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Use local east, north and up axes. At latitude $\lambda$, $\mathbf{\Omega }=(0,\Omega \cos \lambda ,\Omega \sin \lambda )$ and eastward velocity is $(v,0,0)$. The Coriolis acceleration isIts horizontal part is southward, so the lift must acquire a northward component: bank left, with the north wing lower. For the requested local Coriolis correction, let the vertical lift balance $mg$. The Coriolis bank angle for eastward flight isUse $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\Omega =7.2921\times 10^{-5}{\mathrm{s}}^{-1}$. Including the upward Coriolis term replaces $g$ by $g-2\Omega v\cos \lambda$ and changes the answer by only about $0.1\%$. This is the horizontal Coriolis compensation, not a full calculation of a curved constant-latitude flight path. The rotation rate follows Earth's sidereal period.^[ref]

Use a local horizontal plane, constant gravity and first order in Earth's angular speed. Let $x$ point along the initial horizontal shot, $y$ to the gunner's left and $z$ upward. At the North pole, $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. Maximum unperturbed range at equal launch and landing heights requires a $45^{\circ }$ elevation, givingThe leading Coriolis acceleration is $\ddot{y}=-2\Omega v_x$, while vertical motion has no first-order Coriolis correction. With zero initial lateral displacement and velocity,Here $u=500\mathrm{m}{\mathrm{s}}^{-1}$, $g\simeq 9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\Omega \simeq 7.2921\times 10^{-5}{\mathrm{s}}^{-1}$. The miss is to the right of the aiming direction, independent of its azimuth at the pole. The flight time is about $72.1\mathrm{s}$, so $\Omega t_f\simeq 0.0053\ll 1$. This polar projectile Coriolis deflection uses the usual flat, uniform-gravity approximation; planetary curvature and variation of gravity are beyond its accuracy. Using the polar gravity rather than $9.81$ changes the estimate by less than one percent. Earth's sidereal rotation period, rather than the solar day, sets $\Omega$.^[ref]

In the ideal semidiurnal tide model, successive high tides occur after the Earth–Moon relative angle advances by $\pi$. Since the motions have the same sense, subtract their angular speeds:The lunar orbital period is sidereal, so the consistent terrestrial rotation period is $T_E=23.9345\mathrm{h}$. With $T_M=27.3\times 24\mathrm{h}$, this gives about 12 h 25 min. The sidereal terrestrial period is distinct from the 24-hour solar day.^[ref]

If the Earth's rotation is approximated as one revolution in 24 hours, the same calculation instead gives 12 h 27 min. That small difference is the day convention, not a change of sign in the relative angular speed. Actual coastal tide times also depend on ocean dynamics and the solar tide.

For this molecular escape-speed temperature estimate, $v_{\mathrm{rms}}^2=3k_{B}T/m$. Equating this to the surface escape-speed square $v_{\mathrm{esc}}^2=2G{M}_{\oplus }/R_{\oplus }$ givesThis evaluation keeps the paper's $R_{\oplus }=6.4\times 10^6\mathrm{m}$ and $m=28u$, and uses externally documented $G{M}_{\oplus }=3.9860\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, $u=1.660539040\times 10^{-27}\mathrm{kg}$ and $k_B=1.38064852\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$.^[ref][ref] The corresponding escape speed is about $11.2\mathrm{km}{\mathrm{s}}^{-1}$.

This is the requested speed-scale comparison. Atmospheric escape can occur from the high-speed tail at lower temperature; molecular dissociation and ionization also prevent treating an actual gas at this temperature as unchanged neutral nitrogen molecules.