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A uniform thin ring of radius $a$ and mass $M$ has axial gravitational field $g_z=-GMz/(a^2+z^2{)}^{3/2}$. Every ring element is the same distance from the axial point; transverse contributions cancel and axial projection supplies the factor $z/\sqrt{a^2+z^2}$. This also follows by differentiating the axial potential $\Phi =-GM/\sqrt{a^2+z^2}$. Near the centre $g_z\simeq -GMz/a^3$, whereas far away it approaches the point-mass field.

A body force acts throughout a body's or fluid's volume, rather than only through its surface. Gravity is the example used in hydrostatic pressure: a gravitational field $\mathbf{g}$ gives force per unit volume $\rho \mathbf{g}$, where $\rho$ is the mass density. More generally, a body force per unit mass $\mathbf{f}$ contributes $\rho \mathbf{f}$ to momentum conservation. Forces from electromagnetism are another example. Surface pressure forces produce buoyancy on an immersed body, while its body force from gravity acts downward.

Buoyancy is the resultant upward force exerted by a resting fluid on an immersed body in a downward gravitational field. The fluid pushes inward, perpendicular to each part of the body's surface. Because hydrostatic pressure increases with depth, the upward push from below exceeds the downward push from above. Opposite horizontal forces cancel; buoyancy is the remaining vertical force. A sinking body still receives this upward support. ^[ref]

More precisely, pressure $p$ on a small surface of area $dA$ with outward unit normal vector $\mathbf{n}$ exertsThe minus sign says the force points into the body. On a cube, the top force points down, the bottom force points up, and the side forces cancel in pairs because opposite sides have the same pressure at each depth. This is a surface traction; gravity on the body itself is a separate body force. ^[ref]

For a computed example, stipulate resting water with constant mass density ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$ and uniform gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. A fully immersed cube has side $L=0.10\mathrm{m}$, top depth $h_t=0.20\mathrm{m}$ and bottom depth $h_b=0.30\mathrm{m}$. Each horizontal face has area $A=L^2=0.010{\mathrm{m}}^2$, and its volume is $V=L^3=0.001{\mathrm{m}}^3$. Its top and bottom gauge pressures areThe corresponding forces relative to the common surface pressure are $p_{g,t}A=19.62\mathrm{N}$ downward and $p_{g,b}A=29.43\mathrm{N}$ upward. Therefore the net buoyancy isUsing absolute pressures adds the same $p_{\mathrm{surface}}A$ to both face forces, leaving their difference unchanged. Zero gauge pressure at the surface does not mean zero absolute pressure. Increasing both depths by the same amount also leaves buoyancy unchanged in this constant-mass density model.

Archimedes' principle turns this calculation into a rule for bodies of any shape. It explains floating equilibrium and lets us infer an immersed body's volume from the difference between its force from gravity and the force needed to support it. In particular, buoyancy depends on the displaced volume and the surrounding mass density, whereas the body's own mass density decides how its force from gravity compares with that support.

The point through which the resultant gravitational force acts when that force has a single point of application. In a uniform gravitational field, it coincides with the center of mass; a nonuniform field need not give the same point.

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

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

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

Two positive point masses $M_1,M_2$ separated by $D$ have a unique intervening gravitational field zero. At distance $x$ from $M_1$, Newton's law of universal gravitation gives $M_1/x^2=M_2/(D-x{)}^2$, hence the displayed value. The signed offset towards $M_2$ from their centre of mass is $x-D{M}_2/(M_1+M_2)$. This instantaneous cancellation is distinct from a Lagrange point, which includes the centrifugal force of a co-rotating description. Along the joining line the zero is unstable: the derivative of the signed acceleration is $2G{M}_1/x^3+2G{M}_2/(D-x{)}^3>0$.

A circular infinite cylinder of radius $b$ and uniform mass density $\rho$ has linear mass density $\lambda =\pi b^2\rho$. A coaxial Gaussian surface of length $L$ has inward radial gravitational field and no end-cap flux. Gauss's law for gravity gives $2\pi rL{g}_r=-4\pi G\lambda L$ outside, hence the displayed result. Inside, enclosed linear mass density is $\pi r^2\rho$, so $g_r=-2\pi G\rho r$. An exterior circular orbit has speed $\sqrt{2G\lambda }$ independent of radius. The exterior formula presupposes circular axial symmetry; area alone would not specify the near field of a noncircular rod.

The usual Newtonian gravitational potential satisfies $\mathbf{g}=-\nabla \Phi$ and ${\nabla }^2\Phi =4\pi G\rho$. An equally consistent convention sets $\phi =-\Phi$, giving $\mathbf{g}=\nabla \phi$ and ${\nabla }^2\phi =-4\pi G\rho$. With zero potential at infinity, a point mass has $\Phi =-GM/r$ and $\phi =GM/r$. Both conventions describe the same inward gravitational field and the same Gauss's law for gravity. Only the signs of the potential and its differences change. The gravitational potential energy per unit test mass is $\Phi$, or $-\phi$ in the alternative convention. A formula for a potential difference must therefore be read together with its relation to the gravitational field.

Gravity governs falling bodies and orbits. Newtonian gravity describes it through a gravitational field acting on masses; general relativity instead relates matter and radiation to the geometry of spacetime through the Einstein field equations.

For constant mass density $\rho$ on $a<r<b$, the shell theorem gives zero gravitational field for $r\le a$, inward magnitude $4\pi G\rho (r^3-a^3)/(3r^2)$ for $a<r<b$, and $4\pi G\rho (b^3-a^3)/(3r^2)$ outside. These formulas agree continuously at both boundaries. The empty cavity has zero gravitational field but generally nonzero constant gravitational potential.

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 $m$.

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]

For an ordinary stationary laboratory sample, a useful convention isThe bulk kinetic energy $K_{\mathrm{bulk}}$ includes organized whole-body motion; for pure translation it is $M{v}_{\mathrm{cm}}^2/2$, with total mass $M$ and center of mass speed $v_{\mathrm{cm}}$. The external potential energy ${\Phi }_{\mathrm{ext}}$ includes elevation in an imposed gravitational field, approximately $Mgz$ in a uniform field. Accelerating a sealed bottle or lifting it without otherwise changing its state changes these mechanical terms, rather than its internal energy. Molecular kinetic energy is measured relative to the chosen bulk motion.^[ref]

State the convention rather than applying this separation blindly. If an experiment separately tracks a rotating apparatus or an organized flow, its mechanical kinetic energy must not also be counted in $U$. In Joule's paddle-wheel experiment, waiting for the apparatus and fluid to come to rest removes such an endpoint ambiguity. Including the vessel and paddle changes whose internal energy is measured; including the falling weights additionally puts their changing gravitational potential energy inside the enlarged system's total energy balance.

Spatial moments of a mass density distribution organize its far gravitational field. Symmetric trace-free parts isolate angular multipoles; the second moment gives the mass quadrupole moment. In slow-motion linearized gravity, changes of the quadrupole produce the leading gravitational radiation; the conserved monopole and centre-of-mass dipole do not radiate independently.

Neutral buoyancy means a fully immersed body's upward buoyancy equals its downward force from gravity. For a body of fixed volume in a constant-mass density fluid under a uniform gravitational field, this occurs when its mean mass density equals the surrounding mass density. A $0.10\mathrm{m}$ cube of mass density $1000\mathrm{kg}{\mathrm{m}}^{-3}$ in the stipulated water has mass $1.000\mathrm{kg}$, so both forces are $9.81\mathrm{N}$.

Ideally it can remain at rest at any depth for which it stays fully immersed and clear of boundaries. Raising or lowering it changes the absolute pressure, but neither its displaced volume nor the difference between the vertical pressure forces. There is therefore no vertical restoring force toward a preferred depth in this model. Neutral buoyancy is a vertical force balance, not a claim about resistance to tilting or about a compressible body whose volume changes with pressure.

The leading Schwarzschild light deflection of a distant ray passing an isolated spherical mass is $\alpha =2(1+{\gamma }_{\mathrm{PPN}})GM/(b{c}^2)$. Thus measurements of angular displacement can constrain ${\gamma }_{\mathrm{PPN}}$ rather than merely detect a nonzero gravitational field. In general relativity this becomes the Schwarzschild light deflection $4GM/(b{c}^2)$.

By reflection symmetry the vertical gravitational field of the self-gravitating incompressible slab vanishes at its mid-plane. Inside the slab, the gravitational Poisson equation therefore integrates to $d\psi /dz=4\pi G\rho z$. Treat the star as a test particle, so its own mass does not alter that field. Its vertical stellar orbit in a uniform slab obeysHence $z(t)=a\cos (\omega t)$ with $\omega =\sqrt{4\pi G\rho }$. The first mid-plane crossing occurs after one quarter of this simple harmonic motion period:Its independence of $a$ follows from the linear interior restoring force. The calculation neglects drag and changes to the slab, and it uses the infinite planar approximation rather than the field of a finite Galactic disk.

The first argument mixes a material fluid element with a fixed spatial volume. For a material volume element following the fluid, its volume is generally time dependent:Therefore the material derivative cannot be taken outside the product with $\delta V$ as in the proposed cancellation. Instead, mass conservation statesThe momentum balance for a material fluid element, using Newton's second law, is consequentlyFor a fixed control volume, fluid crosses its boundary and carries momentum; a momentum flux term must be included. Its local conservative form is ${\partial }_t(\rho \mathbf{u})+\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=-\nabla p$, equivalent to the displayed Euler equations after using the continuity equation. The erroneous equation omits the term $\rho \mathbf{u}\nabla \cdot \mathbf{u}$ needed to convert a material derivative of momentum density into this conservative form.

For the second task, take a homogeneous fluid with constant mass density $\rho$, prescribed gravitational acceleration $\mathbf{g}$, incompressible flow and inviscid flow. Its equations areUnder the replacement $\rho \mapsto \tilde{\rho }=C\rho$, the same velocity solves these equations if $p\mapsto \tilde{p}=Cp$, up to a spatially constant pressure offset. This is density scaling invariance of incompressible Euler flow:The comparison assumes the same geometry and compatible velocity or gravitational free surface conditions; an independently specified nonuniform boundary pressure would supply additional information. The gravitational field is taken as prescribed. If the fluid's own self-gravity must be included, its field changes with mass density, so the invariance need not hold. A single freely moving particle's density-independent gravitational acceleration does not justify neglecting that distinction.

No. A motionless isolated point mass produces a static gravitational field. All its mass multipole moments are time independent, so there is no outgoing gravitational radiation. In the quadrupole formula, $h_{ij}^{\mathrm{TT}}\propto d^2{Q}_{ij}^{\mathrm{TT}}/d{t}^2=0$.

In a local inertial frame in vacuum, gravitational waves travel at $c$ in general relativity; there is no ordinary material index of refraction analogous to a transparent dielectric. Ordinary matter couples so weakly that laboratory prisms for gravitational waves are implausible. Propagation through a varying background gravitational field nevertheless changes ray directions and travel times, giving gravitational lensing and Shapiro time delay. In weak static coordinates $d{s}^2\simeq -(1+2{\Phi }_N/c^2)c^{2}d{t}^2+(1-2{\Phi }_N/c^2)d{\mathbf{x}}^2$, the null condition gives an effective coordinate index of refraction $n_{\mathrm{eff}}\simeq 1-2{\Phi }_N/c^2$. This useful description depends on coordinates and does not change the local propagation speed. Thus astronomical gravitational refraction is plausible, while a substantial everyday material effect is not.