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Earth is a rocky planet whose surface couples a gaseous atmosphere, liquid ocean, solid crust and regions of ice. Its mean radius is about $6371\mathrm{km}$ and its mean mass density about $5.51\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. These bulk quantities constrain, but do not uniquely specify, its differentiated interior.^[ref]

Seismic waves probe material beneath the surface, plate tectonics describes the moving outer shell, and the geodynamo explains the origin of the large-scale planetary magnetic field. At the surface, radiative transfer and ocean circulation redistribute solar energy. Earth therefore provides a nearby setting in which measurements can test coupled planetary processes that are much harder to separate in an unresolved exoplanet.

Use a circular Earth orbit of radius $a$ about the Sun, neglect Earth's rotation, the other planets and atmospheric drag, and treat the launch as an initial impulse followed by unpowered flight of constant mass. DefineIn the patched conic approximation, a surface launch at speed $u$ leaves Earth's gravitational field with Earth-relative hyperbolic excess speed $w$, whereThus the necessary solar-orbit change must be supplied in addition to escaping Earth's gravity; simply adding the two escape speeds is incorrect. The asymptotic departure direction can be chosen to minimize $w$. Representative values are $v_E\simeq 29.78\mathrm{km}{\mathrm{s}}^{-1}$ and $v_{e,E}\simeq 11.186\mathrm{km}{\mathrm{s}}^{-1}$.^[ref]

Direct fall towards a point Sun. If the Sun is treated as a point target, the projectile must lose its heliocentric angular momentum. The cheapest change is retrograde, cancelling Earth's tangential orbital velocity. It then starts its radial fall with zero heliocentric speed, so $w=v_E$ andThis is the usual radial-fall estimate. Merely launching towards the Sun while retaining Earth's tangential velocity does not remove the angular momentum barrier.

Minimum for impact on a finite Sun. A literal collision with the solar surface does not require zero angular momentum. The limiting solar impact by a retrograde launch instead has apocentre $a$ at Earth's orbit and pericentre $R_{\odot }$ at the solar surface. The vis-viva equation with semi-major axis $(a+R_{\odot })/2$ gives the heliocentric departure speedIt remains prograde about the Sun, but the Earth-relative departure is retrograde with $w=v_E-v_a$. Therefore the surface launch speed for grazing the finite solar surface isHere $a\simeq 149.598\times 10^6\mathrm{km}$ and $R_{\odot }\simeq 695700\mathrm{km}$.^[ref][ref] Slightly reducing the departure speed below $v_a$ makes the trajectory enter the solar surface rather than just graze it. The $31.81$ and $29.15\mathrm{km}{\mathrm{s}}^{-1}$ values correspond to the explicitly different point-target and finite-radius idealizations, not to two ways of adding escape speeds. These are direct-flight estimates; planetary gravity assists and Earth's rotation are outside the model.

Use the supplied planet mass $5.9\times 10^{24}\mathrm{kg}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, and adopt mean Earth radius $R_E=6371\mathrm{km}$. For a geostationary orbit, use one sidereal day, $T\simeq 86164\mathrm{s}$. The adopted radius and rotation period follow NASA's fact sheet.^[ref]The slightly smaller radius than the usual terrestrial value follows from using the question's smaller Earth mass. For coplanar prograde orbits, the preceding formulas giveThe escape transfer has $a\to \infty$, so its first tangential burn reaches the local escape speed $\sqrt{2\mu /r_0}$. No final circularization at infinity is needed:Thus the geostationary transfer requires about $0.687\mathrm{km}{\mathrm{s}}^{-1}$ more total impulse, or about 22 percent more, despite leaving the spacecraft gravitationally bound. Reaching the outer radius and circularizing there are separate requirements. Changing the starting orbit's inclination would add a plane-change cost absent from this calculation.

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.