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For a point mass, $\mathbf{g}(\mathbf{R})=-GM\mathbf{R}/R^3$. Along the outward radial direction, $g_R=-GM/R^2$ andThus a small radial displacement $\mathbf{a}=a\hat{\mathbf{R}}$ has differential accelerationThe farther point falls inward less strongly, so its acceleration relative to the reference point points outward. The formula is first order in $|a|/R$.

For point masses, Newton's force is attractive along their separation with magnitude $G{m}_1{m}_2/r^2$. Treat the narrow rod as uniform, with linear mass density $\mu =M/(2a)$; a mass profile would otherwise be needed beyond total mass and length.

For axial and transverse gravity of a uniform rod, on the axis at $b>a$, integrating the element force givesdirected toward the rod. On the perpendicular bisector, axial components cancel and transverse components add:At $r=b\gg a$,Both approach the point-mass force, but $F_1/F_2=1+3a^2/(2b^2)+O(a^4/b^4)$, so the axial attraction is slightly larger.

For a circular orbit perpendicular to a uniform rod, rotational symmetry about the rod makes a centred circular orbit possible in its perpendicular-bisector plane. With $m\ll M$, centrifugal balance givesFor $r\gg a$, it is Keplerian, $v\sim \sqrt{GM/r}$. For rod thickness $\ll r\ll a$, the force approximates that of an infinite line and $v\to \sqrt{GM/a}$, a flat speed limit rather than a divergence.

For two isolated point masses, introduce total mass $M_T=M+m$, reduced mass $\mu =Mm/M_T$ and relative separation $r$. The relative acceleration is $\ddot{\mathbf{r}}=-G{M}_T\mathbf{r}/r^3$. Conserved specific angular momentum $h=|\mathbf{r}\times \dot{\mathbf{r}}|$ fixes a plane and a constant areal velocity $h/2$. The specific orbital energy isFor $h\ne 0$, the trajectory is a conic:Negative energy gives an ellipse, including the circular limit $e=0$; zero energy gives a parabola; positive energy gives a hyperbola. Bound ellipses obey $\epsilon =-G{M}_T/(2a_o)$ andThus Kepler's laws of planetary motion follow from angular momentum conservation and the inverse-square force. For a circular orbit of radius $r$, $v_c=\sqrt{G{M}_T/r}$, while escape from that radius requires $v_{\mathrm{esc}}=\sqrt{2G{M}_T/r}$. Energy determines whether the orbit is bound; angular momentum also determines its shape. These ideal results neglect extended-body perturbations, drag and relativistic corrections.

For a fixed rotation axis with constant moment of inertia, the axial torque obeys $\tau =I\ddot{\theta }$, with $\tau =\sum (\mathbf{r}\times \mathbf{F})\cdot {\hat{\mathbf{e}}}_{\mathrm{axis}}$. For a point mass at radius $l$, the tangential equation is $F_t=ml\ddot{\theta }$; multiplying by $l$ gives $l{F}_t=m{l}^2\ddot{\theta }$, as required. Radial forces have zero moment about the axis.

Take increasing $\theta$ in the direction of the clapper's initial fall and let the contact force $N$ oppose that tangential motion. Initial equilibrium of the clapper givesWhile contact persists the bell and clapper rotate together. Their respective offsets from the upward vertical are $\theta -\beta$ and $\theta$. Summing external gravitational moments eliminates their internal contact force:The clapper alone satisfies $m{l}^2\ddot{\theta }=mgl\sin \theta -Nl$. Hence for $m\ll M$,A nonadhesive contact cannot exert tension. The first zero of this expression on the released trajectory is the loss-of-contact angle. Write $q=lh/k^2=0.9375$. ThenThe second-quadrant root is reached from $\theta =\beta =30^{\circ }$; selecting the principal negative arctangent would choose the wrong branch. Indeed the exact contact force before taking $m/M\to 0$ is $mMg[k^2\sin \theta -lh\sin (\theta -\beta )]/(M{k}^2+m{l}^2)$, so its zero is unchanged.

A gravitational field is force per unit test mass, $\mathbf{g}=\mathbf{F}/m$. Newton's law of universal gravitation gives the field of a point mass M:The potential is potential energy per unit mass, with zero at infinity for an isolated mass. Fields and potentials superpose. The potential difference fixes the work per unit mass, and a conservative gravitational field has zero circulation around a closed path.

For a continuous mass distribution, the gravitational Gauss law is $\oint \mathbf{g}\cdot d\mathbf{S}=-4\pi G{M}_{\mathrm{enc}}$. A spherical body therefore has the same exterior field as a point mass at its centre. Inside a uniform sphere of density $\rho$,The field is continuous at its surface and vanishes at its centre. An empty spherical shell has zero interior field but a nonzero constant interior potential. Near a small region of Earth's surface one may approximate $\mathbf{g}$ as uniform and the potential change as $g\Delta h$. Farther away its spatial gradient produces tidal relative accelerations. Escape to infinity from radius R requires speed $\sqrt{2GM/R}$ when rotation and other bodies are neglected.

An inverse-square field spreads through area proportional to squared distance. More generally the flux contribution of a point source is proportional to the solid angle subtended by a surface element; the oriented solid angles sum to $4\pi$ for an enclosed source and zero for an exterior one. Superposition extends this to a mass distribution.

Newton's law of universal gravitation gives $\mathbf{g}=-GM\hat{\mathbf{r}}/r^2$ for a point mass. A surrounding sphere therefore has flux $-4\pi GM$, soFor the tunnel, approximate its water as an infinite uniform cylinder with $R=3.5\mathrm{m}$ and density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. The bob is at radial distance $r=R+1.5\mathrm{m}=5.0\mathrm{m}$. A coaxial Gaussian surface givesFor the plumb-line deflection from cylindrical gravity, horizontal equilibrium gives $d/L\simeq |g_r|/g$. Draining removes the attraction toward the tunnel, so the bob shifts away byThe unchanged rock and ordinary gravitational field cancel in the before/after comparison; end effects and mechanical disturbances are neglected.

For the tidal water loading of a spring balance, approximate the extra water at high tide as an infinite plane slab of thickness $h=8\mathrm{m}$, with the suspended mass above it. The gravitational field of an infinite plane sheet, derived with a pillbox Gaussian surface, gives added downward field $\Delta g=2\pi G\rho h$. With initial spring extension $x_0=mg/k=1\mathrm{m}$,The extension increases at high tide. These estimates use $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the stated water-density approximation; salinity, finite estuary geometry and terrestrial tidal deformation alter a real measurement.

A field assigns a physical quantity to each point. Gravitational field is force per unit test mass; electric field is force per unit test charge. In a conservative, time-independent field,Write $U=m\Phi$ for gravitational potential $\Phi$, or $U=q\varphi$ for electric potential $\varphi$, giving $\mathbf{g}=-\nabla \Phi$ and $\mathbf{E}=-\nabla \varphi$. A point mass has $U=-GMm/r$ with zero at infinity, while two point charges have $U=Qq/(4\pi {ϵ}_{0}r)$. Only potential-energy differences matter; a constant can be added freely. The mechanical energy $K+U$ is conserved if no nonconservative forces do work. A magnetic Lorentz force $q\mathbf{v}\times \mathbf{B}$ does no work on a point charge and generally cannot be represented by a position-only scalar potential energy.

Field of discrete masses. Newton's law of universal gravitation and superposition give the gravitational field away from the source points:It is the force per unit test mass, with units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$, and points toward each positive source mass.

Infinite circular cylindrical rod. Let $b$ be the radius of the circular cross-section, so $A=\pi b^2$, and write $\lambda =\rho A$ for the linear mass density. Choose a coaxial cylindrical Gaussian surface with radius $r>b$ and length $L$. Translation and rotation symmetry make $\mathbf{g}=g_r(r)\hat{\mathbf{r}}$. The end caps have zero field flux because their normals are axial; on the curved surface the outward-normal component is the constant negative value $g_r$. Gauss's law for gravity therefore givesThe gravitational field of an infinite uniform cylinder inside the material instead has enclosed mass $\rho \pi r^{2}L$, giving $g_r=-2\pi G\rho r$. The exterior result assumes a circular cylindrical rod; area by itself would not specify the exact near field of a noncircular cross-section.

In the left panel the dashed circle is a transverse section of the coaxial Gaussian surface; the field points inward and the surface normals outward. Its side area is $2\pi rL$.

For an exterior circular orbit, Newton's second law equates the force required for centripetal acceleration to gravity:The orbital speed is independent of radius and of the orbiting object's mass, for orbits outside the infinite rod.

Finite rod. Taking the supplied field as the gravitational field of a finite uniform line segment, the corresponding balance at radius $R$ isSince $M=2a\rho A=2a\lambda$, its two limits areThe first reproduces the infinite-rod result: a nearby observer sees a very long source. The second is the familiar circular orbit speed about a point mass, since a distant observer cannot resolve the rod's length. The dimensionless curve isIt starts at a constant plateau and decreases smoothly, approaching an $R^{-1/2}$ dependence far away. For $R>0$, $dv/dR=-\sqrt{GM}R/[2(R^2+a^2{)}^{5/4}]<0$, consistent with the sketch.

Derivation of the finite-rod field. Put the source along $-a<z<a$ on the $z$ axis. In its central perpendicular plane, an element $dm=\lambda dz$ is at distance $s=\sqrt{r^2+z^2}$. Its field magnitude is $G\lambda dz/s^2$, and its inward radial component has the extra factor $r/s$. The axial components cancel between $z$ and $-z$. ThusReplacing $2a\lambda$ by $M$ givesThe integral treats the source as a line mass. For a finite cylinder of nonzero cross-section, the displayed finite-rod formula is a thin-rod approximation, not an exact field for all $r$. Its near-rod limit should therefore be read as $b\ll R\ll a$, keeping the orbit outside the material; the apparent line-source singularity at $r=0$ does not describe the interior of a real cylinder.

A tidal force is a difference of gravitational acceleration across an extended region. In Newtonian gravity, $g_i=-{\partial }_i\Phi$. Two neighboring freely falling particles separated by ${\xi }^j$ have, to first order,For a point mass, $\Phi =-GM/r$. In a local orthonormal radial/transverse frame,A small freely falling cloud stretches radially and compresses transversely. Outside the source the three eigenvalues sum to zero, consistently with ${\nabla }^2\Phi =0$; within matter the trace is $-4\pi G\rho$. These are accelerations of separations, not the much larger common acceleration of the cloud's centre.

In general relativity, freely falling test particles follow timelike geodesics. A freely falling frame can eliminate the connection at one event, but not the Riemann curvature tensor throughout a finite laboratory. Let $u$ be the geodesic tangent and $\xi$ a connecting vector for neighboring geodesics with $[u,\xi ]=0$. In the explicitly chosen curvature conventionthe geodesic deviation equation isIt follows by commuting ${\nabla }_u{\nabla }_{\xi }u$, using ${\nabla }_{u}u=0$ and zero torsion. In a parallel-propagated orthonormal frame, its spatial components directly describe measured relative acceleration. The weak-field, slow-motion limit recovers the Newtonian Hessian above. Reversing the curvature sign convention reverses the displayed curvature side, without changing the physical stretching. The equivalence principle removes locally uniform gravity, while geodesic deviation measures the residual tidal field. The connection between parallel transport, curvature and neighboring geodesics is developed systematically in the standard geometric construction.^[ref]

Kepler's third law for two point masses in a Newtonian gravitational field isHere $a$ is the semimajor axis of their relative orbit, $P$ the orbital period and $G$ the gravitational constant. Equivalently $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, to the accuracy with which these astronomical units normalize the solar mass.

A free-fall timescale estimate follows from $R/t^2\sim GM/R^2$:Falling to a central point mass gives the more precise factor $\pi /(2\sqrt{2})$, or $1.77\times 10^3\mathrm{s}$; a distributed star changes an order-one coefficient. For any spherical mass distribution with radius $R$,The inequality is strict for an extended star; equality is approached by a thin shell at $R$. For an monatomic gas the virial theorem is $2U+\Omega =0$, so half the gravitational potential energy released by contraction heats the star and half can be radiated. The Kelvin-Helmholtz cooling time is therefore of orderA uniform-density model gives $3G{M}^2/(10RL)\simeq 9.4\times 10^6\mathrm{yr}$. The given bound fixes a scale, rather than an exact contraction history; both estimates are far shorter than the Sun's main-sequence lifetime.

Kepler's third law for two point masses in a Newtonian gravitational field isHere $a$ is the semimajor axis of their relative orbit, $P$ the orbital period and $G$ the gravitational constant. Equivalently $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, to the accuracy with which these astronomical units normalize the solar mass.

A free-fall timescale estimate follows from $R/t^2\sim GM/R^2$:Falling to a central point mass gives the more precise factor $\pi /(2\sqrt{2})$, or $1.77\times 10^3\mathrm{s}$; a distributed star changes an order-one coefficient. For any spherical mass distribution with radius $R$,The inequality is strict for an extended star; equality is approached by a thin shell at $R$. For an monatomic gas the virial theorem is $2U+\Omega =0$, so half the gravitational potential energy released by contraction heats the star and half can be radiated. The Kelvin-Helmholtz cooling time is therefore of orderA uniform-density model gives $3G{M}^2/(10RL)\simeq 9.4\times 10^6\mathrm{yr}$. The given bound fixes a scale, rather than an exact contraction history; both estimates are far shorter than the Sun's main-sequence lifetime.

Neglect disk self-gravity and take $|z|\ll r$ around a central point mass $M_{\star }$. Its vertical gravitational acceleration isVertical hydrostatic equilibrium is $dP/dz=\rho g_z$. For a vertically isothermal ideal gas, $P=\rho c_s^2$ and $c_s$ is independent of $z$, givingHere $v_K=r\Omega =\sqrt{G{M}_{\star }/r}$, $c_s^2=k_{B}T/(\mu m_p)$ is the isothermal sound speed squared, and $h$ is the Gaussian scale height. A different convention writing $e^{-z^2/h^2}$ changes the quoted height by $\sqrt{2}$ but not the physics.

A point mass at the rotation center has zero perpendicular distance to the axis, so $\boxed{C=\int r_{\perp }^{2}dm=0}$. More generally, moving a fixed mass inward lowers its moment even if its total gravitational monopole and outer radius remain fixed.

A uniform rigid rod of length $l$ and mass $M_r$, pivoted at one end with a point mass $m$ at the other, has pivot moment of inertia $I=l^2(m+M_r/3)$. Its gravitational potential energy is $-gl(m+M_r/2)\cos \theta$. The planar Euler-Lagrange equation is $I\ddot{\theta }+gl(m+M_r/2)\sin \theta =0$, giving the displayed small-amplitude oscillation period. The limit $M_r=0$ gives a simple pendulum, whereas $m=0$ gives the uniform-rod physical pendulum. Thinness does not specify a rod's mass, so both contributions must be stated or explicitly neglected.

A perfectly inelastic collision is an inelastic collision in which the bodies stick and share one outgoing velocity. For two isolated point masses, momentum conservation gives $v=(m_1{u}_1+m_2{u}_2)/(m_1+m_2)$. At fixed total momentum, common velocity minimizes the outgoing translational kinetic energy, so sticking gives the greatest loss of that kinetic energy for the specified incoming state.

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.

For a point mass moving as $z=L\cos \omega t$ along one fixed axis, the trace-free mass quadrupole moment is $m{L}^2{\cos }^2\omega t\mathrm{diag}(-1/3,-1/3,2/3)$. Its radiating component has twice the source angular frequency. The quadrupole formula gives $⟨P⟩=16G{m}^2{L}^4{\omega }^6/(15c^5)$ for this contribution in the leading slow-motion weak-field approximation. Any physical driver or support has its own quadrupole contribution; this expression is for the prescribed mass alone.

For two equal point masses $m$ joined by a light tether of length $l$, the central moment of inertia is $m{l}^2/2$. With conserved angular momentum $L$, the rotational kinetic energy is $L^2/(m{l}^2)$. Slow, smooth contraction gives ${\dot{E}}_{\mathrm{rot}}=-T\dot{l}$ with the displayed leading tension. The exact radial correction is $E_{\mathrm{rad}}=m{\dot{l}}^2/4$ and $T=2L^2/(m{l}^3)-m\ddot{l}/2$, so the total work-energy theorem remains exact for non-quasi-static schedules too.