The transfer is a Hohmann transfer orbit. Write $r_0$ for the original circular Kepler orbit radius and $a_t$ for the transfer ellipse's semi-major axis. The diagram measures both radii from Earth's centre, giving $r_1=1.2r_0$, and henceThe Hohmann transfer time is half the transfer orbital period, taking the satellite to the opposite end of the elliptic orbit. By Kepler's third law, relative to the original geostationary orbit's orbital period $T_0$,Using one sidereal day, $T_0=86164\mathrm{s}$, gives $t=4.9703\times 10^4\mathrm{s}=13.806\mathrm{h}$; the usual $24\mathrm{h}$ approximation gives $13.844\mathrm{h}$. The arrival impulse changes the velocity into that of the new circular Kepler orbit without adding coast time in the impulsive model.

There is a real discrepancy between the wording and the diagram: an increase in altitude above Earth's surface is different from the radius increase drawn. If altitude is interpreted literally, let $R_E$ be Earth's radius. Thenusing $R_E\simeq 6371\mathrm{km}$ and $r_0\simeq 42164\mathrm{km}$. Thus $13.8\mathrm{h}$ answers the explicitly labelled geometry; $13.5\mathrm{h}$ answers the literal altitude wording.

Electromagnetic waves drive mobile charges in a conducting medium. Their current changes both the phase and attenuation of the wave. A description in terms of a real, frequency-independent refractive index is therefore generally insufficient: the material response and the propagation constant are complex and frequency dependent.

Use phasors proportional to $e^{i(kz-\omega t)}$, with $k=k^{\prime }+i{k}^{\prime \prime }$ and $k^{\prime \prime }\ge 0$ for a passive medium. In a homogeneous, isotropic, linear material, let ${ϵ}_b$ represent the permittivity associated with bound-charge electric polarization, $\mu$ the magnetic permeability and $\sigma (\omega )$ the mobile-charge conductivity. The macroscopic Maxwell equations giveFor a transverse plane wave, taking another curl yieldsThis division of the response avoids counting mobile electrons both in ${ϵ}_b$ and in $\sigma$. Define the complex refractive index $\tilde{n}=ck/\omega =n+i\kappa$. For a nonmagnetic medium, ${\tilde{n}}^2={ϵ}_{\mathrm{eff}}/{ϵ}_0$. Its real part determines the phase variation, while the positive optical extinction coefficient $\kappa$ gives amplitude attenuation $e^{-\omega \kappa z/c}$ and intensity attenuation $e^{-2\omega \kappa z/c}$. The amplitude penetration length is

A simple microscopic explanation comes from the Drude model. Let electrons of charge $-e$, mass $m$ and number density $n_e$ move against an effectively stationary positive background. Collisions relax their mean velocity at rate $\nu$:ConsequentlyWith a vacuum-like bound-charge background, this givesIts imaginary part is positive for positive $\omega ,\nu$, consistent with dissipation under the chosen phasor convention. Real materials can require an effective electron mass, a nontrivial bound-charge response, or additional absorption associated with the electronic band structure.

A plasma is a conducting ionized medium with mobile electrons and ions, approximately neutral on large scales because of plasma screening. For the electron-frequency transverse waves considered here, the heavy ions can often be treated as stationary. In a dilute weakly collisional plasma, $\nu \ll \omega$, soFor $\omega >{\omega }_p$, $k$ is real in this collisionless model and the wave propagates. Its phase velocity is $v_{\mathrm{ph}}=c/n>c$, while its group velocity is $v_g=cn<c$; their product is $c^2$. A phase crest exceeding $c$ does not imply superluminal information transport. For $\omega <{\omega }_p$, the refractive-index square is negative andThe transmitted field is evanescent rather than a traveling bulk wave. A sufficiently thick collisionless plasma reflects such radiation without Ohmic absorption. Collisions make the response complex and permit absorption. This accounts for a plasma cutoff and, qualitatively, the reflection of some radio frequencies by ionized atmospheric layers. The simple model describes transverse waves; plasmas also support longitudinal collective modes.

In a metal, positive ions usually form a crystal lattice. Mobile conduction electrons occupy states described by the electronic band structure and normally form a degenerate Fermi gas. The conduction-electron density is usually much higher than in a dilute gas plasma. Thus its plasma frequency is commonly in the optical or ultraviolet range, rather than the lower radio-frequency values possible in dilute plasmas. Electron scattering from the lattice, defects and other excitations gives a finite relaxation rate. The distinction is not that every plasma is collisionless and every metal collisional: the controlling ratio for either material is $\omega /\nu$, together with its carrier density and bound-charge response.

At sufficiently low frequency, $\omega \ll \nu$, the metal's conductivity is approximately real, ${\sigma }_0=n_e{e}^2/(m\nu )$. In a good conductor the conduction current dominates the displacement current, ${\sigma }_0\gg \omega {ϵ}_b$. The wave equation then givesFor example, a tangential electric field varies asThe current $\mathbf{J}={\sigma }_0\mathbf{E}$ is therefore concentrated within a few skin depths of the surface: this is the skin effect. As frequency increases within this regime, $\delta \propto {\omega }^{-1/2}$. In a wire whose radius greatly exceeds $\delta$, only a thin outer region carries appreciable alternating current, increasing the effective resistance. In a conducting shield, the same penetration suppresses the field transmitted through a sufficiently thick wall.

The Poynting theorem identifies the loss as Joule heating. The cycle-averaged heating density for a complex conductivity isThe wave impedance is $Z=\mu \omega /k$. For incidence from vacuum at normal incidence, the electric reflection amplitude is $r=(Z-Z_0)/(Z+Z_0)$, where $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$. A good conductor has small $|Z|$ and therefore strong reflection, while the part of the incident power entering it is dissipated over the skin depth.

At high enough frequency that $\omega \gg \nu$, a metal's free-electron response also approaches the collisionless plasma form. The low-frequency formula $\delta \propto {\omega }^{-1/2}$ must not simply be extrapolated into that regime. Real optical behavior additionally depends on the electronic band structure and transitions between bands. Both plasmas and metals are therefore described by the same electromagnetic response framework, but their relevant densities, scattering rates and frequency regimes differ.

For charge sharing between distant connected spheres, use the standard approximation that the spheres are sufficiently far apart for mutual electrostatic effects to be negligible, and that the wire adds negligible capacitance. The capacitance of an isolated conducting sphere is $C=4\pi {ϵ}_{0}r$. ThusThe initial charges are $Q_i=C_i{V}_i$. Connecting the spheres allows charge to flow until their potentials are equal. Charge conservation givesThe initial and final electrostatic energies are $U_i=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}{C}_2{V}_2^2$ and $U_f=\frac{1}{2}(C_1+C_2)V_f^2$. Their difference becomes Joule heating in the resistive wire:Numerically,The resistance determines how fast this happens, not the final energy difference. Sufficient damping suppresses sustained charge-current oscillations; the calculation also neglects energy radiated away. At finite sphere separation the capacitance matrix would be needed, and the radii and initial potentials alone would not specify an exact correction.

For a fixed magnetic dipole moment in an externally imposed uniform magnetic field, the torque is $\mathbf{\tau }=\mathbf{m}\times \mathbf{B}$. Its component in the direction of increasing $\theta$ is $-mB\sin \theta$. Since ${\tau }_{\theta }=-dU/d\theta$, integration giveswhere the arbitrary additive constant has been chosen so that $U=0$ at $\theta =\pi /2$.

The aligned orientation $\theta =0$ has the minimum $U=-mB$ and is stable against small rotations; the anti-aligned orientation $\theta =\pi$ has the maximum $U=+mB$ and is unstable. The graph repeats every $2\pi$. Rotating quasistatically from alignment to anti-alignment requires external work $\Delta U=2mB$.

Assume a hollow, perfectly conducting rectangular electromagnetic waveguide filled with air approximated by vacuum. Let its wider side be $a=0.040\mathrm{m}$ and its narrower side $b=0.020\mathrm{m}$. The longitudinal propagation constant of a mode isso the waveguide cutoff frequency isFor a mode with transverse electric polarization, one of $m,n$ may vanish, but they cannot both vanish. For a mode with transverse magnetic polarization, both must be positive. A single hollow conducting guide has no transverse electromagnetic mode with zero cutoff.

The lowest cutoff therefore belongs to the ${\mathrm{TE}}_{10}$ mode, varying across the wider side:Strictly, a traveling mode transporting energy down the guide requires $f>f_c$. At exactly $f_c$, $k_z=0$ and the axial group velocity vanishes. Below cutoff the field is evanescent. A different dielectric filling would replace $c$ by the wave speed in that material.

The third law of thermodynamics, in its perfect-crystal form, states that the entropy of a pure, perfectly ordered crystal with a unique ground state tends to zero as $T\to 0$. Its more general entropy-difference formulation says that the entropy difference between equilibrium states related by a finite change of a regular external control parameter tends to zero at absolute zero:The common limiting entropy is independent of $X$. Ground-state degeneracy or frozen disorder can produce residual entropy, so zero entropy is not an unconditional assertion about every sample.

In the usual regular thermodynamic setting, absolute zero cannot be attained from positive temperature by a finite sequence of cooling operations with finite resources. An entropy-temperature plot shows why a finite reversible adiabatic step cannot reach the common zero-temperature endpoint from a state with positive entropy above that endpoint: an isentropic process follows a horizontal line, while all the regular entropy curves meet only at $T=0$.

The figure illustrates this with two possible settings having $S-S_0=a_{1}T$ and $S-S_0=a_{2}T$, where $0<a_1<a_2$. First change the setting isothermally from the upper to the lower curve, rejecting heat; then change it reversibly and adiabatically back to the upper curve. Entropy conservation on the latter step givesfor every finite $n$. The staircase approaches the origin but reaches it only after infinitely many such stages. Irreversibility generates entropy and cannot improve on the reversible endpoint.

This is a thermodynamic illustration of unattainability of absolute zero, not a claim that every cooling apparatus has this particular linear entropy law. Its conclusion presumes regular equilibrium states and bounded control parameters; dynamical statements about finite cooling time additionally require physical assumptions about the available interactions and cooling rates.