Squaring the states gives $\boxed{p_0=e^{-x^2/{\sigma }^2}/(\sqrt{\pi }\sigma ),p_1=2x^2{e}^{-x^2/{\sigma }^2}/(\sqrt{\pi }{\sigma }^3)}$. The first peaks at zero; the second has a node there and peaks at $x=\pm \sigma$. For a centered interval of width $\delta x\ll \sigma$, $P_0\simeq \delta x/(\sqrt{\pi }\sigma )$, whereas $P_1\simeq (\delta x{)}^3/(6\sqrt{\pi }{\sigma }^3)$. Its leading linear term vanishes, but the probability in a finite interval is not exactly zero.

For the Gaussian, ${\varphi }_0^{\prime \prime }=(x^2/{\sigma }^4-1/{\sigma }^2){\varphi }_0$. Using $\hslash /(m\omega )={\sigma }^2$, the $x^2$ terms cancel and leave $E_0=\hslash \omega /2$. For ${\varphi }_1\propto x{\varphi }_0$, differentiation gives ${\varphi }_1^{\prime \prime }=(x^2/{\sigma }^4-3/{\sigma }^2){\varphi }_1$, and the same cancellation gives $\boxed{E_1=3\hslash \omega /2}$. The normalization constants do not affect these eigenvalue equations.

Integrating $F=-kx$ gives $V=k{x}^2/2$ up to a constant. Thus $\boxed{-{\hslash }^2{\varphi }^{\prime \prime }/(2m)+m{\omega }^2{x}^2\varphi /2=E\varphi }$, with $\omega =\sqrt{k/m}$. The potential is a symmetric upward parabola and tends to infinity in both directions; normalizable states decay at large $|x|$.

The equation is $i\hslash {\partial }_t\psi =[-{\hslash }^2{\partial }_x^2/(2m)+V(x)]\psi$. The generator of time evolution is the sum of kinetic and potential energy operators. For a sharp stationary energy, insert $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$ to obtain $[-{\hslash }^2{\varphi }^{\prime \prime }/(2m)+V\varphi ]=E\varphi$. The probability in $dx$ is $|\psi {|}^{2}dx$, normalized to one; hence a normalized one-dimensional wavefunction has dimensions length${}^{-1/2}$. Its overall phase is unobservable, but relative phase affects interference.

To distinguish $2^b$ equal transverse bins across width $w$, $\boxed{d=w/2^b}$. Six bits gives one-nanometre pixels; ten gives 0.0625 nm, below the helium wavelength. In the stipulated fixed-energy freely propagating beam, the latter entails transverse diffraction momenta comparable to or exceeding the available total momentum and lies outside the paraxial aperture model. Six-bit localization is less demanding, although fine detector resolution and losses remain practical constraints. This is not a universal theorem that an atom can carry at most six bits in every encoding: accessible modes, energy, aperture, measurements and success probabilities define the channel.

For an aligned narrower slit placed immediately after the first, uniform illumination gives transmission probability $\boxed{d/w}$. Conditional on passage the same normalized formula applies with $w$ replaced by $d$ and central-lobe momentum scale $h/d$. Position rms is $d/\sqrt{12}$, while the exact hard-edge rms $\Delta p_x$ is infinite. Thus the source's finite “uncertainty” can only mean a characteristic diffraction width; the rigorous Heisenberg inequality is not violated, and smoothing the edges restores finite rms moments satisfying it.

A 10 keV X-ray has $\lambda \simeq 0.124$ nm. The ideal position-filter probability and momentum formula at fixed aperture widths are unchanged, since the Fourier transform involves $h$ and the aperture geometry. Its larger total momentum makes the angular diffraction width smaller, approximately $\lambda /d$. Real slit-material transmission and absorption are additional experimental issues.

Immediately after a uniformly illuminated hard slit, the normalized transmitted state is $\psi (x)=w^{-1/2}$ for $|x|<w/2$, zero elsewhere. With Fourier convention $\tilde{\psi }(p)=(2\pi \hslash {)}^{-1/2}\int e^{-ipx/\hslash }\psi (x)dx$,The supplied sinc integral verifies unit normalization. The first zeros occur at $\pm \delta p_x$, but this finite central-lobe width is not the rms uncertainty: $p_x^2\tilde{p}$ fails to have a convergent integral.

Approximate helium mass by $2m_p+2m_n\simeq 6.69\times 10^{-27}$ kg. Nonrelativistic momentum is $p=\sqrt{2mT}$, giving $\boxed{\lambda =h/p\simeq 0.203\mathrm{nm}}$. A transverse width of one micrometre implies a characteristic position uncertainty of that order and a minimum momentum uncertainty of order $\hslash /a\sim 10^{-28}$ kg m s${}^{-1}$, with factors depending on profile convention. For a uniform interval $\Delta x=a/\sqrt{12}$, but an exactly abrupt rectangular wavefunction actually has infinite rms momentum spread. Width alone does not uniquely specify both rms uncertainties; a smooth near-minimum-uncertainty beam is needed for a finite order-of-magnitude estimate.

Blackbody radiation defeated classical equipartition because it predicts excessive short-wavelength emission. Quantized oscillator energy exchange $h\nu$, and ultimately the photon occupation law, gives the Planck spectrum. In the photoelectric effect, electron energy follows light frequency with a threshold, rather than simply increasing with beam intensity; individual photons supply $h\nu$ and the maximum kinetic energy is $h\nu -W$. Intensity chiefly changes the event rate at a fixed frequency. These explain two distinct failures of a purely continuous classical wave-plus-matter model, while retaining wave propagation and interference.

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.

Build a solid uniform sphere from thin shells: $dC=(2/3)r^2(4\pi \rho r^{2}dr)$. Integration gives $C=8\pi \rho R^5/15$, while $M=4\pi \rho R^3/3$. Hence $\boxed{C=2M{R}^2/5}$. This is a spherical uniform-density model; planetary flattening modifies the exact relation.

For a thin uniform spherical shell, $dC=(R\sin \theta {)}^{2}dm$ and angular averaging gives $⟨{\sin }^2\theta ⟩=2/3$. Thus $\boxed{C=2M{R}^2/3}$. The printed formula requires a negligibly thin outer shell. For a finite-thickness hollow sphere of uniform density between $r_0$ and $R$,For example $r_0=R/2$ gives $C/(M{R}^2)=31/70$, not $2/3$. The thin-shell limit $r_0\to R$ restores the printed result. A uniform outer layer of unspecified thickness does not by itself establish it.

Helium becoming partially immiscible in dense hydrogen can form helium-rich droplets that settle inward. Gravitational energy release and associated thermal evolution can supply extra luminosity, while the upper hydrogen envelope becomes helium depleted. Different interior temperature-pressure histories can make separation stronger in Saturn than in Jupiter.

The source's ratio is reversed for this intended explanation. Depleting helium lowers He/H and raises H/He. For a concrete example, keeping ten hydrogen units while reducing one helium unit to one half changes H/He from ten to twenty, not to a lower value. Thus the literal statement that Saturn has lower H/He cannot simultaneously be used as evidence for the stated upper-envelope helium-depletion mechanism. If the intended comparison is lower He/H, the observations and extra heat are mutually consistent with helium rain. Composition measurements and interior modeling are still needed; excess heat alone does not uniquely prove that process.^[ref]

A qualitative sketch has molecular hydrogen/helium outside, increasing pressure into a conducting metallic-hydrogen-rich interior, and centrally enhanced heavy elements, possibly distributed rather than confined to a sharp solid core. Saturn's lower mass produces different pressure/temperature transitions and a larger relative role for central concentration. Both can have compositional gradients and helium separation; the moment alone does not decide whether any boundary is sharp.

All three measured factors are below the uniform-sphere value 0.4, indicating central concentration. Earth at 0.331 is consistent with a dense core and less dense mantle; Jupiter at 0.264 and Saturn at 0.21 have still more centrally concentrated normalized distributions, with Saturn especially concentrated relative to its outer radius. These numbers do not uniquely determine core masses, chemical boundaries or density profiles.

Rotation rate, oblateness and measured external gravitational multipoles provide constraints beyond mass and radius; spin-axis precession/nutation and satellite dynamics can add moment-related information. For a giant planet the deep rotation rate and hydrostatic approximation matter. These measurements require an interior model or supplementary constraints for an accurate moment: a spherically symmetric body's exterior monopole alone cannot reveal its radial density distribution. An exact moment cannot generally be inferred just by labeling a measured $J_2$ as $C$.

Permanently shadowed polar craters can remain cold enough to trap water ice and other volatiles for long periods. LCROSS used an impact plume to expose and spectroscopically sample material from such a cold trap, testing the interpretation of polar hydrogen signatures and the availability/history of lunar volatiles. Shadow was therefore a physical target-selection criterion, not merely an observational inconvenience.^[ref]

At fixed projectile mass and the assumed common quarter-power law, changing diameter from 30 to 75 m increases energy by $(75/30{)}^4$ and speed by $(75/30{)}^2=6.25$. The corresponding primary impact speed is about 15.6 km s${}^{-1}$. Five percent transferred to secondary-forming ejecta gives $v_{ej}\simeq 781$ m s${}^{-1}$. ThereforeThis assumes a 45-degree launch and similar target properties/coupling, as well as the source's fixed mass and velocity-transfer rules. Such a range is a noticeable fraction of the lunar radius, so curvature would produce a correction; the supplied flat-surface formula is an estimate rather than exact lunar orbit dynamics.

In a gravity-dominated excavation model, displaced mass scales as $\rho D^3$ and lifting it through a distance of order $D$ costs order $\rho g{D}^4$. If a comparable fraction of impact energy supplies this work across similar impacts, $\boxed{D\propto E_{\mathrm{impact}}^{1/4}}$. This is a scaling justification, not a universal crater law: strength-dominated small craters or varying coupling efficiencies can have different exponents.

For a locally flat airless surface, constant gravity and equal launch/landing heights, $x=v\cos \theta t$ and $z=v\sin \theta t-g{t}^2/2$. The positive flight time is $2v\sin \theta /g$. ThusNeglected effects include curvature, varying gravity, rotation and terrain. At ranges comparable to the body's radius a spherical ballistic calculation is necessary.