Impulsive maneuvers from Lesson 4 of Orbital Dynamics. Define the central body and the initial orbit once (including the spacecraft's current true anomaly \(\theta_0\)), then explore each maneuver family. Every orbit plot can be switched to an interactive 3D view (drag to rotate, wheel to zoom) and played as a time animation from \(\theta_0\) through each impulse. Velocities in km/s, distances in km, angles in degrees.
Phasor solution: \(V_f\angle\varphi = \Delta V\angle\psi + V_i\angle 0^\circ\). Both \(\psi\) and \(\varphi\) are measured counterclockwise from \(\vec V_i\) (\(\psi = 0^\circ\) is a prograde tangential burn, \(\psi = 180^\circ\) retrograde). Equivalent trigonometric form: \(\Delta V^2 = V_i^2 + V_f^2 - 2V_iV_f\cos\varphi\), \(\sin\psi = V_f\sin\varphi/\Delta V\). The new orbit follows from \((r, V_f, \gamma_f)\) with \(\gamma_f = \gamma_i - \varphi\). The spacecraft coasts from \(\theta_0\) to \(\theta_b\) before the impulse.
Rule: to change the apogee, burn tangentially at perigee; to change the perigee, burn tangentially at apogee. With vis-viva at the burn radius, \(\Delta V = \lvert V_f - V_i\rvert\), e.g. at perigee \(\Delta V = \sqrt{\tfrac{2\mu}{r_p}}\,\Bigl\lvert \sqrt{1-\tfrac{1}{1+r_{af}/r_p}} - \sqrt{1-\tfrac{1}{1+r_{ai}/r_p}}\Bigr\rvert\). Circularization at perigee: \(\Delta V = \sqrt{\tfrac{\mu}{a}}\,\tfrac{\sqrt{1+e}-1}{\sqrt{1-e}}\).
The burn must happen at a point common to both orbits: \[\frac{a_i(1-e_i^2)}{1+e_i\cos\theta_i} = \frac{a_f(1-e_f^2)}{1+e_f\cos\theta_f},\qquad \theta_i+\omega_i = \theta_f+\omega_f .\] Writing \(\theta_f = \theta_i - \Delta\omega\) this becomes \(A\sin\theta_i + B\cos\theta_i = C\), solvable only if \(A^2+B^2 \ge C^2\). With \(D=\sqrt{A^2+B^2}\), \(\sin\theta_0 = A/D\), \(\cos\theta_0 = B/D\): \(\theta_i = \theta_0 \pm \arccos(C/D)\), two solutions in general. Then \(\varphi = \gamma_i - \gamma_f\) with \(\tan\gamma = \dfrac{e\sin\theta}{1+e\cos\theta}\).
Initial radius \(r_i = a\) from the defined orbit. Two tangential impulses through an ellipse with \(a_H = (r_i + r_f)/2\): \[\Delta V_1 = \sqrt{\frac{2\mu}{r_i}-\frac{\mu}{a_H}} - \sqrt{\frac{\mu}{r_i}},\qquad \Delta V_2 = \sqrt{\frac{\mu}{r_f}} - \sqrt{\frac{2\mu}{r_f}-\frac{\mu}{a_H}},\qquad T_H = \pi\sqrt{\frac{a_H^3}{\mu}} .\] With \(\lambda = r_f/r_i\): \(\dfrac{\Delta V_T}{V_i} = \sqrt{\dfrac{2}{\lambda(1+\lambda)}}\,(\lambda-1) + \sqrt{\dfrac{1}{\lambda}} - 1\), maximum at \(\lambda \approx 15.58\), escape limit \(\sqrt2 - 1\) for \(\lambda\to\infty\). The first impulse is applied at the chosen \(\theta_b\) (position angle on the circular orbit), which becomes the perigee of the transfer ellipse.
Three tangential impulses through two ellipses, \(a_1 = (r_i + r_t)/2\) and \(a_2 = (r_f + r_t)/2\): \[\Delta V_1 = \sqrt{\tfrac{2\mu}{r_i}-\tfrac{\mu}{a_1}} - \sqrt{\tfrac{\mu}{r_i}},\quad \Delta V_2 = \sqrt{\tfrac{2\mu}{r_t}-\tfrac{\mu}{a_2}} - \sqrt{\tfrac{2\mu}{r_t}-\tfrac{\mu}{a_1}},\quad \Delta V_3 = \sqrt{\tfrac{2\mu}{r_f}-\tfrac{\mu}{a_2}} - \sqrt{\tfrac{\mu}{r_f}} .\] With \(\beta = r_t/r_i\) the cost decreases as \(\beta\) grows; for \(\beta\to\infty\) (bi-parabolic) \(\Delta V_T \to V_i(\sqrt2-1)\bigl(1+\sqrt{1/\lambda}\bigr)\). Time: \(T_B = \pi\bigl(\sqrt{a_1^3/\mu}+\sqrt{a_2^3/\mu}\bigr)\). It only beats Hohmann for \(\lambda > 11.94\); for \(\lambda > 15.58\) it always does when \(\beta > \lambda\).
Initial orbit = the defined orbit \((a, e)\). The final orbit is coaxial (same \(\omega\)). The two Hohmann-like options connect the apogee of one orbit with the perigee of the other; the optimal rule is to choose the trajectory through the larger apogee.
Keep \((a, e, \omega, \theta)\) and rotate the plane, so \(V_f = V_i\). The burn happens on the intersection line of both planes, at latitude \(\phi\): \[\cos\Delta A = \cos i_i\cos i_f + \sin i_i\sin i_f\cos(\Omega_f-\Omega_i),\qquad \sin\phi = \frac{\sin i_i\sin i_f\sin(\Omega_f-\Omega_i)}{\sin\Delta A},\qquad \Delta V = 2V\sin\frac{\Delta A}{2}.\] Special cases: \(\Omega\) only, \(\sin(\Delta A/2) = \sin i\,\sin\bigl((\Omega_f-\Omega_i)/2\bigr)\); \(i\) only, \(\Delta A = i_f - i_i\) at the nodes (\(\phi = 0\)).
Velocity triangle: \(\lvert\vec V_f\rvert = \lvert\vec V_i\rvert\) with angle \(\Delta A\) between them; \(\Delta V\) is the base of the isosceles triangle.
Raise apogee to \(r_2 = \lambda r_1\), change the plane there where the speed is lower, and come back: \[\Delta V_1 = \Delta V_3 = \sqrt{\frac{2\mu}{r_1}-\frac{2\mu}{r_1+r_2}} - \sqrt{\frac{\mu}{r_1}},\qquad \Delta V_2 = 2\sqrt{\frac{2\mu}{r_2}-\frac{2\mu}{r_1+r_2}}\,\sin\frac{\Delta i}{2},\qquad \frac{\Delta V_T}{V_1} = 2\Bigl[\sqrt{\tfrac{2\lambda}{1+\lambda}}\Bigl(1+\tfrac{\sin(\Delta i/2)}{\lambda}\Bigr)-1\Bigr].\] Optimum \(\lambda_{\mathrm{opt}} = \dfrac{\sin(\Delta i/2)}{1-2\sin(\Delta i/2)}\). Rule: \(\Delta i \le 38.94^\circ\) not worth it; \(38.94^\circ < \Delta i \le 60^\circ\) use \(\lambda_{\mathrm{opt}}\); \(\Delta i > 60^\circ\) use \(\lambda\to\infty\).
The plane change at \(r_2\) must happen on the line of nodes, so the transfer apogee is placed at the ascending node; the first impulse is at the descending node.
Combine the radius change and the plane change. The plane change \(\Delta i\) can be applied at the first impulse, the second, or split (fraction \(\alpha\) at the first burn). Each impulse follows the law of cosines: \[\Delta V_1^2 = V_{c1}^2 + V_{t1}^2 - 2V_{c1}V_{t1}\cos(\alpha\,\Delta i),\qquad \Delta V_2^2 = V_{c2}^2 + V_{t2}^2 - 2V_{c2}V_{t2}\cos\bigl((1-\alpha)\Delta i\bigr).\] In practice the whole change is applied at the farther impulse; the exact optimum over \(\alpha\) is also computed.
Drawn and animated with the whole plane change at the farther impulse, which is placed on the line of nodes.
Rotate \(\omega\) by \(\Delta\omega\) keeping \(a\) and \(e\), so \(V_f = V_i = \sqrt{2\mu/r-\mu/a}\). The burn is at \(\theta_i = \Delta\omega/2\) (then \(\theta_f = -\Delta\omega/2\)) or at \(\theta_i = \pi + \Delta\omega/2\) (\(\theta_f = \pi - \Delta\omega/2\)). Since \(\gamma_f = -\gamma_i\), \(\varphi = 2\gamma_i\) with \(\tan\gamma_i = \dfrac{e\sin(\Delta\omega/2)}{1+e\cos(\Delta\omega/2)}\), and \[\Delta V = 2V_i\sin\gamma_i = 2e\sqrt{\frac{\mu}{p}}\,\Bigl\lvert\sin\frac{\Delta\omega}{2}\Bigr\rvert ,\] identical for both solutions.
Interceptor at \(\theta_0\); target on the same orbit, trailing or leading by \(\nu\). The interceptor coasts to \(\theta_b\) (the phase angle is preserved), then enters the phasing orbit. New period \(T_{ph} = T\bigl(1 \mp \tfrac{\nu}{2\pi k}\bigr)\) (− trailing, + leading), then \[a_{ph} = \Bigl(\mu\bigl(\tfrac{T_{ph}}{2\pi}\bigr)^2\Bigr)^{1/3},\qquad \Delta V_1 = \sqrt{\frac{2\mu}{r}-\frac{\mu}{a_{ph}}} - \sqrt{\frac{\mu}{r}},\qquad \Delta V_T = 2\lvert\Delta V_1\rvert ,\] with total maneuver time \(k\,T_{ph}\).
Angles are not swept uniformly, so phasing is done in time: with the interceptor at \(\theta_A = \theta_0\) and the target at \(\theta_B\), \(t_{AB} = \Delta t(\theta_B) - \Delta t(\theta_A)\) from Kepler's equation and \(T_{ph} = T - t_{AB}/k\). The burn is tangential, preferably at perigee (largest \(V\), so the same \(\Delta V\) produces the largest change in \(a\) and \(T\)): \[\Delta V_1 = \sqrt{\frac{2\mu}{r}-\frac{\mu}{a_{ph}}} - \sqrt{\frac{2\mu}{r}-\frac{\mu}{a}},\qquad \Delta V_T = 2\lvert\Delta V_1\rvert .\]
\(\Delta V = V_e\ln\dfrac{m_0}{m_0-m_F}\) with \(V_e = I_{sp}\,g_0\), hence \(m_F = m_0\bigl(1-e^{-\Delta V/(I_{sp}g_0)}\bigr)\). For several maneuvers with the same motor the \(\Delta V\) simply add: \(m_{F,\mathrm{total}} = m_0\bigl(1-e^{-\Delta V_T/V_e}\bigr)\). Use the “→ fuel” buttons next to any result to send its \(\Delta V\) here.