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The kick changes the tangential velocity by $\gamma v_K\cos (\varphi -f)$. With $s=\sqrt{1-e^2}$ and $r/a=s^2/(1+e\cos f)$, the new specific angular momentum isEvery Kepler orbit satisfies $h^{\prime 2}=\mu a^{\prime }(1-e^{\prime 2})$. Combining this identity with the value of $a/a^{\prime }$ found in part (d) giveswhere $a/a^{\prime }$ is the explicit function of $e,f,\gamma ,\varphi$ in part (d).

In steady state, mass conservation makes the outward flux constant, $\mathcal{F}=\dot{M}$. Angular-momentum conservation then givesThe zero-torque condition at $r_{\mathrm{out}}$ fixes the constant to $\dot{M}{h}_{\mathrm{out}}$, soFor a Kepler orbit, $h=\sqrt{GMr}$ and $d\Omega /dr=-3\Omega /(2r)$. The viscous torque is consequently $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Equating the two expressions yields the steady decretion disk structure

The first term, $m{\dot{r}}^2/2$, is radial kinetic energy. The second, $L^2/(2m{r}^2)$, is tangential kinetic energy, since conservation of angular momentum gives $L=m{r}^2\dot{\phi }$. The final term, $-GMm/r$, is the gravitational potential energy, with zero at infinite separation. ThusThe second and third terms together form the radial effective potential for a Kepler orbit. At an apsis the radial velocity vanishes, so $E=U_{\mathrm{eff}}$; for a bound noncircular orbit the two accessible roots give the nearest and furthest radii.

Adding tangential velocity along the existing motion increases both orbital energy and angular momentum:For a sufficiently small kick the Kepler orbit remains an ellipse. At the kick point its radial velocity vanishes and its tangential speed exceeds the circular orbital speed, so that point becomes the pericentre. The ellipse extends farther outward than the original circle.

Choose the orientation $J>0$. To leading order in the radial disturbance, the azimuthal angular frequency is ${\omega }_{\theta }=J/(m{r}_0^2)$, soFor either orientation the positive frequency ratio is ${\omega }_r/|{\omega }_{\theta }|$. At $k=-1$, the ratio is $1$: there is one radial oscillation per revolution. The bound Kepler orbit is an ellipse with the force centre at a focus, with one closest and one furthest approach per revolution. Its exact polar coordinates equation isAt $k=2$, the ratio is $2$: there are two radial oscillations per revolution. The isotropic harmonic oscillator trajectory is an ellipse centred on the force centre. An exact parametrization isHere $r^2=a^2{\cos }^2\Omega t+b^2{\sin }^2\Omega t$ repeats twice during one full circuit. The two different placements of the force centre distinguish these ellipses. These are the two all-bound-orbit cases of Bertrand's theorem.

Use a circular Kepler orbit to eliminate the terrestrial gravitational parameter: $T^2/r^3$ is constant for both satellites. The reference orbital radius is $r_1=(6400+150)\mathrm{km}=6550\mathrm{km}$, measured from Earth's centre. A geostationary orbit is circular, equatorial and prograde, with the sidereal rotation period, approximately $86164\mathrm{s}$. ThusIts altitude is about $3.61\times 10^4\mathrm{km}$ above the stated surface. The approximate low-orbit input sets the accuracy; replacing it by a modern measured gravitational parameter would answer a different numerical estimate. A $24$-hour approximation changes the radius by less than $0.2\%$. Matching the period alone would not make an inclined or eccentric orbit stationary above one point.

Classification of attractive inverse-square orbits. Write the potential energy as $U=-\kappa /r$, with $\kappa =GMm>0$ and zero at infinity. For nonzero angular momentum $J$, the Kepler orbit is a conic,Negative total orbital energy gives an elliptic orbit, $0\le e<1$. Its special case $e=0$ is the circular Kepler orbit, at the minimum energy permitted for that $J$, $E=-m{\kappa }^2/(2J^2)$. Zero energy gives a parabolic Kepler orbit, $e=1$, and positive energy a hyperbolic Kepler orbit, $e>1$. For $J=0$, the motion is radial: these are degenerate limiting trajectories and may strike the centre or the finite planetary surface.

The scattering geometry. The impact parameter $b$ is the perpendicular offset of the undeflected incoming asymptote from the centre, not the actual periapsis radius $r_{\min }$. Gravity bends the incoming path toward the planet.

The grazing threshold. At infinity, conservation of energy and angular momentum give $E=m{v}_{\infty }^2/2$ and $J=mb{v}_{\infty }$. At closest approach the radial speed is zero, so the velocity is tangential andEliminating $v_p$ givesThis increases monotonically with $r_{\min }>0$. Setting $r_{\min }=R$ gives the grazing impact parameter in gravitational focusing,Here $v_{\mathrm{esc}}=\sqrt{2GM/R}$ is the surface escape speed. The expression has units of length. Equality is the tangency limit; a strictly noncolliding trajectory requires $b>b_{\min }$. At high speed, $b_{\min }\to R$, while at low speed $b_{\min }\sim \sqrt{2GMR}/v_{\infty }$. Thus gravitational focusing enlarges the collision cross-section to $\pi b_{\min }^2$.

Deflection angle. For this hyperbola,The asymptotes occur at $\cos {\varphi }_{\infty }=-1/e$. The change of velocity direction is $\chi =2{\varphi }_{\infty }-\pi$, giving $\sin (\chi /2)=1/e$. Hence the gravitational scattering angle satisfiesFor the grazing trajectory this isIt is the attractive counterpart of the Rutherford impact-parameter relation; the sign of the transverse bending changes, but the positive deflection magnitude has this same form.

Range of angles. Along the grazing family, varying $v_{\infty }>0$ givesThe limits are $\chi \to 0$ as $v_{\infty }\to \infty$ and $\chi \to \pi$ as $v_{\infty }\to 0^{+}$. Neither endpoint is reached at finite positive speed. The low-speed limiting orbit still has a finite periapsis, but an increasingly distant incoming asymptote.

Powered vertical flight. Take upward velocity $v$ as positive, and let $v_e>0$ be the downward exhaust speed relative to the rocket. During $dt$ the rocket's mass changes by $dm<0$, and exhaust mass $-dm$ leaves with Earth-frame velocity $v-v_e$. Compare the momentum of the rocket plus this newly expelled exhaust before and after the interval:Discard the second-order product $dmdv$. The result is $mdv+v_{e}dm=-mgdt$, orThe positive rocket thrust is $-v_e\dot{m}$. Integrating with $v(0)=0$ and $m(0)=m_0$ gives the rocket equation with gravitational loss:Using $d(mv)/dt$ for the rocket alone without the exhaust momentum flux would give the wrong equation for this open system.

Orbital speed under the stated gravity assumption. A circular orbit of radius $r_o=R_E+h_o=7.90\times 10^6\mathrm{m}$ requires centripetal acceleration $v_o^2/r_o$. If the magnitude of gravitational acceleration is the stipulated constant $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ even at the orbit,This treats gravity as centrally directed for the orbital calculation, while retaining its constant magnitude. The subsequent flat-surface approximation treats its direction as locally vertical.

Burnout height. During the unpowered projectile motion, horizontal velocity is constant. At the top of the trajectory the vertical component vanishes, so the horizontal component must equal the orbital speed $v_o$. At burnout, therefore,These follow either by integrating $\ddot{y}=-g$ or by the vertical work-energy theorem. Substituting the preceding constant-$g$ orbital result givesThe vertical component is initially positive, as required to reach the orbit after burnout.

Alternative inverse-square interpretation. A real Earth's exterior Newtonian gravitational field is not constant over a $1500\mathrm{km}$ altitude change. If $9.8\mathrm{m}{\mathrm{s}}^{-2}$ instead means the surface value $g_0$, then Newton's law of universal gravitation gives $GM=g_0{R}_E^2$ andUsing this orbital speed but the paper's constant-$g$ flat projectile motion approximation for the coast would giveThis alternative changes the global constant-gravity assumption; it is not the same model as the $183\mathrm{km}$ result. An exact curved Kepler orbit coast would also abandon the prescribed flat trajectory. The printed assumptions support the constant-magnitude answers above, while the inverse-square values explain the different answer obtained from the usual physical Earth-orbit formula.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

For two bodies of masses $M_1,M_2$, the relative Kepler orbit has semimajor axis $a$ and orbital period $P$ satisfying Kepler's third lawHere $a$ is the semimajor axis of the separation vector, not either star's individual barycentric orbit. In conventional stellar units the same relation is approximately $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, with year, astronomical unit and solar mass as the units.

The patched conic approximation models different portions of a spacecraft trajectory by separate two-body Kepler orbits, matching position and velocity where one attracting body gives way to another. For a surface departure from a nonrotating planet, conservation of energy gives $u^2=v_e^2+w^2$, where $u$ is surface launch speed, $v_e$ is the planet's surface escape speed, and $w$ is the planet-relative hyperbolic excess speed. Adding that excess velocity vector to the planet's orbital velocity supplies the heliocentric initial condition. It neglects the simultaneous full gravitational action of both bodies near the matching region.

The specific angular momentum of an elliptic Kepler orbit is $h=\sqrt{\mu a(1-e^2)}$. At periapsis, $r=a(1-e)$, soThe mean motion understates instantaneous angular speed near periapsis. A particle in an exterior mean-motion resonance can briefly overtake the planet in angle despite completing full orbits more slowly.

For a Kepler orbit with pericentre distance $q$, apocentre distance $Q$, and semi-major axis $a=(q+Q)/2$, a uniform mean anomaly givesThe two radial passages per orbital period are included, and ${\int }_q^Q{p}_r(r)dr=1$.

The dimensionless coefficient $J_2$ describes the leading axisymmetric departure of a planet's external Newtonian gravitational potential from a monopole: $\Phi =-G{M}_p/r+[G{M}_p{J}_2{R}_p^2/r^3]P_2(\cos \vartheta )$, where $\vartheta$ is colatitude from the spin axis. An oblate mass distribution has positive $J_2$. The disturbing function is minus the nonspherical potential term. On a circular inclined Kepler orbit, $\cos \vartheta =\sin I\sin M$, so orbit averaging gives ${\overline{\mathcal{R}}}_{J_2}=G{M}_p{J}_2{R}_p^2{a}^{-3}(1/2-3{\sin }^{2}I/4)$. The exact circular nodal rate is $-3n{J}_2(R_p/a{)}^2\cos I/2$; its small-tilt limit is $-3n{J}_2(R_p/a{)}^2/2$.

For a bound Kepler orbit, specific orbital energy $v^2/2-\mu /r$ is negative. Therefore $\mu >v^{2}r/2$. A projected separation $R_{\perp }\le r$ and a systemic-motion-corrected line-of-sight speed $|v_{\mathrm{los}}|\le v$ give the displayed lower bound. One position and one spectral velocity already constrain the mass in this model; a fitted orbit gives much more. The bound assumes a bound orbit in a point-mass potential; it is not an unrestricted enclosed-mass formula for an arbitrary extended potential.

The semimajor axis $a_1$ of the pulsar's orbit about the binary barycenter, projected along the line of sight and divided by the speed of light. The orbital inclination $i$ is zero for a face-on orbit. Thus $x$ is a time, often quoted as the numerical number of light-seconds in $a_1\sin i$, not the true binary separation. The relative-orbit semimajor axis is instead $a=a_1+a_2$; conservation of momentum gives $a_1=a{m}_2/(m_1+m_2)$.

Combining Kepler's third law with that barycenter relation gives the timing form of the binary mass function, also valid for eccentric Kepler orbits:Choose $x=2\mathrm{s}$ and $P_b=8\mathrm{h}$ for an original example: $a_1\sin i=599,585\mathrm{km}$ and $f\simeq 0.0773M_{\odot }$, where $M_{\odot }$ is a solar mass. For a specified pulsar mass $m_1=1.4M_{\odot }$, the smallest allowed companion mass occurs at $\sin i=1$ and is approximately $0.698M_{\odot }$. A different assumed $m_1$ changes this bound. One measured $f$ does not determine both masses and orbital inclination.

An impulse directed along the position vector relative to a specified origin. Its angular impulse vanishes because $\Delta \mathbf{J}=\mathbf{r}\times \Delta \mathbf{p}=0$. It can nevertheless change kinetic energy; if an initially circular Kepler orbit receives radial velocity $w$, the added specific orbital energy is $w^2/2$.

On a nonradial zero-energy Kepler orbit, conserved specific angular momentum $h$ gives periapsis $r_p=h^2/(2GM)$ and ${\dot{r}}^2=2GM(r-r_p)/r^2$. Integrating with $s=\sqrt{r-r_p}$ yields $t-t_p=\pm \sqrt{2/(GM)}(r+2r_p)\sqrt{r-r_p}/3$. Squaring gives the displayed version of the Barker equation, valid on both incoming and outgoing branches. The time origin $t_p$ is passage through periapsis, not an arbitrary point on the orbit.

The distance from a focus to a conic section along a line perpendicular to its symmetry axis. In focus-centred polar coordinates, an ellipse may be written $r(1-e\cos \theta )=\ell$, where $e$ is its eccentricity of a conic.

The distance from a focus of a conic section to the curve measured perpendicular to its principal axis. In the focus-based polar coordinates equation $r=p/(1+e\cos \varphi )$ it is the length $p$, where $e$ is the eccentricity. For an ellipse, $p=a(1-e^2)=b^2/a$. For a Newtonian Kepler orbit, $p=j^2/(GM)$ with $j$ the specific angular momentum.

A spacecraft trajectory is the path traced by a spacecraft's position over time. An unpowered exterior arc can approximate a Kepler orbit; other bodies' gravity, radiation pressure, and propulsion can change it.