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For a Hohmann transfer orbit from radius $r_0$ to $\eta r_0$, with $\eta >1$ and initial circular speed $v_0=\sqrt{GM/r_0}$, conservation of orbital energy and angular momentum gives transfer endpoint speeds $v_p/v_0=\sqrt{2\eta /(1+\eta )}$ and $v_a/v_0=\sqrt{2/[\eta (1+\eta )]}$. The required prograde impulses areIf the spacecraft mass is constant and fuel use is proportional to impulse magnitude, these also give the fuel fractions relative to a budget with total speed change $v_0$. Their sum is a sum of magnitudes, even though the burns occur at different positions and directions.

A Hohmann transfer orbit between coplanar circular Kepler orbits of radii $r_1,r_2$ traverses half an elliptic orbit of semi-major axis $a=(r_1+r_2)/2$. Kepler's third law therefore gives $t=\pi \sqrt{a^3/(GM)}$. Relative to the initial orbital period $T_1$, $t=(T_1/2)[(r_1+r_2)/(2r_1){]}^{3/2}$. Radii must be measured from the attracting centre, rather than from its surface.

For the Hohmann transfer orbit, the inner and outer circular radii are the ellipse's pericentre and apocentre distances, so $a=(r_0+r_1)/2$. The vis-viva equation gives the transfer speed at the inner tangent point,Both old and new velocities are tangential there, so the required first impulse is prograde, with

For a satellite of mass $m$ moving in the Newtonian gravitational field of a much larger, effectively fixed mass $M$, put $\mu =GM$. Its conserved mechanical energy, with gravitational potential energy zero at infinity, isHere $r$ is the instantaneous centre-to-centre distance, $v$ the satellite's speed, $G$ the gravitational constant, and $a$ the semi-major axis of its bound elliptic orbit. The kinetic energy and potential energy vary around the orbit, while their sum remains constant. Dividing by $m$ and solving for $v$ gives the vis-viva equation, $v^2=\mu (2/r-1/a)$.

For this Hohmann transfer orbit, the periapsis and apoapsis distances are $r_1=1\mathrm{AU}$ and $r_2=1.52\mathrm{AU}$, soThe vis-viva equation gives the endpoint speedsAs a check, conservation of angular momentum at the two endpoints gives $r_1{v}_{\mathrm{p}}=r_2{v}_{\mathrm{a}}$ because each velocity is perpendicular to its radius. Using $1\mathrm{AU}=1.49598\times 10^{11}\mathrm{m}$ and $\mu =1.32712\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ gives $v_1=29.78\mathrm{km}{\mathrm{s}}^{-1}$ andThe departure impulse must increase the speed in the direction of the initial circular orbit:These are Sun-centred speeds; leaving the Earth's own gravitational field would require an additional calculation beyond this idealized solar orbit transfer.

At $r_2$, the desired circular orbit speed is $v_2=\sqrt{\mu /r_2}=24.16\mathrm{km}{\mathrm{s}}^{-1}$. The arrival impulse is thereforeThe required thrust on the spacecraft is forward, tangent to its orbit, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's velocity; this distinguishes the direction of thrust from the direction in which the exhaust is fired.

The Hohmann transfer time is half the orbital period of the transfer ellipse. From Kepler's third law, relative to the Earth's one-year orbital period,Meanwhile Mars's orbital period is $T_M=1.52^{3/2}\mathrm{yr}=1.8740\mathrm{yr}$. The spacecraft arrives $\pi$ radians from its departure position. For a rendezvous, Mars must reach that same position after advancing through $2\pi t_H/T_M$. Thus its initial lead angle over the departing spacecraft isMars must be ahead by this angle in the common direction of motion around their orbits. At departure the spacecraft and the Earth share the inner orbital position; arrival occurs on the opposite side of the Sun.

The Hohmann transfer orbit has periapsis $r$ and apoapsis $2r$, so its semi-major axis is $3r/2$. Kepler's third law gives a period $(3/2{)}^{3/2}$ times the original orbital period. Only half the ellipse is traversed:The relevant radii are measured from Earth's centre, not its surface.

A Hohmann transfer orbit connects two coplanar circular orbits using two tangential, effectively instantaneous velocity changes. Take outward transfer radii $r_2>r_1$, and write ${\mu }_g=GM$ for the central body's gravitational parameter. The intervening ellipse has periapsis $r_1$, apoapsis $r_2$, semimajor axis $a_t=(r_1+r_2)/2$ and orbital eccentricity $e=(r_2-r_1)/(r_1+r_2)$.

The circular speed is $\sqrt{{\mu }_g/r}$. Conservation of orbital energy gives the vis-viva equation $v^2={\mu }_g(2/r-1/a_t)$. Therefore the transfer speeds at the two impulses areThe first prograde impulse raises the far side of the orbit; the second prograde impulse raises the near side to circularize at $r_2$. Their magnitudes areThe transfer takes half an ellipse period. Kepler's third law givesFor inward transfer the same ellipse is traversed from apoapsis to periapsis and both burns are retrograde. A rendezvous also needs the correct initial phase of the target: for outward transfer its initial angular lead is $\pi -\sqrt{{\mu }_g/r_2^3}{t}_t$ modulo $2\pi$. The method assumes a two-body gravitational field, negligible atmospheric drag and burns short compared with an orbit. It is an economical two-impulse transfer between these circles; it is not universally the least-$\Delta v$ transfer if additional impulses and very large intermediate radii are permitted.

Use a Hohmann transfer orbit. At the initial radius $r_0$, make a short prograde burn parallel to the existing velocity. The increased speed changes the circular Kepler orbit into an elliptic orbit whose periapsis is the burn point and whose apoapsis lies on the desired larger circle. Coast for half the transfer orbital period to apoapsis.

At that point the transfer velocity is again tangent to the target circle, but its magnitude is smaller than the circular speed there. A second short prograde burn raises it to the required circular speed. Thus both burns increase the instantaneous speed and preserve its direction at the burn, even though the final circular speed is lower than the original one. During the coast the spacecraft loses kinetic energy while gaining gravitational potential energy.

The figure illustrates $3r_0$ for the subsequent parts, but the same construction works for any larger target radius. The impulsive approximation neglects the distance travelled during each burn and keeps both circles and the transfer ellipse coplanar.

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.