\mathrm{kg}$ and rocket dry mass $M_{\mathrm{final}}=250\mathrm{kg}$ after all rocket propellant is spent, the rocket mass ratio is $4$ andBurning from $1000$ to $500\mathrm{kg}$ gains $2\ln 2km/s\simeq 1.3863km/s$. Burning from $500$ to $250\mathrm{kg}$ gains exactly the same velocity increment, using only $250\mathrm{kg}$ of rocket propellant instead of $500\mathrm{kg}$. Both burns halve the remaining mass. The second burn accelerates a lighter rocket; equal fractional mass losses give equal velocity gains.

The natural logarithm makes the rocket equation both useful and revealing: multiplying successive rocket mass ratios adds their velocity gains. Conversely, the required rocket mass ratio is the exponential function $M_{\mathrm{initial}}/M_{\mathrm{final}}=e^{\Delta v/u}$. Each extra $u\ln 2$ of ideal velocity gain doubles the required initial mass for fixed final mass. This explains why a larger rocket exhaust speed is valuable and why carrying ever more rocket propellant has diminishing returns. Changing the burn duration alone does not change this ideal endpoint result.

For a short proof outline, follow the rocket together with a tiny parcel about to be expelled. Before ejection, their total momentum is $Mv$. After ejection, the remaining rocket has mass $M+dM$ and velocity $v+dv$, while the parcel has positive mass $-dM$ and, to first order, inertial velocity $v-u$. Momentum conservation givesDiscarding the second-order product $dMdv$ gives $Mdv=-udM$; taking the integral gives $\Delta v=-u{\int }_{M_{\mathrm{initial}}}^{M_{\mathrm{final}}}dM/M$. Evaluating the integral yields the natural logarithm above. This is a variable-mass system calculation: the rocket alone is not a closed system, and its momentum changes as the expelled material carries away momentum. The fixed collection consisting of the rocket and all its expelled material has conserved total momentum when external forces vanish.

Fixed rocket exhaust speed means fixed relative speed at release, not fixed inertial velocity. As the rocket's velocity $v$ changes, newly emitted material has changing inertial velocity $v-u$; previously released parcels coast at their own release velocities in this ideal model. If $v>u$, the exhaust even moves forward in the inertial frame, although it is emitted rearward relative to the rocket. No surrounding medium is needed for this exchange of momentum.

Real flight adds gravity, drag force and possibly a pressure correction to rocket thrust. A varying rocket thrust due solely to a changing mass loss rate still gives the same ideal rocket equation if $u$ is constant; it changes the acceleration and burn duration. A changing rocket exhaust speed instead requires $\Delta v=-\int udM/M$, and external forces require their own contribution to the velocity change. The existing constant mass-loss propulsion with quadratic drag and constant-burn rocket with ground friction illustrate such additional forces; the forward-exhaust rocket equation reverses the ejection direction and therefore the sign of the recoil.

A secular tidal torque is the torque remaining after averaging the changing tidal interaction over fast orbital and rotational motion. The misaligned tidal deformation lets the companion's gravity change the spin angular momentum; the orbit receives the opposite torque. Tidal dissipation converts mechanical energy into heat, while total angular momentum is conserved in an isolated system. Thus the spin-orbit coupling exchanges angular momentum rather than destroying it. ^[ref].

For intuition near synchronous rotation on a circular orbit, hold mean motion $n$ and moment of inertia $I$ fixed and adopt the illustrative linear dissipative lawHere $\omega$ is spin angular velocity, $\theta$ is a marked surface direction in an inertial frame, and $K$ is a response coefficient, not a universal material constant. A faster spin is slowed and a slower prograde spin is accelerated. The solution and the associated mechanical power loss areThe orbital contribution to this power is $-n⟨\tau ⟩$, and the spin contribution is $\omega ⟨\tau ⟩$; their sum becomes heat. Holding $n$ fixed is a local approximation to the coupled evolution. Actual secular tidal torque depends on the deformation response and orbital eccentricity, so this simple law does not establish which spin-orbit resonance an eccentric body captures into.

A spacecraft trajectory is the path traced by a spacecraft's position over time. An unpowered exterior arc can approximate a Kepler orbit; other bodies' gravity, radiation pressure, and propulsion can change it.

For a sphere of radius $a$ injecting an incompressible flow into an infinite homogeneous isotropic porous medium, Darcy's law and conservation of mass give $r^2{q}_r=C$. Integrating $p_r=-\mu C/(\Pi r^2)$ with $p(a)-p(\infty )=\Delta p$ gives $C=\Pi a\Delta p/\mu$. The hydraulic resistance is $\Delta p/Q=\mu /(4\pi \Pi a)$. A common background hydrostatic pressure cancels gravity if the imposed excess pressure is measured relative to that background.

Clemson's documented apparatus is a book held tightly closed with several rubber bands. Toss it freely with spin about each principal axis in turn: the stated visual readout is steady orientation for the smallest and largest principal moments of inertia, but a flip before the catch for the intermediate one. The same book and the two extreme-axis throws provide comparison controls; the bands support the rigid body approximation by restraining the pages. This is a documented qualitative demonstration, not a measurement of ${\lambda }_2$.^[ref]

For a rectangular book with length $a$, width $b$ and thickness $c$, where $a>b>c$, a uniform mass model gives principal moments of inertia $M(b^2+c^2)/12$, $M(a^2+c^2)/12$ and $M(a^2+b^2)/12$. The intermediate principal axis runs across the width through the centre of mass; the maximum is normal to the cover. A marked cover or end helps distinguish ordinary spin from reversal of a directed body axis.

Interpret only the interval after release and before the catch: hand contact applies torque, whereas gravity in a uniform gravitational field has zero net torque about the centre of mass because $\int \mathbf{r}dm=0$. Air resistance, page motion, an imperfect release and short flight limit the approximation. A toss ending before the perturbation grows need not reveal instability. Comparable initial angular speeds and flight durations would improve the comparison; these are suggested controls, not quantified controls reported by Clemson.

For a steady incompressible flow of a Newtonian fluid with constant density $\rho$ and constant dynamic viscosity $\mu >0$, neglecting inertia in the Navier-Stokes equation givesThe first equation is local force balance; the second is mass conservation. The term $\mu {\nabla }^2\mathbf{u}$ is the divergence of the viscous part of the Newtonian fluid stress tensor. The body force $\mathbf{f}$ has units of force per unit volume, $\mathrm{N}{\mathrm{m}}^{-3}$, and equals $\rho \mathbf{g}$ for uniform gravity. Appropriate boundary data for velocity or traction, together with suitable distant conditions for an exterior flow, complete the problem; a rigid surface normally imposes the no-slip boundary condition.

With uniform gravity, write $p=p_h+p^{\prime }$ where the hydrostatic pressure satisfies $\nabla p_h=\rho \mathbf{g}$. Then the disturbance equation is $-\nabla p^{\prime }+\mu {\nabla }^2\mathbf{u}=0$. This removes gravity from the fluid disturbance equation, while the hydrostatic pressure still supplies buoyancy on an immersed body. Omitting it from the total force balance would give the wrong Stokes settling velocity.

Take downward as positive and let gravity have magnitude $g$. For a rigid sphere of density ${\rho }_p$ in otherwise stationary fluid of density ${\rho }_f$, its weight minus buoyancy is $\frac{4}{3}\pi a^3({\rho }_p-{\rho }_f)g$. At terminal velocity, this balances the Stokes drag law:This signed downward settling velocity is negative for a buoyantly rising sphere and zero for equal densities. For fixed density contrast and dynamic viscosity, the submerged weight grows as $a^3$ while linear drag at fixed speed grows as $a$; their balance gives the settling-velocity radius scaling $U_s\propto a^2$. Meanwhile,Thus doubling the radius multiplies the predicted Stokes settling speed by four and the particle Reynolds number by eight. This makes the formula useful for separating particles by size, while also providing its own check against extrapolation to large particles.

For a computed water example, take ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $\mu =1.00\times 10^{-3}\mathrm{Pa}\mathrm{s}$, ${\rho }_p=2500\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and $a=10.0\mu \mathrm{m}$. These are specified rounded inputs rather than a claim of exact water properties at every temperature. They giveThe diameter is $20.0\mu \mathrm{m}$, and both the net downward weight and the upward drag force are $6.16\times 10^{-11}\mathrm{N}$. At this terminal velocity, falling $1.00\mathrm{mm}$ takes $3.06\mathrm{s}$. At $a=20.0\mu \mathrm{m}$ the prediction is $1.308\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.05232$; at $a=50.0\mu \mathrm{m}$ it is $8.175\mathrm{mm}{\mathrm{s}}^{-1}$ with ${\mathrm{Re}}_d=0.8175$, which is no longer a convincing small-Reynolds number case. The figure labels these as predictions of the Stokes settling velocity formula and shows where its inertia check fails.

The calculation describes the eventual terminal velocity, not the release transient governed by particle acceleration and possibly the Unsteady Stokes equation. Walls, hydrodynamic interactions, departures from a rigid sphere, or a different surface condition require a different resistance. For sufficiently small particles, Brownian motion can obscure individual settling trajectories; the formula still specifies the gravitational drift within its assumptions. The balancing argument already appears in article 43 of ^[ref].

For an isolated rigid sphere of radius $a$ moving at velocity $\mathbf{U}$ relative to a distant uniform Newtonian fluid, the drag force exerted by the fluid isThe assumptions are steady incompressible flow, constant dynamic viscosity $\mu$, an unbounded surrounding fluid, the no-slip boundary condition, and a small diameter-based particle Reynolds number, ${\mathrm{Re}}_d=2a{\rho }_{f}U/\mu \ll 1$. Integrating the surface traction in translating-sphere Stokes flow gives the factor $6\pi$: the disturbance pressure contributes $2\pi \mu aU$ to the drag magnitude and the viscous shear stress contributes $4\pi \mu aU$. See Richard Fitzpatrick, Axisymmetric Stokes Flow Around a Solid Sphere, equations 10.120–10.129.

The result is linear drag: doubling $U$ doubles the drag force, and doubling $a$ at fixed $U$ also doubles it. The required power is $6\pi \mu a{U}^2$, all supplied to viscous dissipation. Density enters the check of the particle Reynolds number, although it does not appear in the leading drag formula.

Nearby walls or other particles change the hydrodynamic resistance matrix and invalidate the isolated-sphere coefficient, even at very small Reynolds number. Finite Reynolds number introduces inertia corrections; the Oseen approximation resolves the distant region where the leading Stokes flow approximation ceases to be uniform. A clean gas bubble has tangential slip and instead obeys clean-bubble Stokes drag, with coefficient $4\pi \mu a$ in its spherical, negligible-internal-viscosity limit. See H. K. Moffatt, Six Lectures on General Fluid Dynamics, section 3.4 for the sphere solution and its small-Reynolds number limitation. The continuum mechanics and boundary assumptions must also remain valid as the particle becomes very small.

Balancing this Stokes drag law against gravity and buoyancy gives the Stokes settling velocity of an isolated rigid sphere.

Real atoms have multiple internal quantum states. Optical pumping and spatially varying electromagnetic polarization can remove kinetic energy by additional mechanisms, producing sub-Doppler cooling below the ideal Doppler cooling limit. In polarization-gradient cooling, state-dependent energy shifts from the Stark effect and delayed optical pumping make atoms repeatedly climb potential energy hills. This mechanism needs the internal-state structure omitted by the independent-beam two-level system above. ^[ref].

William D. Phillips's team tested unexpectedly cold sodium molasses with time-of-flight atom detection. A probe above the cloud initially saw no atoms; deliberately heating them verified that the detector worked. Moving the probe below the cloud produced arrival signals consistent with temperatures below the Doppler cooling limit. Independent release/recapture checks used sharply apertured beams and included gravity; uncertain cloud and probe geometry still limited the inferred temperature. These controls tested the thermometer as well as the cooling theory. ^[ref].