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For a lossless ideal Hertzian dipole driven by an incident wave, use its induced open-circuit voltage and its feed impedance $R_{\mathrm{rad}}+iX$. A conjugately matched load has impedance $R_{\mathrm{rad}}-iX$. The same induced current therefore delivers equal power to the load and to radiation resistance: $P_{\mathrm{sca}}=P_{\mathrm{load}}$. The scattered wave has the usual dipole angular pattern. In direction of gain $G$, its scattering cross-section is $G{A}_{\mathrm{eff}}$, since the load receives $S_{\mathrm{inc}}{A}_{\mathrm{eff}}$.

This ideal single-mode scattering is not the same as a signal that travels through an already matched antenna port, reflects from a mismatched terminating load, and returns. A zero port reflection coefficient removes that latter echo but does not make the physical dipole invisible to the incident field. Additional structural scattering, ohmic loss or a different load changes the result.

A Hertzian dipole is an ideal localized oscillating source much smaller than its radiation wavelength. Its electric version is an infinitesimal current element; its magnetic version is a small current loop with moment $\mathbf{m}=IA\mathbf{n}$. Such elements model the far-field radiation and radiation resistance of electrically small antennas. The approximation requires source dimensions much smaller than $c/\omega$.

The supplied magnetic field has a quasistatic $[m]/r^3$ term, an induction $[\dot{m}]/(c{r}^2)$ term and a transverse radiation $[\ddot{m}]/(c^{2}r)$ term. The electric field has induction and radiation terms. Brackets mean evaluation at the retarded time $t_r=t-r/c$, not a time average. Near terms mainly exchange stored energy; only the $r^{-1}$ fields give a nonvanishing outgoing power integral as $r\to \infty$.

For a fixed dipole axis the radiation fields satisfy $E_{\varphi }=-c{B}_{\theta }$. The radial Poynting flux is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ yieldsFor a general changing dipole direction, replace the squared scalar acceleration by $|\ddot{\mathbf{m}}{|}^2$.

Radiation resistance is the equivalent input resistance whose mean power equals radiated power, not ohmic heating. For a fixed small loop carrying $I(t)=I_0\cos \omega t$, $m=AI$, soThis uses peak current $I_0$; with rms current the power is $I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$.

The satellite loop needs its changing orientation as well as its induced current. Put the spin axis along z, the external field along x, and let $\mathbf{n}(t)=(\cos \omega t,\sin \omega t,0)$. With $A=\pi b^2$ and resistance $R=10^{-4}\Omega$, the resistance-only approximation givesWrite $m_0=B{A}^2\omega /R$. The full magnetic moment isIt has a static component and a rotating component at $2\omega$. Consequently $|\ddot{\mathbf{m}}{|}^2=4m_0^2{\omega }^4$, andHolding the dipole axis fixed while inserting the sinusoidal current would instead give $B^2{A}^4{\omega }^6/(12\pi {ϵ}_0{c}^5{R}^2)$, eight times smaller. That is the fixed-axis-loop result; it does not describe the stated rotating loop. This is frequency doubling of an induction-driven rotating loop; the changing vector direction is essential.

The resistance-only model is conditional: loop inductance is not supplied. If self-inductance L matters, solve $L\dot{I}+RI=BA\omega \sin \omega t$. In the steady state $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$. Its phase changes the static moment direction but not the twice-frequency component's magnitude, so the power becomesThe first expression is the requested resistive idealization and an upper estimate within this passive RL correction. Radiation reaction is negligible compared with the stipulated ohmic resistance. The radiated energy comes from rotational kinetic energy: the induced magnetic torque brakes the loop. Most lost rotational energy is Joule heating; a small part escapes as radiation. Maintaining a constant spin would require external mechanical power.

For scale only, take $B=50\mu \mathrm{T}$, a representative near-Earth field rather than a supplied orbital measurement. The terrestrial surface range reported by NOAA is roughly $25$–$65\mu \mathrm{T}$; the actual orbital value depends on position.^[ref] With $b=1\mathrm{m}$ and $\omega =2\pi (500{\mathrm{s}}^{-1})$, the resistance-only model givesInductance would reduce this power. The kilohertz signal is exceptionally weak and a metre-scale receiving antenna is electrically tiny. Ionospheric absorption/reflection and strong natural or instrumental low-frequency backgrounds make ordinary ground reception impractical. This is not a theorem that the ionosphere blocks every low-frequency mode: field-guided propagation and condition-dependent leakage are real, as the VIPER mission's measurements were designed to test.^[ref]

At a separation $r=100\mathrm{km}$, $k_{\mathrm{rad}}r\simeq 2.1$, so the observer is not securely in the far field. The radiation term alone has at most the characteristic magnetic amplitudein this optimistic resistance-only vacuum model. Near and induction fields must be retained for a full prediction; a simple far-field inverse-square power estimate is not an exact received signal here. Another satellite avoids the whole ground ionospheric path, but plasma propagation, receiving sensitivity, bandwidth, orientation, integration time and spacecraft noise remain unspecified. Detection is not forbidden in principle, but routine practical detection is implausible at this scale; a definitive instrument-level yes/no does not follow from the supplied data.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]

The receiving dipole remains matched. To model its actual reradiation, let its feed impedance be $R_{\mathrm{rad}}+iX$ with no ohmic loss, and its matched load be $R_{\mathrm{rad}}-iX$. The induced current dissipates equal powers in the load and in radiation resistance. Thus matched-load scattering of a Hertzian dipole has $P_{\mathrm{sca}}=P_2$, even though its port is matched.

The scattered wave has the same dipole gain toward the original antenna. Applying the reciprocal link factor a second time gives the reciprocal Hertzian-dipole echo:Each leg supplies an inverse-square spreading factor. This is the carrier-averaged power during a returned pulse with the stated transmitted power during its pulse. The pulse-train average additionally depends on duty cycle, which is not given. Treating the pulses as narrowband keeps the same gain and wavelength approximation.

A matched feed does not make a physical dipole a perfect absorber with zero external scattering. A model retaining only an antenna load-mismatch echo would set its reflected travelling-port wave to zero and would not determine this reradiation. Conversely, an arbitrary antenna structure can add scattering not fixed by gain and effective area alone. The numerical answer uses the ideal lossless Hertzian-dipole scattering model just stated.

Radiation resistance represents emitted electromagnetic power as a real input resistance of an antenna. It does not describe conversion to heat inside the conductor. If peak current amplitude is used, $P_{\mathrm{rad}}=|I_0{|}^2{R}_{\mathrm{rad}}/2$. Real ohmic loss adds a separate resistance and lowers radiation efficiency. For an ideal short uniform-current element of length $d$, the Hertzian dipole gives $R_{\mathrm{rad}}=80{\pi }^2(d/\lambda {)}^2$ in ohms, using the usual vacuum impedance approximation $120\pi \Omega$.