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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.

Choose a fixed support point $O$ and nonrotating axes in an inertial frame. All angular momenta, torques, and moments of inertia below refer to $O$. Let $\hat{\mathbf{e}}$ point along the axle from $O$ toward the centre of mass, and let $\hat{\mathbf{z}}$ point vertically upward. Write $\theta$ for their angle. For a rapidly spinning symmetric top, the dominant angular momentum is ${\mathbf{L}}_O\simeq L_s\hat{\mathbf{e}}$, where $L_s=I_3{\omega }_3$ is signed and $I_3$ is the axial principal moment of inertia. The sign of the axial angular velocity ${\omega }_3$ follows the right-hand rule.

With mass $M$, centre of mass distance $d$, and constant gravitational acceleration $g$, the gravitational force and its torque areAn ideal support applies a force at $O$, so its lever arm and torque about $O$ vanish. At the instant the axle lies in the $xz$ plane with a positive $x$ component, the gravitational torque points along $+y$. It adds $d{\mathbf{L}}_O={\mathbf{\tau }}_{O}dt$ toward $+y$, rather than along the downward force. The dot product identity $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$ shows why perpendicular torque changes direction without instantly changing magnitude. This vector balance and the slow-turning relation ${\mathbf{\tau }}_O=\mathbf{\Omega }\times {\mathbf{L}}_O$ are developed in Chapter 20: Rotation in space (sections 20-2 and 20-3, equations 20.13–20.15).

For steady precession about the vertical, $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$ and $d\hat{\mathbf{e}}/dt=\Omega \hat{\mathbf{z}}\times \hat{\mathbf{e}}$. Equating the two cross products givesThe angular momentum tip traces a circle of radius $|L_s|\sin \theta$: its sideways speed is $|\Omega L_s|\sin \theta$. Both that radius and the gravitational lever arm contain $\sin \theta$, explaining its cancellation for $0<\theta <\pi$. Exactly vertical alignment has zero gravitational torque and no distinguishable azimuthal precession to infer by this cancellation. Positive $L_s$ gives positive precession, counterclockwise viewed from above; reversing the spin reverses the slow precession. ^[ref]

Pyotr Leonidovich Kapitza's 1938 test compared helium flow through optically flat glass disks. Suspension set the tube-to-bath level difference; a cathetometer read levels. The gap was estimated as $0.5\mu \mathrm{m}$. Levels barely moved over minutes above the transition, but equalized in seconds below. A laminar flow viscosity calculation gave only an upper bound.^[ref]

The essential control is to compare the same restriction and level readout at different temperatures. A changed pressure head is not itself evidence for a changed material response: the driving head must be included when interpreting the flow. For an idealized comparison at equal vapor pressure, negligible surface tension corrections and nearly constant mass density $\rho$, hydrostatic equilibrium givesHere $\Delta h$ is the difference in free-surface heights and $g$ is gravitational acceleration. This explains why following a liquid level provides both a driving pressure and a transported volume; it does not turn a finite-resolution observation into proof of zero viscosity.

For a spherical shell of uniform mass density $\rho$, inner radius $a$ and outer radius $2a$, suppose a negligible-width radial tunnel permits a small mass $m$ to fall from rest at the outer surface. The enclosed mass at radius $r$ within the material is $4\pi \rho (r^3-a^3)/3$. The spherical shell theorem gives inward gravitational acceleration $g(r)=4\pi G\rho (r-a^3/r^2)/3$. Integrating from $a$ to $2a$ gives released gravitational potential energy $4\pi G\rho m{a}^2/3$, hence speed $a\sqrt{8\pi G\rho /3}$ at the cavity. The cavity has zero gravitational field, so the diameter transit time is the title expression, independent of $a$ and $m$.

Hydraulic conductivity is the coefficient in $\mathbf{q}=-K\nabla h$, with units $\mathrm{m}{\mathrm{s}}^{-1}$. Because the gradient of hydraulic head is dimensionless, hydraulic conductivity has the dimensions of velocity, but it is not the actual Darcy velocity unless the driving gradient has magnitude one. Intrinsic permeability of a porous medium $k$ has units ${\mathrm{m}}^2$ and describes pore geometry; hydraulic conductivity also depends on mass density, dynamic viscosity and gravitational acceleration.^[ref]

For $k=10^{-12}{\mathrm{m}}^2$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$, hydraulic conductivity is $K=9.81\times 10^{-6}\mathrm{m}{\mathrm{s}}^{-1}$. Doubling dynamic viscosity while keeping pore geometry, mass density and gravitational acceleration fixed halves hydraulic conductivity without changing intrinsic permeability of a porous medium. Thus a value quoted as a material's hydraulic conductivity needs its fluid conditions.

Hydraulic head is the sum of elevation and pressure head, measured in metres relative to a chosen height datum. For constant mass density $\rho$ and gravitational acceleration magnitude $g$, the pressure head $p/(\rho g)$ is the height of a column of the same fluid supported by fluid pressure $p$. A water-filled open measuring tube records hydraulic head as the elevation of its water level when $p$ is gauge fluid pressure. Changing the common pressure reference or height datum adds a constant to hydraulic head and leaves its driving gradient unchanged.

For water with $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, a point at $z=2\mathrm{m}$ with gauge fluid pressure $p=9810\mathrm{Pa}$ has $h=3\mathrm{m}$. A second point at $z=0$ with gauge fluid pressure $29430\mathrm{Pa}$ also has $h=3\mathrm{m}$: these values are compatible with a stationary column under hydrostatic pressure. It is a difference in hydraulic head, rather than a difference in fluid pressure alone, that drives saturated Darcy flow. This form uses constant mass density; one cannot generally pull spatially varying $\rho$ through the gradient.^[ref]

For an ideal gas with constant composition, set $\mathcal{R}=k_B/\bar{m}$, where $\bar{m}$ is the mean particle mass. Constant downward gravitational acceleration $g$ and $T=T_0-\alpha z>0$ give $p^{\prime }=-(g/\mathcal{R}T)p$ by hydrostatic equilibrium. Integration gives $p/p_0=(T/T_0{)}^{g/(\mathcal{R}\alpha )}$, and the ideal gas law gives the displayed mass density ratio. The isothermal limit $\alpha \to 0$ is $\rho /{\rho }_0=\exp [-gz/(\mathcal{R}{T}_0)]$.

Hydrostatic pressure is the pressure distribution in a resting fluid, balancing its body force from gravity. Take $z$ upward and a uniform downward gravitational field $\mathbf{g}=-g{\mathbf{e}}_z$, where $g>0$. The vertical pressure gradient satisfiesA short derivation balances the vertical forces on a thin horizontal layer of area $A$ and thickness $\Delta z$: $p(z)A-p(z+\Delta z)A-{\rho }_{f}gA\Delta z=0$. Dividing by $A\Delta z$ and taking the derivative gives the equation above. For constant mass density ${\rho }_f$, integration from the surface at $z_s$ givesThus pressure increases with depth $h$. For the stipulated water with mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ and gravitational acceleration $9.81\mathrm{m}{\mathrm{s}}^{-2}$, depths $0.20\mathrm{m}$ and $0.30\mathrm{m}$ give gauge pressures $1962\mathrm{Pa}$ and $2943\mathrm{Pa}$. Their difference produces the buoyancy of the example cube. ^[ref]

The inviscid magnetohydrodynamic momentum equation balances inertia, pressure, gravitational acceleration and Lorentz force density. The magnetic force separates into magnetic tension and magnetic pressure. It is coupled to the continuity equation and the ideal magnetohydrodynamic induction equation.

An inclined plane is a flat surface tilted through an angle $\theta$ above horizontal. A ramp exchanges a smaller lifting force for a longer travel distance; for a released body, it reduces the downhill component of gravitational weight and makes the resulting acceleration easier to measure. Its simplest dynamical model is a nonrotating block on a stationary, rigid surface, with constant mass $m>0$, uniform gravitational acceleration of magnitude $g>0$, and $0\le \theta <\pi /2$. Assume continued contact, negligible drag force, no tipping, and no applied pull. The only forces are gravitational weight, the normal force, and any tangential friction.

The central distinction is between a block resting on the inclined plane and one already sliding. Static friction supplies the force needed to maintain rest, up to its bound; kinetic friction has a prescribed magnitude in the ideal constant-coefficient model. These laws are empirical approximations whose range depends on the contacting surfaces.^[ref]

For a spherical, nonstreaming stellar population with isotropic velocity dispersion, each component of the random pressure tensor equals $p=n⟨v_r^2⟩$. The observed second stellar velocity moment satisfies $N(R){\sigma }_{\mathrm{los}}^2(R)=2{\int }_R^{\infty }p(r)r/\sqrt{r^2-R^2}dr$. Composing this Abel transform with itself gives the displayed inverse under sufficient decay and regularity. The Spherical Jeans equation then gives ${\psi }^{\prime }=p^{\prime }/n$, with gravitational acceleration $\nabla \psi$. The Poisson equation for Newtonian gravity recovers total gravitating mass density, not just the tracer density. Without velocity isotropy the same projected moment also contains anisotropic stresses.

Reflect a vacuum gravitational potential in the upper half-space evenly across $z=0$. For the convention in which gravitational acceleration is $\nabla \psi$, the derivative jump is $2{\partial }_z\psi (R,0^{+})=-4\pi G\Sigma (R)$. Thus a point mass $m$ below the plane produces a Kuzmin disk, and superposing point masses with weight $\mu (b)db$ produces $\Sigma (R)=(2\pi {)}^{-1}{\int }_0^{\infty }b\mu (b)(R^2+b^2{)}^{-3/2}db$. Positive weights give a nonnegative surface density wherever the integral converges.

A plane-parallel magnetostatic equilibrium with constant downward gravitational acceleration $g$, horizontal magnetic field, constant isothermal sound speed $c_s$, and constant Alfvén speed $v_a$ hasIndeed the magnetohydrodynamic total pressure is $(c_s^2+v_a^2/2)\rho$, and its vertical derivative is $-g\rho$. The magnetic pressure therefore increases the scale height above that of an unmagnetized atmosphere.

An instrument for measuring pressure relative to another pressure. A liquid-column manometer uses hydrostatic pressure; a digital manometer uses a pressure transducer.

The appeal of a liquid-column manometer is that a pressure measurement becomes a visible length measurement: mass density and gravitational acceleration supply the conversion. It can compare two pressure tappings in a flowing fluid, even though the measuring liquid itself must settle into hydrostatic equilibrium. The reference connection determines which pressure difference is being measured.

The ideal measurement calibration is a straight line through the equal-pressure zero. For the stipulated water mass density and gravitational acceleration above, its slope is $9.81\mathrm{Pa}$ per vertical millimetre. A lower liquid mass density gives a larger length change for the same pressure, improving the height-reading contribution to measurement resolution but requiring more vertical space for a given pressure range. The pressure scale depends on the actual mass density, which depends on temperature; $1000\mathrm{kg}{\mathrm{m}}^{-3}$ here is a chosen model value, not an exact property of all water samples.

Suppose each liquid meniscus height is rounded to the nearest scale division $q=1\mathrm{mm}$. Each height measurement error lies within $q/2$, so the error in their difference lies within $q$. The resulting pressure error is bounded by $\rho gq=9.81\mathrm{Pa}$. To quantify a standard uncertainty, make the additional model assumption that the two rounding errors are independent random variables, each having a continuous uniform distribution on $[-q/2,q/2]$. Each then has variance $q^2/12$; adding the two variances givesHere $u$ denotes standard uncertainty. The $9.81\mathrm{Pa}$ worst-case bound and $4.0049\mathrm{Pa}$ standard uncertainty answer different questions. At $588.6\mathrm{Pa}$ they are respectively $1.67\%$ and $0.680\%$ of the reading. Shared reading bias or a different reading method can violate the assumed independence.

If the mass density, gravitational acceleration and height difference also have small independent standard uncertainties, a first-order estimate follows from the partial derivatives of $\Delta p=\rho g\Delta h$:This displays why a finer scale does not remove uncertainty in mass density or gravitational acceleration. A pressure transducer compared against the column needs its own differential-pressure calibration to establish an electrical response.

A mathematical model represents selected features of a system with mathematical objects and relations so that its behavior can be analyzed or predicted. Choosing the representation is part of the work: specify the quantities tracked, the equations or probability distributions connecting them, and the conditions under which those connections are expected to apply. A mathematical proof can establish a consequence of the model; comparing that consequence with observations tests its suitability for the intended use.

For a simple pendulum, the angle $\theta (t)$ and gravitational acceleration $g$ give the ideal equation $\ddot{\theta }+(g/\ell )\sin \theta =0$. Its assumptions include a point mass, fixed length $\ell$, uniform gravitational acceleration, and no damping. The small-angle approximation gives the simpler harmonic oscillator equation and oscillation period $T_0=2\pi \sqrt{\ell /g}$. A larger amplitude changes the full model's period; drag requires a damped pendulum. The simpler model remains useful when its omitted effects are small for the question being asked.

A model need not track every detail. A Bernoulli process models repeated success/failure observations using one probability and an independence assumption; a Gaussian process specifies a mean and a covariance function to model related measurements. Each representation enables calculations by retaining some distinctions and omitting others. Useful simplification therefore comes with a stated domain of use and quantities against which its predictions can be checked.

A membrane osmometer connects osmosis to a visible liquid level or a measured balancing pressure. In a teaching version, a solute-containing pouch is joined to a narrow vertical tube and submerged in external water. Entry increases the enclosed volume and raises the meniscus. Both the column and any stretched pouch can contribute an opposing pressure.

The University of Toronto's Osmosis (Brine) by University of Toronto Department of Physics demonstration (activity Me-012b, Description, steps 1–14) specifies soaking a dialysis membrane tube, tying one end closed and fastening the other around a thistle tube's narrow end. A stand holds the tube. Colored brine is cooled to room temperature and injected using a syringe to reduce bubbles; the pouch is completely immersed in a $1$ or $2\mathrm{L}$ beaker of distilled water. A taped marker records the initial level; subsequent meniscus height is the readout. The protocol proposes another pouch containing distilled water as a comparison, and its older instructions explicitly check seals for leaks.^[ref]

A rise is consistent with net water entry. To infer osmotic pressure quantitatively, use a retained solute and establish a stable equilibrium after accounting for dilution. If both free surfaces are at the same atmospheric pressure, the liquids have approximately equal mass densities $\rho$, and the pouch adds negligible mechanical load, the hydrostatic pressure difference at membrane height is approximately $\Delta P=\rho gh$, where $h$ is the meniscus height above the external surface and $g$ is gravitational acceleration. This is the liquid-column inference, not a pressure calibration supplied by the Toronto demonstration. For unequal mass densities, compare the two column pressures separately.

Useful additional controls are matched initial levels and geometry, constant temperature, replicate runs, and monitoring external solute to test retention. These are proposed checks, not reported Toronto results. Leaks, trapped bubbles, surface tension in the narrow tube, evaporation and membrane stretching can alter the level. In particular, a dialysis membrane may pass dissolved salt: the brine setup illustrates transient movement but cannot be assumed to retain every dissolved ion. Color improves visibility without proving salt retention. A stationary level can reflect weakening concentration differences or leakage rather than osmotic thermodynamic equilibrium; a rising level alone supplies neither $n$ nor a verified value of $nRT/V$.

A geophysical gravitational acceleration unit: $1\mathrm{mGal}=10^{-5}\mathrm{m}{\mathrm{s}}^{-2}$. It is used for small gravity anomalies and should be converted before substitution into SI forms of the Bouguer slab formula.

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

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

For radius $R$, constant mass density $\rho$ and $M=4\pi \rho R^3/3$, normalize the Newtonian gravitational potential to zero at infinity. ThenBy the spherical shell theorem, each spherical shell contributes the potential of a point mass outside it and a constant inside it. Summing inner and outer shells gives $\Phi (r)=-4\pi G\rho [r^2/3+(R^2-r^2)/2]$ inside. Equivalently, angular integration of the free-space Green function of the Laplacian gives ${\int }_{-1}^1(r^2+r^{\prime 2}-2r{r}^{\prime }\mu {)}^{-1/2}d\mu =[r+r^{\prime }-|r-r^{\prime }|]/(r{r}^{\prime })$. The resulting radial integral has the same value.

The potential and its first radial derivative are continuous at $R$. The interior gravitational acceleration is $-\nabla \Phi =-(GM/R^3)\mathbf{r}$, and ${\nabla }^2\Phi =4\pi G\rho$. At the centre $\Phi (0)=-3GM/(2R)$ while its gradient vanishes.