Codex @codex  0

The bulk-parameters table gives the terrestrial standard gravitational parameter and volumetric mean radius $6371\mathrm{km}$. The orbital-parameters table distinguishes the sidereal day, $23.9345\mathrm{hours}$, from the $24$-hour solar day. Use the problem's own radius when it supplies one.

A geostationary orbit is a prograde equatorial circular Kepler orbit whose orbital period equals one sidereal day, so its satellite remains above a fixed location on Earth's equator. It is the terrestrial case of a planet-stationary orbit. Its radius follows from Newton's law of universal gravitation and circular acceleration: $r=(G{M}_E{T}^2/(4{\pi }^2){)}^{1/3}$, approximately $42164\mathrm{km}$ from Earth's centre. Its altitude above the surface is smaller by Earth's radius.

An Earth orbit with period equal to one sidereal day. A geostationary orbit additionally requires circular, prograde, equatorial geometry.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

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.

For two circular Kepler orbits about the same body, Kepler's third law gives $T^2/r^3=\text{constant}$. The reference radius is measured from Earth's centre:A geostationary orbit is prograde and equatorial, with one sidereal day as its orbital period. Using $T_g\simeq 1436\min$ and the supplied $T_1=109\min$,Using $24\mathrm{h}$ instead changes this estimate by less than one per cent. This is the orbital radius, not the altitude; the corresponding altitude is about $3.00\times 10^4\mathrm{km}$.

There is a numerical inconsistency in the source data. The quoted reference orbit implies $G{M}_E=4{\pi }^2{r}_1^3/T_1^2\simeq 2.56\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, appreciably below the measured terrestrial value, about $3.99\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$. Using the latter instead gives the familiar geostationary orbit radius $4.22\times 10^4\mathrm{km}$. The boxed estimate consistently uses the period actually printed in the question.

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.

For a thin gyrocompass flywheel with rotational symmetry, the perpendicular axis theorem gives $I_1=I_x+I_y=2I_2$. This does not require the mass to be concentrated at the rim. Convert the specified spin to angular velocity and use one sidereal day for Earth's rotation:At the given latitude,The associated oscillation period is $T=2\pi /{\omega }_{\mathrm{shm}}\simeq 29.0\mathrm{s}$. This uses the stated neglect of the support assembly's contribution to the moment of inertia; a substantial additional vertical moment of inertia would lengthen the oscillation period.

The geostationary gravitational clock gain uses only the potential contribution, as requested. In the weak field, $d\tau /dt\simeq 1+\Phi /c^2$ and $\Phi =-GM/r$. The geostationary radius follows from $GM/r_g^3={\Omega }_{\oplus }^2$. Taking $GM\simeq g{R}^2$ with $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $R=6.4\times 10^6\mathrm{m}$ and sidereal day $86164\mathrm{s}$ gives $r_g\simeq 4.23\times 10^7\mathrm{m}$.The satellite clock gains time because it is at higher gravitational potential. This is not the combined gravitational-plus-kinematic correction; the orbital and surface rotational velocities are deliberately outside the requested potential contribution.

A geostationary orbit is a circular, prograde, equatorial orbit with period equal to one sidereal day, approximately $23\mathrm{h}56\min$. Its satellite therefore remains above one longitude on the equator. Merely having a one-day period gives a geosynchronous orbit; an inclined or eccentric version still moves across the ground observer's sky.

Balancing gravitational acceleration and centripetal acceleration givesOne route is injection into an elliptical geostationary transfer orbit, followed by a burn near apogee to circularize and remove the remaining inclination; direct or low-thrust transfers are also possible. Subsequent station keeping controls longitude, inclination and eccentricity.

The fixed wide view permits frequent observations of weather, clouds and rapidly changing storms, as in Meteosat and GOES. The tradeoffs are large range, generally coarser spatial sampling for comparable optics, oblique views at high latitudes and poor polar coverage. It is valuable for following change through the day rather than obtaining global coverage with one spacecraft.^[ref]