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A prolate spheroid is an ellipsoid with two equal shorter semiaxes and a longer symmetry-axis semiaxis. Its meridional section is an ellipse rotated about its major axis. In the coordinates $R^2/(\lambda +\alpha )+z^2/(\lambda +\beta )=1$ with $\beta >\alpha$, a fixed regular $\lambda$ has equatorial radius $\sqrt{\lambda +\alpha }$ and polar radius $\sqrt{\lambda +\beta }$; its focal separation is independent of $\lambda$, giving a confocal family.

For a two-dimensional symmetric positive-definite matrix $M$ and $R>0$, the region ${\mathbf{x}}^{T}M\mathbf{x}\le R$ is a filled ellipse of area $\pi R/\sqrt{\det M}$. By the spectral theorem for real symmetric matrices, an orthogonal change of coordinates gives positive eigenvalues ${\lambda }_1,{\lambda }_2$ and semiaxes $\sqrt{R/{\lambda }_1}$ and $\sqrt{R/{\lambda }_2}$. Their product and the area of an ellipse prove the formula. Translating the center of an ellipse leaves the area unchanged.

The real case of the finite-dimensional spectral theorem says that a self-adjoint operator on a finite-dimensional real inner product space has a real orthonormal eigenbasis. In orthonormal coordinates this is the spectral theorem for real symmetric matrices, $A=Q\Lambda Q^T$. The worked ellipse under finite-dimensional spectral theorem shows why both orthogonal diagonalization and orthogonality matter: the new coordinates separate the quadratic form into squares without distorting lengths or angles.

A real normal matrix need not have real eigenvalues. For example, the quarter-turn $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ is an orthogonal matrix, but its characteristic polynomial is $z^2+1$ and its eigenvalues are $i,-i$. Over the complex numbers it has an orthonormal eigenbasis; over the real numbers it has no eigenvector. Self-adjointness supplies the reality conclusion in the real spectral theorem.

An ellipse is a constant sum of distances to its foci. Its outward normal vector is parallel to the sum of the two unit vectors pointing from the foci to a boundary point. The tangent is therefore equally inclined to the two focal lines. Reflecting an incident unit ray from one focus across this normal gives the direction toward the other focus. Every such one-reflection path has length $2a$, giving equal travel time in a homogeneous medium. Finite-wavelength wave focusing is limited by diffraction.

A right singular vector $v_j$ is a unit vector in the input space satisfying $A^{\ast }A{v}_j={\sigma }_j^2{v}_j$. For positive singular values, $A{v}_j={\sigma }_j{u}_j$ determines the matching left singular vector. In the ellipse example, $A{v}_1=(5,0{)}^T$ is not parallel to $v_1$: a right singular vector need not be an eigenvector of $A$. Indeed, the characteristic polynomial of $A$ is ${\lambda }^2-(18/5)\lambda +5$, with eigenvalues $(9\pm i\sqrt{44})/5$ and no real eigenvectors. Nevertheless, its real right singular vectors give a complete orthonormal basis. Eigenvectors describe preserved directions; right singular vectors describe orthogonal input channels of stretching.^[ref]

An arbitrary input mixes these channels. For $x=c_1{v}_1+c_2{v}_2$, orthogonality gives $\Vert Ax{\Vert }_2^2=25|c_1{|}^2+|c_2{|}^2$, while $\Vert x{\Vert }_2^2=|c_1{|}^2+|c_2{|}^2$. Equal contributions, $c_1=c_2=1/\sqrt{2}$, produce output Euclidean norm $\sqrt{13}$ from a unit vector. Concentrating the entire input along $v_1$ produces the maximum $5$. This is why the right singular vectors give a direct reading of how any input will behave. For a compact operator, the same definition describes the positive terms of its singular value system.

The distance from a focus to a conic section along a line perpendicular to its symmetry axis. In focus-centred polar coordinates, an ellipse may be written $r(1-e\cos \theta )=\ell$, where $e$ is its eccentricity of a conic.

The distance from a focus of a conic section to the curve measured perpendicular to its principal axis. In the focus-based polar coordinates equation $r=p/(1+e\cos \varphi )$ it is the length $p$, where $e$ is the eccentricity. For an ellipse, $p=a(1-e^2)=b^2/a$. For a Newtonian Kepler orbit, $p=j^2/(GM)$ with $j$ the specific angular momentum.

The semi-major and semi-minor axes are the larger and smaller half-axis lengths of an ellipse. In principal-axis coordinates the ellipse has equation $x^2/a^2+y^2/b^2=1$ with $a\ge b>0$. The same terminology is used for a hyperbola $x^2/a^2-y^2/b^2=1$, where $a$ is half its transverse axis and $b$ half its conjugate axis. Orbital semiaxes apply these geometric quantities to conic sections.

Signed swept area counts the area traversed by a position vector, positively or negatively according to its direction of turning. For a continuously differentiable planar trajectory, its accumulated value relative to $t_0$ is the line integralA short segment contributes the oriented triangle area $\frac{1}{2}\mathbf{n}\cdot (\mathbf{r}\times d\mathbf{r})$; taking its time derivative gives areal velocity. Repeated turns count repeated area; a reversal subtracts it. Thus this quantity need not equal the area of the union of all previously swept regions. The origin matters: the relevant origin for Kepler's second law is the attracting focus of the ellipse, rather than its geometric centre.

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

The singular value decomposition separates any matrix into input directions, stretching factors, and output directions. Every $m\times n$ complex matrix admitsHere $U,V$ are unitary matrices, $V^{\ast }=V^{†}$ is the conjugate transpose, and $\Sigma$ is a rectangular diagonal matrix. Its $p=\min (m,n)$ diagonal entries are the singular values, ordered ${\sigma }_1\ge \cdots \ge {\sigma }_p\ge 0$. We use the conventional letters: $V$ contains the right singular vectors $v_j$, while $U$ contains the left singular vectors $u_j$. Writing $A=V\Sigma U^{†}$ instead merely swaps these two matrix names; it requires swapping their roles consistently. For real matrices, choose orthogonal matrices and replace the conjugate transpose with the transpose.^[ref]

Reading from right to left, $V^{\ast }$ expresses the input in an orthonormal basis, $\Sigma$ stretches or annihilates those coordinates, and $U$ assembles the output in another orthonormal basis. Unlike matrix similarity, the input and output bases can differ, and even have different sizes. A $3\times 2$ matrix has full factors of sizes $3\times 3$, $3\times 2$, and $2\times 2$. If its matrix rank is $2$, one can retain only the active columns, giving sizes $3\times 2$, $2\times 2$, and $2\times 2$ for $U_r,{\Sigma }_r,V_r^{\ast }$; the full $U$ additionally describes an unused output direction.

For a fully computed nonsymmetric example, takeMultiplication gives $U\Sigma V^T=A$. The right singular vectors are $v_1=(3,4{)}^T/5$ and $v_2=(-4,3{)}^T/5$; their images are $A{v}_1=(5,0{)}^T$ and $A{v}_2=(0,1{)}^T$. Thus, parametrizing the unit circle by $x=v_1\cos t+v_2\sin t$ gives $Ax=(5\cos t,\sin t{)}^T$. Its image is the ellipse $y_1^2/25+y_2^2=1$. The two singular values are its semiaxis lengths, and the left singular vectors point along its axes. The singular value decomposition finds precisely the perpendicular input directions whose images stay perpendicular.

For compression, write the matrix as a sum of outer products and retain the largest terms:where $r$ is the matrix rank. Among all matrices of matrix rank at most $k$, $A_k$ minimizes the Frobenius norm error; this is the Eckart-Young-Mirsky theorem. The operator norm error is ${\sigma }_{k+1}$ for $k<r$. These statements concern specified norms, rather than guaranteeing that discarded details are visually or scientifically unimportant.^[ref]

Consider an $8\times 10$ brightness matrix $B=p{q}^T+a{b}^T/2$, withand $a_i=(-1{)}^{i-1}$, $b_j=(-1{)}^{j-1}$. The broad bright patch $p{q}^T$ and alternating checkerboard $a{b}^T/2$ have orthogonal row and column directions: $p^{T}a=q^{T}b=0$. Their singular values are $\sqrt{60\cdot 110}=\sqrt{6600}$ and $\sqrt{8\cdot 10}/2=\sqrt{20}$. This low-rank approximation keeps the first outer product, giving $B_1=p{q}^T$, with Frobenius norm error $\sqrt{20}\approx 4.472$ and relative error $\sqrt{20/6620}\approx 5.50\%$. For example, $B_{11}=1.5$ becomes $1$, and $B_{12}=1.5$ becomes $2$: every entry changes by $0.5$ in magnitude. The discarded pattern is small but entirely lost.

For this illustrative image compression, storing one normalized left singular vector, one normalized right singular vector, and one singular value requires $8+10+1=19$ scalars instead of $80$. This counts scalars at equal precision, before storage overhead. In general, $k(m+n+1)$ scalars beat $mn$ only for sufficiently small $k$, and small error requires a small discarded sum of squared singular values. The usefulness comes from measuring the tradeoff before throwing information away.^[ref]

In the circular-Earth-orbit, unpowered patched conic approximation, the least Earth-relative excess speed for direct solar-surface contact lowers the heliocentric tangential speed to that of an ellipse with apocentre $a$ and pericentre $R_{\odot }$. The vis-viva equation gives $v_a=v_E\sqrt{2R_{\odot }/(a+R_{\odot })}$, so the retrograde excess magnitude is $v_E-v_a$ and the terrestrial surface launch speed is the title expression. Cancelling all heliocentric angular momentum instead produces radial fall to a point Sun and the larger estimate $\sqrt{v_{e,E}^2+v_E^2}$. Rotation, planetary gravity assists and atmospheric drag are outside this direct-flight model.

A spheroid is an ellipsoid of revolution obtained by rotating an ellipse about a principal axis. Rotation about the shorter axis gives an oblate spheroid; rotation about the longer axis gives a prolate spheroid. Thus the oblate form is a member of the general family.

The Steiner inellipse of a nondegenerate triangle is the unique ellipse of largest area contained in the triangle. Its center is the centroid, and it touches each side at its midpoint.

For $0<a<1$, the Joukowski map $z=\zeta +a^2/\zeta$ maps the exterior unit circle to an ellipse of horizontal and vertical semiaxes $1+a^2$ and $1-a^2$. The Milne-Thomson circle theorem gives $W(\zeta )=U(e^{-i\alpha }\zeta +e^{i\alpha }/\zeta )$ for a uniform flow inclined by $\alpha$. The chain rule produces the displayed complex velocity in the physical plane. Since $z\sim \zeta$, the far-field flow retains its speed and direction. Its two stagnation points areobtained from $\zeta =\pm e^{i\alpha }$. The map derivative is nonzero on and outside the unit circle, so these zeros are preserved without introducing a boundary singularity.

The University of New South Wales documents an undriven, visitor-started Foucault pendulum. A fixed vertical wire and wall line define its initial north–south swing plane. The documented comparison one hour later shows reduced amplitude and about nine degrees of anticlockwise precession. The apparatus swings for several hours without an electromagnetic drive.^[ref]

To turn that demonstration into a quantitative experiment, the following is a proposed measurement protocol. Establish a rigid support with no preferred horizontal swing direction, measure the wire length to the bob's centre, and mark two perpendicular horizontal axes fixed to the building. Start with a small displacement along the sighting plane, then release with as little transverse velocity as possible. Record the actual launch method; burning a fine restraining thread is one possible proposed release, not a claim about the documented UNSW launch. Keep people clear of the bob and avoid touching the suspension during the observation.

Record the horizontal position from above with a stationary, calibrated camera, or sight the bob against the wire and wall line at recorded times. Fit the long axis of each short group of swings to estimate its orientation $\psi$; use the extrema on both sides rather than following the bob's rapidly reversing velocity. A swing line has orientation modulo $\pi$, so unwrap successive small direction changes continuously. Rotate no camera or reference between measurements. Record the fast oscillation period separately, and compare the slope of $\psi (t)$ with the Latitude law of Foucault precession. Repeat launches along different directions and monitor ellipses, pivot motion, and air currents before attributing the slow change to Earth's rotation.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.

A symmetric positive-definite matrix $M$ defines an inner product on real coordinate vectors byFor diagonal $M$, positive diagonal entries weight the squared coordinate errors. This is the finite-coordinate version of a weighted inner product; the positive-definite matrix example appears in Francis Narcowich, Inner Products and Norms, standard inner products, example 8.

For $M=\mathrm{diag}(4,1)$, the squared norm of $(a,b{)}^T$ is $4a^2+b^2$. A first-coordinate error of one costs as much as a second-coordinate error of two. The unit sphere is the ellipse $4a^2+b^2=1$, and orthogonality means $4u_1{v}_1+u_2{v}_2=0$. Thus “closest” and “perpendicular” depend on the chosen inner product. The orthogonal projection residual example shows how this changes a best approximation.