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Let $\psi ={\lambda }_d-{\lambda }_p$ be azimuth in the frame rotating with the planet, with the planet fixed at $\psi =0$. Since $n_d/n_p=4/5$, one synodic circuit contains four radial oscillations. To first order in $e_d$, choosing conjunction at apoapsis givesThe requested sketch is therefore a four-lobed closed curve centred on the star. For $e_d=0.2$, its radii range from $0.8a_d$ to $1.2a_d$; an outer lobe points along the star-planet line, and the planet lies at $(a_p,0)$. At that radial alignment $\varphi =\pi$, because ${\lambda }_d={\lambda }_p$ and ${\lambda }_d={\varpi }_d+\pi$ at apoapsis.

At exact conjunction the planet's force is radial and has no instantaneous torque, but the approach and departure do not cancel when conjunction is displaced from apoapsis. Write the conjunction longitude as ${\lambda }_c={\varpi }_d+\pi +\delta$. At conjunction,The resonant part of the given evolution law has ${\dot{a}}_d=-8e_d{a}_{d}D\sin \varphi$. Thus a conjunction after apoapsis, $\delta >0$, gives $\sin \varphi <0$ and transfers angular momentum to the dust, increasing $a_d$. A conjunction before apoapsis removes angular momentum, while one exactly at apoapsis has zero secular transfer by symmetry.

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.

Assume the station's mass remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's Newtonian gravitational field. Put $\kappa =G{M}_E$, and work per unit station mass. The initial circular orbit has specific angular momentum $j$ and specific orbital energy ${\epsilon }_0$ given byLet $\delta$ denote the fractional increase of kinetic energy. The position and potential energy are unchanged during the kick, soUnchanged angular momentum at unchanged position fixes the tangential velocity; the added kinetic energy must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new orbital eccentricity follows fromAt either radial turning point, the Kepler radial energy equation gives $\epsilon =j^2/(2r^2)-\kappa /r$. Its two roots, equivalently the periapsis and apoapsis of the elliptic orbit, areFor $\delta =0.01$, $e=0.1$ andAn inward or outward kick gives the same pair of radial turning points, although it places the station at a different phase of the orbit. These two finite positive radii require $0\le \delta <1$.

Kepler's third law for negligible planet mass gives $a_P/a_E=(P_P/P_E{)}^{2/3}$. Therefore $a_P=1.5\times 10^8(248{)}^{2/3}\simeq 5.92\times 10^9\mathrm{km}$. The furthest point is apoapsis, soThe supplied orbital period and eccentricity are sufficient; no present-day ephemeris is needed.

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.

Choose zero gravitational potential energy at infinite separation. The spacecraft's conserved total orbital energy isAt periapsis and apoapsis, the radial velocity vanishes, so the velocity is perpendicular to the radius. The conserved angular momentum magnitude is thereforeThe angular momentum vector is normal to the fixed orbital plane. The respective units are joules and $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. These expressions use Newtonian gravity and the approximation $M\gg m$.

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.

At transfer apoapsis, angular momentum conservation givesThe desired circular Kepler orbit has speed $v_c=\sqrt{GM/(3r_0)}=v_0/\sqrt{3}$. Thusof the original fuel supply is used in the second burn. This is about $21.8\%$ of the fuel remaining after the first burn. The total transfer consumes $f_1+f_2\simeq 39.4\%$ of the original supply, leaving about $60.6\%$. The two positive impulse magnitudes are added for the fuel budget, even though their directions at opposite ends of the orbit differ.

A higher orbital line density at apoapsis does not always give a higher surface density of a disk: the local ring width also changes. In a narrow aligned eccentric ring, neglecting disc twist and keeping first-order orbital eccentricity, ${\Sigma }_{\mathrm{peri}}\propto 1+ade/da$ and ${\Sigma }_{\mathrm{apo}}\propto 1-ade/da$. Constant eccentricity therefore gives constant surface density to this order.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.