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For a satellite of mass $m$ moving in the Newtonian gravitational field of a much larger, effectively fixed mass $M$, put $\mu =GM$. Its conserved mechanical energy, with gravitational potential energy zero at infinity, isHere $r$ is the instantaneous centre-to-centre distance, $v$ the satellite's speed, $G$ the gravitational constant, and $a$ the semi-major axis of its bound elliptic orbit. The kinetic energy and potential energy vary around the orbit, while their sum remains constant. Dividing by $m$ and solving for $v$ gives the vis-viva equation, $v^2=\mu (2/r-1/a)$.

For this Hohmann transfer orbit, the periapsis and apoapsis distances are $r_1=1\mathrm{AU}$ and $r_2=1.52\mathrm{AU}$, soThe vis-viva equation gives the endpoint speedsAs a check, conservation of angular momentum at the two endpoints gives $r_1{v}_{\mathrm{p}}=r_2{v}_{\mathrm{a}}$ because each velocity is perpendicular to its radius. Using $1\mathrm{AU}=1.49598\times 10^{11}\mathrm{m}$ and $\mu =1.32712\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ gives $v_1=29.78\mathrm{km}{\mathrm{s}}^{-1}$ andThe departure impulse must increase the speed in the direction of the initial circular orbit:These are Sun-centred speeds; leaving the Earth's own gravitational field would require an additional calculation beyond this idealized solar orbit transfer.

At $r_2$, the desired circular orbit speed is $v_2=\sqrt{\mu /r_2}=24.16\mathrm{km}{\mathrm{s}}^{-1}$. The arrival impulse is thereforeThe required thrust on the spacecraft is forward, tangent to its orbit, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's velocity; this distinguishes the direction of thrust from the direction in which the exhaust is fired.

The Hohmann transfer time is half the orbital period of the transfer ellipse. From Kepler's third law, relative to the Earth's one-year orbital period,Meanwhile Mars's orbital period is $T_M=1.52^{3/2}\mathrm{yr}=1.8740\mathrm{yr}$. The spacecraft arrives $\pi$ radians from its departure position. For a rendezvous, Mars must reach that same position after advancing through $2\pi t_H/T_M$. Thus its initial lead angle over the departing spacecraft isMars must be ahead by this angle in the common direction of motion around their orbits. At departure the spacecraft and the Earth share the inner orbital position; arrival occurs on the opposite side of the Sun.

Let $D$ be the separation of the Earth and Moon, and let $x$ be the spacecraft's distance from the Earth's centre towards the Moon. On $0<x<D$, the two gravitational forces oppose each other. Their equality, by Newton's law of universal gravitation, isTaking the positive square root givesThe centre of mass of the two bodies is at $x_G=D{M}_M/(M_E+M_M)$ from the Earth's centre. Thus the gravitational field null between two point masses lies a distancefrom their centre of mass, on the Moon's side. With $M_E/M_M=81$, $x=9D/10$ and $x_G=D/82$, soThe point is $3.84\times 10^7\mathrm{m}$ from the Moon. It is an instantaneous zero of the physical gravitational field, not a co-rotating Lagrange point: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero centripetal acceleration. With zero net force it cannot follow that circular co-rotation.

A gravitational assist changes a spacecraft's velocity relative to the Sun by scattering it in the gravitational field of a moving planet. In the planet's rest frame, ignore other forces during the encounter and take the spacecraft mass negligible. The incoming and outgoing asymptotic velocities have equal magnitude,because the encounter is conservative, but their directions differ along a hyperbolic orbit. If the planet has approximately constant heliocentric velocity ${\mathbf{V}}_p$, the heliocentric spacecraft velocities are ${\mathbf{v}}_{\mathrm{in},\mathrm{out}}={\mathbf{V}}_p+{\mathbf{u}}_{\mathrm{in},\mathrm{out}}$. Its specific energy change isThus a flyby can increase or decrease heliocentric energy depending on its geometry. The planet exchanges momentum and orbital energy with the spacecraft; its speed change is extremely small because its mass is much greater. There is no energy creation and no increase of asymptotic speed in the planet's own frame.

For a point-mass planet, conservation of energy and angular momentum gives the hyperbolic eccentricity and deflection anglewhere $r_p$ is periapsis distance and $b$ the impact parameter. The velocity-vector change has magnitude $2u_{\infty }\sin (\delta /2)$, so the maximum energy-change magnitude for this fixed deflection is $2V_p{u}_{\infty }\sin (\delta /2)$. A small periapsis increases the turning angle, but collision, atmosphere and tidal forces limit how close a real spacecraft may pass. Multiple encounters allow large changes without expending the propellant that equivalent direct velocity changes would require.

A Hohmann transfer orbit connects two coplanar circular orbits using two tangential, effectively instantaneous velocity changes. Take outward transfer radii $r_2>r_1$, and write ${\mu }_g=GM$ for the central body's gravitational parameter. The intervening ellipse has periapsis $r_1$, apoapsis $r_2$, semimajor axis $a_t=(r_1+r_2)/2$ and orbital eccentricity $e=(r_2-r_1)/(r_1+r_2)$.

The circular speed is $\sqrt{{\mu }_g/r}$. Conservation of orbital energy gives the vis-viva equation $v^2={\mu }_g(2/r-1/a_t)$. Therefore the transfer speeds at the two impulses areThe first prograde impulse raises the far side of the orbit; the second prograde impulse raises the near side to circularize at $r_2$. Their magnitudes areThe transfer takes half an ellipse period. Kepler's third law givesFor inward transfer the same ellipse is traversed from apoapsis to periapsis and both burns are retrograde. A rendezvous also needs the correct initial phase of the target: for outward transfer its initial angular lead is $\pi -\sqrt{{\mu }_g/r_2^3}{t}_t$ modulo $2\pi$. The method assumes a two-body gravitational field, negligible atmospheric drag and burns short compared with an orbit. It is an economical two-impulse transfer between these circles; it is not universally the least-$\Delta v$ transfer if additional impulses and very large intermediate radii are permitted.

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.

For uniform Earth density, the shell theorem says that only mass within radius $r$ contributes to the gravitational field there. Since enclosed mass is proportional to $r^3$,At depth $h$, $g_h/g_0=1-h/R$. A simple pendulum has small-amplitude period $P=2\pi \sqrt{l/g}$, so the pendulum clock below a uniform-density surface runs at the fraction $P_0/P_h=\sqrt{1-h/R}$ of its surface-calibrated rate. After a surface-clock hour, its lag isThe mine clock is slower because gravity is weaker under the stated uniform-density assumption. The expansion $\Delta t\simeq 3600h/(2R)$ gives the same result. This is a change in the mechanical pendulum period, not a relativistic clock effect; a realistic Earth density profile would require a different interior-gravity calculation.

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.

The tunnel is taken thin enough that removal of its material does not significantly disturb spherical symmetry. For $a\le r\le 2a$, the enclosed mass isBy the spherical shell theorem, the inward gravitational acceleration has magnitudeThe released gravitational potential energy is the positive work done by gravity while the mass falls from radius $2a$ to $a$:By conservation of mechanical energy,Inside the cavity, the spherical shell theorem gives zero gravitational field, so the velocity stays constant. The original figure places B and C at opposite ends of a cavity diameter, a distance $2a$ apart. ThereforeThe quantity $G\rho$ has units ${\mathrm{s}}^{-2}$, so this is a time. There are no numerical values for $a$ or $\rho$ in the paper. The transit time is independent of $a$ and the falling mass $m$. The cavity's gravitational potential is constant, not necessarily zero; only its gradient vanishes. The calculation is a hollow-sphere gravitational transit.

In general relativity, light follows null geodesics. A weak gravitational field outside an isolated spherical mass gives the Schwarzschild light deflectionfor impact parameter $b\gg GM/c^2$. At the solar limb this is approximately $1.75$ arcseconds. A Newtonian corpuscular calculation gives half this value; the full result depends on spatial curvature as well as the gravitational time dilation. The parametrized post-Newtonian formalism predicts $\alpha =2(1+{\gamma }_{\mathrm{PPN}})GM/(b{c}^2)$, with ${\gamma }_{\mathrm{PPN}}=1$ in general relativity.

The Eddington experiment compared photographic star positions near the eclipsed Sun with reference exposures taken without the Sun in that field. Rays bend toward the Sun, but their backward extrapolations place the apparent stellar images farther from it. Plate scale, focus changes, atmospheric refraction and comparison-field alignment must be separated from a radial displacement proportional to $1/b$. The original report discusses the instruments, plate reductions and competing predicted deflections; the test is a quantitative comparison with uncertainty, not a unique proof of a gravitational theory.^[ref]

radio interferometry permits an independent test without an eclipse. A 2005 Very Long Baseline Array experiment observed four compact sources at 43, 23 and 15 GHz. Differential astrometry suppressed common antenna errors, while a fit $\mathbf{p}(\nu )={\mathbf{p}}_0+\mathbf{I}{\nu }^{-2}$ separated refraction in the solar corona from achromatic gravitational deflection. Tropospheric errors and radio-source structure also required calibration. The reported ${\gamma }_{\mathrm{PPN}}=0.9998\pm 0.0003$ was consistent with general relativity.^[ref]

The three classic tests probe orbital motion, the propagation of light, and clock rates in a gravitational field.

Schwarzschild perihelion precession. For an orbit with semi-major axis $a$ and orbital eccentricity $e$, the leading advance per revolution isThe relevant observation for Mercury is the residual advance after the much larger ordinary perturbations have been included, approximately $43$ arcseconds per century.^[ref] An orbital fit must account for the other planets and reference-frame effects before interpreting that residual as a relativistic contribution.

Schwarzschild light deflection. A ray with large impact parameter $b$ passing a nearly spherical mass is deflected throughAt the Sun's limb this is about $1.75$ arcseconds. The Eddington experiment used the 1919 solar eclipse to photograph background stars near the obscured Sun, then compared their positions with reference photographs. The measured effect is an apparent outward displacement of the background stars, not a direct photograph of a curved light path. Telescope focus, plate scale and atmospheric effects must be controlled when extracting such small shifts.^[ref]

Gravitational redshift. For a photon emitted lower in a weak static field and received a height $H$ above it,The Pound-Rebka experiment used the narrow recoil-free nuclear resonance supplied by the Mössbauer effect. A controlled source motion supplies a small Doppler shift that compensates the gravitational shift, allowing a change in resonant absorption to reveal it across a laboratory tower.^[ref] The leading weak-field redshift follows from the equivalence principle and therefore does not, by itself, distinguish every competing metric theory of gravity.

Local equivalence and tides. In Newtonian mechanics, inertial mass $m_{\mathrm{i}}$ measures the response to a force, whereas passive gravitational mass $m_{\mathrm{g}}$ couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.

The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.

In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing $\mathbf{g}=-GM\mathbf{r}/r^3$ across a small separation $\mathbf{\xi }$ gives the tidal tensor predictionA radial separation stretches with coefficient $2GM/r^3$, whereas a transverse separation is compressed with coefficient $-GM/r^3$. Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.

The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere $D/d\tau$ is the covariant derivative along the reference geodesic, and $U^{\mu }$ its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to ${\ddot{\xi }}^i=-{\partial }_i{\partial }_j\Phi {\xi }^j$, recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.

Weak-field connection and potential. Take $x^0=ct$ and the $+---$ metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be $g_{\nu 0}$, not $g_{\mu 0}$. The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ and treat both $h$ and its derivatives as first order. To that order $g^{00}=1-h_{00}$, $g^{0i}=O(h)$, and $g^{ij}=-{\delta }^{ij}+O(h)$. A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that $dt/d\tau$ is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel $(dt/d\tau {)}^2$:Comparison with the Newtonian equation $d^2\mathbf{x}/d{t}^2=-\nabla \Phi$ identifies $h_{00}=2\Phi /c^2+C$. For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving $C=0$. ThusThe quantity $\Phi$ is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$. For a point source, $\Phi =-GM/r$. A static clock has $d\tau =\sqrt{g_{00}}dt\approx (1+\Phi /c^2)dt$, connecting the Newtonian limit with gravitational time dilation.

Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of $R_S/r$ possible.

Put $f=1-R_S/r$ and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere $v$ is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has $v^2=GM/r+O((GM/r{)}^2/c^2)$ in the weak-field limit, so $v^2/c^2=R_S/(2r)+O((R_S/r{)}^2)$. Expanding both factors,Solving for the emitted frequency givesThe gravitational part is $R_S/(2r)$ and the transverse kinematic time dilation contributes another $R_S/(4r)$; the absence of line-of-sight motion does not remove this second effect.

As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies $v^2/c^2=\mu /(r-2\mu )$, where $\mu =GM/c^2=R_S/2$. Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead ${\nu }_{\mathrm{em}}\approx {\nu }_{\mathrm{ob}}[1+GM/(r{c}^2)+v^2/(2c^2)]$, with $v$ supplied independently.

Treat the gravitational field as approximately spherical. The circular orbit condition givesIf every star had a representative mass $m$ and obeyed $L/L_{\odot }=(m/M_{\odot }{)}^3$, its mass-to-light ratio would be $(m/M_{\odot }{)}^{-2}$, giving $\boxed{m\simeq 0.47M_{\odot }}$. This is a one-mass estimate, not an actual average inferred uniquely from the data: a galactic stellar mass function, stellar remnants, dust and nonstellar mass alter the relation. The flattening of the galactic disk also changes the spherical conversion by an order-one factor.

Treat the gravitational field as approximately spherical. The circular orbit condition givesIf every star had a representative mass $m$ and obeyed $L/L_{\odot }=(m/M_{\odot }{)}^3$, its mass-to-light ratio would be $(m/M_{\odot }{)}^{-2}$, giving $\boxed{m\simeq 0.47M_{\odot }}$. This is a one-mass estimate, not an actual average inferred uniquely from the data: a galactic stellar mass function, stellar remnants, dust and nonstellar mass alter the relation. The flattening of the galactic disk also changes the spherical conversion by an order-one factor.

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.

Rough elastic collisions can exchange rotational and translational kinetic energy. For three-dimensional rigid balls with all three rotations thermalized, these six quadratic terms contribute $3k_{B}T$ per particle. A uniform gravitational height distribution adds $k_{B}T$, giving total mechanical mean energy $4k_{B}T$.

Extra propagation time of a light or gravitational-wave signal through a gravitational field compared with an appropriate flat-space reference. In a weak static field with Newtonian potential of a point mass ${\Phi }_N<0$, the leading delay along an approximately straight ray is $\Delta t=-2\int {\Phi }_{N}dl/c^3>0$. The local speed of light remains $c$; the integrated geometry changes the relation between emission and reception times.

Model the assembly as a heavy symmetric top with a fixed frictionless support, a uniform gravitational field, and centre of mass on its symmetry axis. Assume constant spin, negligible damping, and initial conditions prepared for uniform precession of a heavy symmetric top. Large spin means the transverse angular momentum from precession is small: $I_1|\Omega |\sin \theta \ll |I_3{\omega }_3|$. Also require $|\Omega |\ll |{\omega }_3|$. A large spin rate alone does not select steady motion after an arbitrary release.

For an original illustrative calculation, stipulateThe transverse principal moments of inertia $I_1,I_2$ are about the pivot, including the parallel axis theorem contribution. These inputs giveThe ratio ${\Omega }_0/{\omega }_3=0.00368$ is small; the more direct angular momentum ratio $I_1{\Omega }_0\sin 60^{\circ }/L_s=0.00358$ is small too. These are model calculations, not experimental results.

Check the approximation against the existing exact uniform precession of a heavy symmetric top condition. Its total angular momentum isAt constant $\theta ,\Omega ,L_s$, differentiating this expression and equating it to the gravitational torque givesFor $L_s>0$ the slow solution can be evaluated without subtracting two nearly equal numbers:It exceeds ${\Omega }_0$ by about $0.208\%$. The other solution is about $177.409rad/s$ and violates the assumed slow precession; the quadratic's two roots must not be conflated. Its discriminant must be nonnegative for these steady solutions to exist.

The original curves hold $I_1,I_3,\theta$ fixed and vary spin and the imposed torque coefficient $K$ in ${\mathbf{\tau }}_O=K\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. For gravity alone, $K=Mgd$; the other curves are hypothetical changes in the transverse torque, not observations of this wheel. Solid curves use the slow quadratic root; dashed curves use $K/(I_3{\omega }_3)$. The second panel displays their relative difference. Increasing spin improves this approximation for the parameters shown. It does not remove nutation caused by inappropriate initial conditions.