Assuming that (7) is negative, we can simplify the solution to (6) by choosing $a$ so that $\frac{{\mathbb{U}}^{\prime }{\bar{\mathbb{U}}}^{\prime }-2a}{r}$ is equal to −1, i.e.,

$a={\left(\frac{2}{r_0}-\frac{v_0^2}{\mu }\right)}^{-1}$(9)

where $v_0$ and $r_0$ are the initial values of $v$ and $r$, respectively.

The general solution to ${\mathbb{U}}^{″}+\mathbb{U}=0$ is then given by

$\mathbb{U}=\mathbb{A}\exp \left(\text{i}s\right)-\mathbb{B}\exp \left(-\text{i}s\right)$(10)

where $\mathbb{A}$ and $\mathbb{B}$ are complex numbers. To find their values, we need the rocket’s initial position $r_0$ and velocity $v_0$. Using zero as the initial value of $s$, solving for these coefficients is then quite routine (note that $v_0$ needs to be multiplied by the initial value of the RHS of (3) to convert it from ordinary velocity $\frac{\text{d}r}{\text{d}t}$ to $\frac{\text{d}r}{\text{d}s}=2\mathbb{U}{\mathbb{U}}^{\prime }$). The resulting equations then read

$\mathbb{A}-\mathbb{B}=\sqrt{r_0}$(11)

$\mathbb{A}+\mathbb{B}=\frac{v_0\overline{\sqrt{r_0}}}{\text{i}}\sqrt{\frac{a}{\mu }}=-\text{i}\frac{v_0}{v_0}\overline{\sqrt{r_0}}\sqrt{\frac{2a}{r_0}-1}$

where we have replaced $\sqrt{\frac{a}{\mu }}$ in (3) by $\frac{1}{v_0}\sqrt{\frac{2a}{r_0}-1}$, utilizing (7) being equal to −1.

The last two equations are easily solved for $\mathbb{A}$ and $\mathbb{B}$; note that we can choose either sign of $\sqrt{r_0}$. Using (2) and the first two lines of (4), we can then compute the location and velocity of the rocket at any future value of $s$; to convert $s$ into actual time $t$, we use

$t=2\sqrt{\frac{a}{\mu }}\left(as+\text{Im}\left(\mathbb{A}\bar{\mathbb{B}}\left(1-\text{exp}\left(2\text{i}s\right)\right)\right)\right)$(12)

assuming $t_0=0$.

The technique is sufficient for the purpose of this article.

Orbital elements (optional) Nevertheless, to prove that the resulting trajectory is an ellipse [8] , and to make contact with the usual description of this kind of motion, we express $\mathbb{A}$ in its polar form of $A\exp \left(-\text{i}\alpha \right)$, and $\mathbb{B}$ similarly as $B\exp \left(\text{i}\beta \right)$, so that (10) now reads

$\begin{array}{c}\mathbb{U}=A\exp \left(\text{i}\left(s-\alpha \right)\right)-B\exp \left(-\text{i}\left(s-\beta \right)\right)\\ =\exp \left(\text{i}\frac{\psi }{2}\right)\left(A\exp \left(\text{i}\left(s-s_p\right)\right)-B\exp \left(-\text{i}\left(s-s_p\right)\right)\right)\end{array}$

where $s_p:=\frac{\alpha +\beta }{2}$ is the value of $s$ at the trajectory’s pericenter [1] , and $\psi :=\beta -\alpha$.

Note that $\frac{{\mathbb{U}}^{\prime }{\bar{\mathbb{U}}}^{\prime }-2a}{r}=-1$ then yields

$\begin{array}{c}2a=A^2+B^2+2AB\cos \left(2\left(s-s_p\right)\right)\\ +A^2+B^2-2AB\cos \left(2\left(s-s_p\right)\right)\\ =2\left(A^2+B^2\right)\end{array}$

implying that $a=A^2+B^2$.

For the rocket’s location at any future value of $s$, we get

$\begin{array}{c}r=\exp \left(\text{i}\psi \right)\left(A^2\exp \left(2\text{i}\left(s-s_p\right)\right)+B^2\exp \left(-2\text{i}\left(s-s_p\right)\right)-2AB\right)\\ =\exp \left(\text{i}\psi \right)\left(A^2+B^2\right)\left(\cos 2\left(s-s_p\right)+\text{i}\frac{A^2-B^2}{A^2+B^2}\sin 2\left(s-s_p\right)-\frac{2AB}{A^2+B^2}\right)\end{array}$(13)

This confirms that the actual path is an ellipse with the length of its semi-major axis [1] equal to $a=A^2+B^2$, its eccentricity equal to $\epsilon :=\frac{2AB}{A^2+B^2}$ and one (the rightmost) of its foci at the origin (where the primary’s center is) further rotated (around the origin) by angle $\psi$; these are the usual orbital elements.

Proof: ignoring the $\text{exp}\left(\text{i}\psi \right)$ rotation, the magnitude of the remaining expression (i.e., distance to the origin) is $\left(A^2+B^2\right)\left(1-AB\cos 2\left(s-s_p\right)\right)$, while the distance to the second focus (located at $-4AB$) is $\left(A^2+B^2\right)\left(1+AB\cos 2\left(s-s_p\right)\right)$; the sum of the two distances is $2\left(A^2+B^2\right)$, which is the usual way of describing an ellipse.

We can also compute the velocity of the rocket at the modified time $s$ by

$v=\text{i}\sqrt{\frac{\mu }{a}}\exp \left(i\psi \right)\frac{A^2\exp \left(2\text{i}\left(s-s_p\right)\right)-B^2\exp \left(-2\text{i}\left(s-s_p\right)\right)}{A^2+B^2-2AB\cos 2\left(s-s_p\right)}$

To convert $s$ to actual time $t$, we integrate (3), getting

$\begin{array}{c}t=2\sqrt{\frac{a}{\mu }}{\displaystyle \int \left(A^2+B^2-2AB\cos 2\left(s-s_o\right)\right)\text{d}s}\\ =\sqrt{\frac{a^3}{\mu }}\left(2s-\epsilon \sin 2\left(s-s_p\right)-\epsilon \sin 2s_p\right)\end{array}$

assuming the initial value of $t$ is also set to zero.

Note that when $s$ advances by π, the rocket returns to its original location; this implies that $T=2\pi \sqrt{\frac{a^3}{\mu }}$ is the time it takes the rocket to complete a full orbit.

Also note that, when $B=0$, the motion is circular, when $A=B$, it becomes rectilinear (running directly towards or away from the Sun), and when $B>A$, the rocket then orbits the primary in the retrograde (clockwise) direction.

When (7) is equal to zero, the value of $a$ no longer affects the equation; we make it equal to 1/2 to simplify the solution. Solving ${\mathbb{U}}^{″}=0$ results in

$\mathbb{U}=\mathbb{A}+\mathbb{B}s$

Starting at a specific initial location $r_0$ requires that $\mathbb{A}=:A\exp \left(-\text{i}\alpha \right)=\sqrt{r_0}$; to make the initial velocity equal to $v_0$, we also need

$\frac{\mathbb{U}{\mathbb{U}}^{\prime }}{r}\sqrt{\frac{\mu }{a}}=\frac{\mathbb{B}}{A\exp \left(\text{i}\alpha \right)}\sqrt{2\mu }=v_0$

Since (7) is equal to 0, the magnitude of $v_0$ must equal to $A\sqrt{2\mu }$, implying that $\left|\mathbb{B}\right|=1$. By choosing $\mathbb{B}:=\text{i}\exp \left(\text{i}\beta \right)$, the previous equation can be easily solved for $\beta$.

Orbital elements (optional) The solution can be expressed in the following form

$\mathbb{U}=\text{exp}\left(\text{i}\beta \right)\left(A\cos \left(\alpha +\beta \right)+\text{i}\left(s-s_v\right)\right)$

where $s_v=A\sin \left(\alpha +\beta \right)$ is the value of $s$ at the closest (to primary) approach (i.e., the path’s vertex).

The resulting trajectory is then

$r=\exp \left(2\text{i}\beta \right)\left(A^2{\cos }^2\left(\alpha +\beta \right)+2A\text{i}\left(s-s_v\right)\cos \left(\alpha +\beta \right)-{\left(s-s_v\right)}^2\right)$

which is a parabola [9] with its focus at the origin and its vertex at $A^2{\sin }^2\left(\alpha +\beta \right)$, further rotated by $2\beta$. To prove this, we show that the distance to the origin, namely

$r={\left(s-s_v\right)}^2+A^2{\cos }^2\left(\alpha +\beta \right)$

is the same as the pre-rotation distance from $r$ to the directrix, namely a vertical line with its $x$ coordinate equal to $2A^2{\cos }^2\left(\alpha +\beta \right)$, as can be easily verified.

This time

$t=\sqrt{\frac{2}{\mu }}\left(\frac{s^3}{3}-s_v{s}^2+\left(A^2{\cos }^2\left(\alpha +\beta \right)+s_v^2\right)s\right)$

When $\text{cos}\left(\alpha +\beta \right)<0$, the motion is retrograde.

Finally, when (7) is positive, we choose the value of $a$ so that $\frac{{\mathbb{U}}^{\prime }{\bar{\mathbb{U}}}^{\prime }-2a}{r}=1$ and (6) thus reads ${\mathbb{U}}^{″}-\mathbb{U}=0$; its general solution is

$\mathbb{U}=\mathbb{A}\exp \left(s\right)+\mathbb{B}\exp \left(-s\right)$

where

$\mathbb{A}+\mathbb{B}=\sqrt{r_0}$

$\mathbb{A}-\mathbb{B}=\frac{v_0{r}_0}{\sqrt{r_0}}\sqrt{\frac{a}{\mu }}=\frac{v_0}{v_0}\overline{\sqrt{r_0}}\sqrt{\frac{2a}{r_0}+1}$

with

$a=\frac{\mu }{2E_0}$

based on the initial value of (7).

Using (2) and (4), we can then easily compute the location and velocity of the rocket at any future value of $s;$ the following expression converts this value into the corresponding real time $t$ (assuming that its initial value is also 0)

$t=\sqrt{\frac{a}{\mu }}\left({\left|\mathbb{A}\right|}^2\left(\exp \left(2s\right)-1\right)-{\left|\mathbb{B}\right|}^2\left(\exp \left(-2s\right)-1\right)+2\left(\bar{\mathbb{A}}\mathbb{B}+\mathbb{A}\bar{\mathbb{B}}\right)s\right)$

Orbital elements (optional) To show that the resulting trajectory is one branch of a hyperbola [10] , we introduce $\mathbb{A}:=A\exp \left(\text{i}\alpha \right)$ and $\mathbb{B}:=B\exp \left(-\text{i}\beta \right)$; this enables us to write

$\mathbb{U}=\sqrt{AB}\exp \left(\text{i}\frac{\psi }{2}\right)\left(\exp \left(s-s_v+\text{i}\frac{\gamma }{2}\right)+\exp \left(-s+s_v-\text{i}\frac{\gamma }{2}\right)\right)$(14)

where $\gamma =\alpha +\beta$, $\psi =\alpha -\beta$ and $s_v=\text{ln}\sqrt{\frac{B}{A}}$ is the value of $s$ at the hyperbola’s vertex.

Based on $\frac{{\mathbb{U}}^{\prime }{\bar{\mathbb{U}}}^{\prime }-2a}{r}=1$, we get

$\begin{array}{c}2a=AB\left(\exp \left(2\left(s-s_v\right)\right)+\exp \left(-2\left(s-s_v\right)\right)-2\cos \gamma \right)\\ -AB\left(\exp \left(2\left(s-s_v\right)\right)+\exp \left(-2\left(s-s_v\right)\right)+2\cos \gamma \right)\\ =-4AB\cos \gamma \end{array}$

implying that $a=-AB\cos \gamma$.

(14) further implies that

$r=\exp \left(\text{i}\psi \right)AB\left(2+\exp \left(\text{i}\gamma \right){\text{e}}^{2\left(s-s_v\right)}+\exp \left(-\text{i}\gamma \right){\text{e}}^{-2\left(s-s_v\right)}\right)$

which represents the left branch of a hyperbola with its focus at the origin, its vertex at $2AB\left(1+\cos \gamma \right)$, its center at $2AB$, and two asymptotes going through the center at angles $\pm \gamma$, further rotated by $\psi$. This can be shown by comparing $r=AB\left(\exp \left(2\left(s-s_v\right)\right)+\exp \left(-2\left(s-s_v\right)\right)+2\cos \gamma \right)$ with the pre-rotational distance of $r$ to the second focus at $4AB$, which turns out to be $AB\left(\exp \left(2\left(s-s_v\right)\right)+\exp \left(-2\left(s-s_v\right)\right)-2\cos \gamma \right)$; the constant difference, namely $-4AB\cos \gamma$, then yields $2a$ (where $a$ now represents the apex-to-center distance). Note that $\text{cos}\gamma <0$ as $\gamma$ must be within the $\pi /2$ and $\pi$ range; the lower limit results in only a small deviation from the rocket’s original direction, while, at the upper limit, its direction is nearly reversed. When $\pi <\gamma <3\pi /2$, the motion becomes retrograde (note that $\gamma$ is measured from the $+x$ axis).

It can be shown that, to maximize an increase in total energy and thus enable it to reach higher planetary orbits, the extra velocity imparted to the rocket upon leaving Earth must be in the direction of the Earth’s own motion; the resulting orbit is called tangential. Denoting the total (Earth-inherited plus the extra boost) speed by $V$, the radius of Earth’s orbit by $a_0$ and the radius of the target orbit by $a_1$ (do not forget: both are assumed to be circular), our first task is to establish when and at what point will the rocket reach the target orbit.

Placing the Sun at the origin of our coordinates and orienting the $x$ axis so that the rocket’s initial location is $a_0$ (real and positive), and its initial velocity $v_0$ equals $\text{i}V$ (thus tangential to Earth’s orbit), we can routinely compute the transfer orbit’s $a$ value, based on (9), by

$a={\left(\frac{2}{a_0}-\frac{V^2}{\mu }\right)}^{-1}$

and construct the corresponding solution $\mathbb{U}$. In this section, we assume that $V<\sqrt{\frac{2\mu }{a_0}}$, making the orbit elliptical; for larger velocities, the transfer orbit becomes hyperbolic (to be dealt with next). We also need to verify that $a\ge \frac{a_0+a_2}{2}$, otherwise, the resulting trajectory never reaches the target orbit.

Deriving the next set of formulas is quite routine; we skip the details.

First, we get

$\mathbb{U}=\sqrt{a_0}\cos s+\text{i}\sqrt{2a-a_0}\sin s$

implying that the rocket’s distance from the Sun is

$r=a-\left(a-a_0\right)\cos 2s$

The target orbit is reached as soon as $r$ becomes equal to $a_1$, yielding the following solution for the corresponding value of $s$ (denoted $s_1$)

$\cos 2s_1=\frac{a-a_1}{a-a_0}$

We already know that this solution exists only when $a\ge \frac{a_0+a_1}{2}$; in the case of exact equality, the solution yields $2s_1=\pi$ and the target orbit is also reached tangentially (the corresponding transfer orbit is then called cotangential). For this orbit, the initial value $V$ needs to be equal to

$\sqrt{\frac{2a_1\mu }{a_0\left(a_0+a_1\right)}}$(15)

while the target orbit is approached with the final speed of

$\sqrt{\frac{2a_0\mu }{a_1\left(a_0+a_1\right)}}$(16)

We now return to a general solution: the location of where the transfer and target orbits intersect is clearly at distance $a_1$ from the Sun; we thus need only the corresponding polar-coordinate angle (denoted $\theta$). It is found based on the real component of $r={\mathbb{U}}^2$ evaluated at $s=s_1$, which turns out to be

$a_0-\frac{a_1-a_0}{a-a_0}a$

The last expression, divided by $a_1$ provides the value of $\cos \theta$, and consequently of $\theta$.

To establish the rocket’s velocity at this point, we evaluate

$v=\sqrt{\frac{\mu }{a}}\frac{\mathbb{U}{\mathbb{U}}^{\prime }}{r}$

at modified time $s_1$, thus getting

$v_1:=\sqrt{\frac{\mu }{a}}\left(-\frac{a}{a_1}\sin 2s_1+\text{i}\frac{\sqrt{\left(2a-a_0\right)a_0}}{a_1}\cos 2s_1\right)$

And finally, the real time of the same event (setting the initial value of time to 0) is given by

$t=\sqrt{\frac{a}{\mu }}\left(2a{s}_1-\left(a-a_0\right)\sin 2s_1\right)$

Note that the velocity of the target planet (to be encountered at that location) is

$v_2:=\text{i}\sqrt{\frac{\mu }{a_1}}\exp \left(\text{i}\theta \right)$

As mentioned already, the rocket now needs to use its engines to reduce the difference between the last two velocities to zero before it can land safely on the target planet.

Example The value of Sun’s gravitational mass $\mu$ is, to a sufficient approximation, equal to 1.33 × 10²⁰ meter³/second²; to simplify our notation, we make 10¹⁰ meters our unit of distance, and 10⁶ seconds (about 11.6 days) our unit of time, implying that 10 km/sec is our velocity unit.

Let us now explore transfer orbits taking a rocket from Earth to Mars; the Sun’s $\mu$ then equals 133 units, and the distance of Earth (Mars) from the Sun is 15 (22.5) units (the final reminder: both orbits are assumed to be circular and in the same plane). The two escape velocities (to leave Earth and to land on Mars; both are major contributors to fuel consumption) are 1.12 and 0.50 units, respectively (same for all transfer orbits). For simplicity (and maximum efficiency), we keep the initial velocity $v_0$ (after adding an extra boost $\text{Δ}{v}_0$ to its Earth-inherited value) tangential to the Earth’s orbit.

In the following table, we vary the $a$ value of the (always elliptic, by our choice) transfer orbit, from the smallest (cotangential), up to its $a\to \infty$ limit (effectively parabolic), while quoting the corresponding magnitude of the change in velocity needed at each end of the journey (after escaping Earth’s gravity and before landing on Mars), the total travel time $t$, and the location of the actual arrival (in terms of $\theta$).

Due to a substantial increase in fuel consumption (an exponential function of velocity changes), only solutions close to cotangential (the first column) are practical.

Going (more quickly) over the same steps when $V>\sqrt{\frac{2\mu }{a_0}}$, we get

$a={\left(\frac{V^2}{\mu }-\frac{2}{a_0}\right)}^{-1}$

where $a$ is the orbit’s center-to-apex distance, and

$\mathbb{U}=\sqrt{a_0}\cosh s+\text{i}\sqrt{2a+a_0}\sinh s$

implying

$r=-a+\left(a+a_0\right)\cosh 2s$

Setting $r=a_1$ and solving for $2s_1$, we get

$\cosh 2s_1=\frac{a+a_1}{a+a_0}$

resulting in an angle $\theta$ of the point of encounter, established from

$\text{cos}\theta =\frac{a_0}{a_1}-\frac{a}{a_1}\left(\frac{a_1-a_0}{a+a_0}\right)$

and the corresponding rocket’s velocity of

$v_1=\sqrt{\frac{\mu }{a}}\left(-\frac{a}{a_1}\sinh 2s_1+\text{i}\frac{\sqrt{\left(2a+a_0\right)a_0}}{a_1}\cosh 2s_1\right)$

The time of the encounter is

$t=\sqrt{\frac{a}{\mu }}\left(-2a{s}_1+\left(a+a_0\right)\sinh 2s_1\right)$

while the velocity of the target planet (being in that location at that time) is given by

$v_2=\text{i}\sqrt{\frac{\mu }{a_1}}\exp \left(\text{i}\theta \right)$