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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.

A geostationary orbit is a circular, prograde, equatorial orbit with period equal to one sidereal day, approximately $23\mathrm{h}56\min$. Its satellite therefore remains above one longitude on the equator. Merely having a one-day period gives a geosynchronous orbit; an inclined or eccentric version still moves across the ground observer's sky.

Balancing gravitational acceleration and centripetal acceleration givesOne route is injection into an elliptical geostationary transfer orbit, followed by a burn near apogee to circularize and remove the remaining inclination; direct or low-thrust transfers are also possible. Subsequent station keeping controls longitude, inclination and eccentricity.

The fixed wide view permits frequent observations of weather, clouds and rapidly changing storms, as in Meteosat and GOES. The tradeoffs are large range, generally coarser spatial sampling for comparable optics, oblique views at high latitudes and poor polar coverage. It is valuable for following change through the day rather than obtaining global coverage with one spacecraft.^[ref]

Physics investigates matter, motion, and interactions by building mathematical models and testing their consequences against observations. A physical law supplies a relation; a model specifies the objects, conditions, and approximations used to apply it. Classical mechanics, electromagnetism, thermodynamics, and quantum mechanics organize different aspects of these investigations. Their usefulness comes from turning measurements into predictions that can be checked and refined.

Consider a simple pendulum: how does its oscillation period change when its length changes? For small amplitude, the ideal model gives $T_0=2\pi \sqrt{\ell /g}$, where $g$ is gravitational acceleration. Taking $g=9.81m/{\mathrm{s}}^2$ as the model parameter, a one-metre pendulum has predicted period $2.0061\mathrm{s}$. Increasing its length to four metres gives $4.0121\mathrm{s}$: quadrupling the length doubles the period. The point mass cancels from the equation, so changing it alone does not change this model's answer.

The prediction identifies a mechanism and a way to challenge it. The small-angle approximation replaces $\sin \theta$ by $\theta$; the full ideal equation predicts a longer period at large amplitude. At $60^{\circ }$, its period is about $1.07318T_0$, already a $7.3\%$ change. Measurements at such amplitudes can reveal the limit of the approximation, while appreciable drag requires a damped pendulum. These distinctions are developed in simple pendulum and dimensional analysis. The small-amplitude model is also derived in Ling, Sanny and Moebs, University Physics, pendulums.

Several different experiments can become one calculation. Nondimensionalization rewrites the pendulum equation using $\tau =t\sqrt{g/\ell }$, leaving $d^2\theta /d{\tau }^2+\sin \theta =0$. All lengths share that dimensionless equation; the amplitude remains a parameter. Similarly, conservation laws and symmetry reveal features that many models share. This is one source of the subject's beauty: a compact relation can organize a large range of phenomena.

The connection to evidence remains essential when a model is elegant. In his introductory lectures, Richard Feynman emphasizes testing proposed relations by experiment and using idealizations whose accuracy depends on the conditions. ^[ref] A discrepancy can motivate a better measurement, a better approximation, or a different physical model; the mathematical calculation and the empirical assessment are both part of the explanation.

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.

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]

A uniform spherical cloud with negligible pressure and velocity $\mathbf{u}=(\dot{R}/R)\mathbf{r}$ remains uniform: the continuity equation gives $\rho R^3={\rho }_{00}{R}_0^3$. The shell theorem gives interior gravitational acceleration $-(4\pi G\rho /3)\mathbf{r}$, so its outer radius satisfies the displayed equation, with fixed mass $M=4\pi {\rho }_{00}{R}_0^3/3$. Collapse from rest has ${\dot{R}}^2=2GM(1/R-1/R_0)$. Writing $C^2=2GM/R_0$, its cycloid parametrization is $R=R_0(1+\cos \varphi )/2$ and $Ct=R_0(\varphi +\sin \varphi )/2$, for $0\le \varphi \le \pi$. Complete collapse occurs after $\sqrt{3\pi /(32G{\rho }_{00})}$. This Newtonian dust solution ceases to be a regular fluid configuration at zero radius.

Motion after launch under a specified gravitational field, with negligible drag force and no thrust. In a constant downward gravitational acceleration $g$, horizontal velocity is constant and vertical position obeys $y(t)=y_0+v_{y0}t-g{t}^2/2$. Eliminating time gives a parabola when horizontal velocity is nonzero. Over larger distances the changing direction and magnitude of gravity require an orbital model instead.

The horizontal displacement between launch and landing in a projectile motion model. With equal launch and landing heights, uniform gravitational acceleration $g$, no air resistance, launch speed $u$ and angle $\theta$, it is $R=u^2\sin (2\theta )/g$. Different landing heights or forces change the result.

A projectile launched from and returning to the floor beneath a horizontal ceiling of height $H$ has maximum projectile range $\sqrt{8H(u^2/g-2H)}$ when $0\le H\le u^2/(4g)$. For higher ceilings the maximum is $u^2/g$. This assumes uniform gravitational acceleration, no air resistance and fixed launch speed $u$.

Indeed, with $z={\sin }^2\theta$, the apex condition is $z\le 2gH/u^2$ and $R^2=4u^{4}z(1-z)/g^2$. This concave quadratic function is maximized at the admissible value nearest $1/2$. Thus the lower-ceiling expression must not be extrapolated past its regime. If touching the ceiling is excluded, the limiting value is a supremum instead of an attained maximum.

A falling free surface does not instantaneously dry previously wetted rigid walls. Thin films remain above the bulk liquid and drain under the component of gravitational acceleration parallel to the walls. For a film of thickness $b$ over a wetted distance $\ell$, falling film flow gives a speed of order $g{b}^2/\nu$, hence $b\sim (\nu \ell /(gt){)}^{1/2}$ at late times for an order-one wall angle. These films can be visible in a cross-section while holding asymptotically negligible volume. In a V-shaped channel gravity current, local bulk depth falls only as $t^{-1/7}$; at a fixed nonzero station with a finite previous maximum wetted height, film volume decays as $t^{-1/2}$ while bulk-sectional volume decays as $t^{-2/7}$. Thus the film does not change the leading spreading law.

The motivating observation is a bob going around at almost constant height while its string traces a cone. The University of Iowa's demonstration places a camera above a short pendulum to reveal this motion. ^[ref] The following proposed measurement procedure develops that observation into a test of the conical pendulum model; it is not a report of collected data.

Suspend a compact bob from a rigid, fixed support using a light string, and measure $\ell$ from the pivot to the bob's centre. Put a stationary marker beside the orbit as a timing reference. Displace the bob sideways and give it a horizontal push tangent to the intended circle, then let go. For a chosen starting angle $\theta$, the predicted initial speed is $v=\sqrt{g\ell {\sin }^2\theta /\cos \theta }$. Adjust the launch between trials until a short observation interval has nearly constant orbit radius and height. Keep the pivot fixed while recording: circling the support by hand changes the problem, and a maintained drive would introduce another force.

Record from above with the camera looking approximately along the vertical axis. Check that the bob's centre follows a circle and that successive returns to the marker take nearly equal times. A separate side recording checks whether the bob's height stays constant. A markedly elliptical orbit or visibly rising and falling bob is not the uniform circular motion assumed by the conical pendulum equations. Measure the orbit diameter with a scale at the orbit's height; in a level side view, use the two extreme horizontal positions, not one arbitrary frame. Halve the diameter to obtain $r$, or measure the vertical drop $h$ directly in a calibrated side view. An oblique camera view and a scale at a different depth distort apparent lengths, so arrange the geometry before interpreting the images.

Time $N$ complete returns to the same marker in the same direction and estimate $P=\Delta t/N$. Counting ten revolutions, when the orbit stays sufficiently steady, spreads an endpoint timing error over ten periods. Repeat the timing and compare the first and last few revolutions to detect drift; using a longer interval helps only if the motion remains close to uniform. From independently measured geometry, compute $\sin \theta =r/\ell$ or $\cos \theta =h/\ell$, and check $r^2+h^2\simeq {\ell }^2$ if both were measured. Then compare the measured oscillation period with $P^2=4{\pi }^{2}h/g$. MIT's 8.01X experimental assignment explicitly asks students to compare the conical pendulum's period, length, and orbit radius with theory. ^[ref]

Repeat at several launch radii with the same $\ell$. Larger $r$ means smaller $h=\sqrt{{\ell }^2-r^2}$, so the ideal oscillation period decreases. A plot of measured $P^2$ against independently measured $h$ should follow a straight line through the origin with slope $4{\pi }^2/g$; conversely, $g=4{\pi }^{2}h/P^2$ estimates gravitational acceleration. Do not infer $h$ from the timing equation and then use that same equation as a test. Compare discrepancies with the spread of repeated timings and the resolution of the length measurements. String stretch, a moving pivot, and appreciable bob size challenge the stated geometry; drag force and pivot friction challenge the assumption of undriven uniform motion. Observe a short interval and record any shrinking or drifting orbit rather than silently treating it as an exact conical pendulum.

For an illustrative calculation, assume $\ell =1.00\mathrm{m}$, $m=0.050\mathrm{kg}$, $g=9.81m/{\mathrm{s}}^2$, and $\theta =40^{\circ }$. The model gives $r\approx 0.643\mathrm{m}$, $h\approx 0.766\mathrm{m}$, $P\approx 1.756\mathrm{s}$, $v\approx 2.300m/s$, and $T\approx 0.640\mathrm{N}$. Its vertical tension component is $0.4905\mathrm{N}$, equal to the gravitational weight; its inward component is approximately $0.412\mathrm{N}$. These are predictions from chosen parameters, not measured data or measurements of local $g$. The next figure uses the same stipulated $\ell$ and $g$, including illustrative angles $15^{\circ }$, $40^{\circ }$, and $60^{\circ }$.

In an Earth-fixed rotating reference frame, the horizontal displacement of a freely falling body receives contributions from Coriolis acceleration and centrifugal acceleration. A short-fall approximation substitutes the unperturbed downward velocity into the Coriolis term and the Earth's surface position into the centrifugal term. If gravitational acceleration excludes centrifugal acceleration, both contributions must be included explicitly.

Self-gravity is the gravitational acceleration produced by a system's own mass density. It is obtained by solving the Poisson equation for Newtonian gravity for that density, rather than treating the Newtonian gravitational potential as an externally fixed field. It competes with pressure and rotational support in gravitational instability of an astrophysical disk.

A point mass constrained to a fixed distance $r$ from a pivot and free to move in three dimensions under uniform gravitational acceleration. With the vertical axis downward and polar angle $\theta$ measured from it, the Lagrangian is the expression in the title. The azimuth $\varphi$ is a cyclic coordinate, so $m{r}^2{\sin }^2\theta \dot{\varphi }$ is conserved angular momentum about the vertical axis. A rigid massless rod enforces the constraint without the loss-of-tension limitation of a string.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

For steady flow through a fixed control volume with one inlet and one outlet, the first law of thermodynamics and conservation of mass giveHere $\dot{m}$ is the mass flow rate, $\dot{Q}$ incoming heat per unit time, ${\dot{W}}_s$ outgoing non-flow work per unit time, $w$ the flow speed, $z$ elevation and $g$ gravitational acceleration. The specific enthalpy $h=u+p/\rho$ includes both the transported internal energy per unit mass $u$ and the pressure work needed to push volume through the boundary; $\rho$ is mass density. The inlet supplies that pressure work, while the outlet consumes it to displace the downstream fluid.^[ref]

For an insulated stationary throttle, require negligible heat transfer, no shaft or other non-flow work, and negligible changes in bulk kinetic energy and gravitational potential energy. Then $h_2=h_1$. With unchanged composition this is also equality of molar enthalpy. The upstream and downstream states must admit equilibrium pressure, temperature and property values, although the constriction itself is irreversible and can be far from thermodynamic equilibrium. Start-up storage of energy in the walls and fluid, heat leakage, or a large outlet speed invalidates the reduced balance. A nozzle can convert specific enthalpy into bulk kinetic energy even when it is insulated.

The same accounting for a fixed transported amount is $U_2-U_1=p_1{V}_1-p_2{V}_2$, hence $U_1+p_1{V}_1=U_2+p_2{V}_2$. This explains why dropping the two pressure work terms would give the wrong conserved quantity.^[ref]

For a sphere with constant mass density $\rho$ collecting mass at rate $\beta$ per unit surface area, $\dot{m}=4\pi r^2\beta$ and $m=4\pi \rho r^3/3$ give $\dot{r}=\beta /\rho$. If it falls through stationary material under constant gravitational acceleration $g$ without drag, its velocity satisfies $\dot{v}+3\dot{r}v/r=g$. With $v(0)=0$, an integrating factor $r^3$ gives $v=g[r-r_0^4/r^3]/(4\dot{r})$.

In a plane-parallel polytropic atmosphere of index $m$, let $z$ measure depth from a zero-pressure boundary and take constant gravitational acceleration $g$. The stellar hydrostatic equation and $p=K{\rho }^{1+1/m}$ integrate to $(m+1)p/\rho =gz$. Consequently the adiabatic sound speed satisfies $c_s^2=\gamma p/\rho =\gamma gz/(m+1)$, so $c_s\propto z^{1/2}$. This local relation does not justify extrapolating a surface model to the stellar core.

Consider a point mass of fixed mass $M>0$ released from rest in a stationary medium. Choose downward as positive, assume constant gravitational acceleration $g>0$, neglect buoyancy, and impose linear drag $F_d=-bv$ with constant linear friction coefficient $b>0$. Here $v$ is the signed downward velocity component; it equals speed during this descent. Newton's second law becomesThe terminal velocity balances the two forces, and the time constant $\tau$ controls how quickly that balance is approached:For the illustrative parameters $M=1\mathrm{kg}$, $b=2kg/s$ and $g=9.8m/{\mathrm{s}}^2$, the time constant is $0.5\mathrm{s}$ and the terminal velocity is $4.9m/s$ downward. At $t=1\mathrm{s}=2\tau$, the downward velocity is $4.9(1-e^{-2})m/s\simeq 4.23686m/s$ and the downward acceleration is $9.8e^{-2}m/{\mathrm{s}}^2\simeq 1.32629m/{\mathrm{s}}^2$. This is a toy linear drag model, not a claim that every $1\mathrm{kg}$ falling object has this resistance law.

For a proof outline, subtract the equilibrium value: $u=v-v_T$ obeys $du/dt=-u/\tau$. The exponential decay solution and the initial condition give $u=-v_T{e}^{-t/\tau }$; one derivative gives the acceleration. Every time constant multiplies the remaining gap $v_T-v$ by $e^{-1}$. The exact limit is reached only as $t\to \infty$. This makes the model useful for estimating a settling duration without mistaking vanishing acceleration for vanishing velocity.

Newton already analyzed vertical motion under uniform gravity and resistance proportional to velocity in Book II, Section I, Proposition III of the Principia. His geometric construction includes a greatest downward velocity, illustrating that a limiting rate can emerge from balancing a driving force and resistance. ^[ref]

The relative acceleration of neighboring freely falling bodies is measured by geodesic deviation, comparing their infinitesimal separation $\xi$ in a parallel transport frame along a reference timelike geodesic with tangent $U$. With $R^a{}_{bcd}$ defined by the covariant derivative commutator, it has the displayed form. In the Newtonian limit the corresponding tidal tensor is the spatial gradient of the gravitational acceleration. A single freely falling body can remove local gravitational acceleration by its frame choice, whereas relative tidal acceleration measures curvature.