Astronomy Answers: Space Travel with Gravity

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You can do a lot of calculations about gravity, planetary orbits, and space travel.

## 1. Newton's Law of Gravity

Newton's Law of Gravity is suitable for calculations involving everyday gravity. If speeds close to the speed of light (about 300 000 kilometers per second) are involved, or if very high accuracy is required, then you should use the General Theory of Relativity of Albert Einstein, which is much more complex but (as far as we know) correct, whereas Newton's Law of Gravity yields noticeably different (and hence noticeably wrong) results in those circumstances. But for everyday cases Newton's law is sufficiently accurate, and it is a lot simpler than the General Theory of Relativity.

Newton's Law of Gravity yields the strength of the force of gravity between two objects with mass:

$$F_g = \frac{GMm}{r^2}$$

Here $$F_g$$ is the force of gravity, $$G$$ is the universal constant of gravity, $$M$$ is the one mass, $$m$$ is the other mass, and $$r$$ is the distance between the centers of mass of the objects. The universal constant of gravity is equal to 6.672 × 10−11 if the masses are measured in kilograms (kg), the distance in meters (m), and the force in newtons (N).

For example: What is the (strength of the force of) gravity between two persons of 50 kg each at 10 m from one another? Then $$M = m = 50 \text{ kg}$$ and $$r = 10 \text{ m}$$, so

$F_g = \frac{6.672×10^{−11} × 50 × 50}{10^2} = 1.668×10^{−9} \text{ N}$

That force is comparable to a weight on Earth of a very small grain of sand.

Often, the acceleration that you get thanks to the force of gravity is more interesting than the strengthe of the force of gravity itself. The acceleration is equal to the force divided by the mass that the force acts on (according to Newton's Second Law). The gravitational acceleration $$g$$ is therefore equal to

$$g = \frac{F_g}{m} = \frac{GM}{r^2} = \frac{γ}{r^2} \label{eq:g}$$

$$γ ≡ G M \label{eq:mu}$$

Here $$γ$$ is the gravity parameter of the system.

For example: How great is the gravitational acceleration at the surface of the Earth? The mass of the Earth is equal to $$M = 5.976×10^{24} \text{ kg}$$, so the gravity parameter of the Earth is

$γ = 6.672×10^{−11} × 5.976×10^{24} = 3.987×10^{14} \text{ m}^3/\text{s}^2$

The radius (the average distance between the center and the surface) of the Earth is $$r = 6378 \text{ km} = 6 378 000 \text{ m}$$, so the gravitational acceleration at the surface of the Earth is on average

$g = \frac{3.987×10^{14}}{(6.378×10^{6})^2} = 9.80 \text{ m}/\text{s}^2$

The acceleration of each of the persons from the previous example due to their gravity is

$g = \frac{1.668×10^{−9}}{50} = 3.336×10^{−11} \text{ m}/\text{s}^2$

which is very much smaller than the gravitational acceleration at the surface of the Earth.

If you divide the gravitational acceleration at the surface of another planet by that at the surface of the Earth, then you find

$$\frac{g}{g_\text{earth}} = \frac{M}{M_\text{earth}} \left( \frac{r_\text{earth}}{r} \right)^2$$

and if you measure the radius or distance $$r$$ in kilometers, then you find

$$\frac{g}{g_\text{earth}} = 40 678 880 \frac{M/M_\text{earth}}{r^2}$$

For example, the Sun has 332 946 times as much mass as the Earth and has a radius of 695 990 km. What is the gravitational acceleration at the surface of the Sun? It is

$\frac{g_\text{sun}}{g_\text{earth}} = \frac{40 678 880 × 332 946}{695 990^2} = 27.96$

## 2. The Law of Conservation of Energy

The Law of Conservation of Energy is a physical law that says that the total amount of energy in a closed system (i.e., a system for which there is no exhange of anything with the outside) cannot change. Energy can change its form, and can be exchanged with other parts of the system, but cannot disappear or appear. Different kinds of energy are recognized: kinetic energy (due to speed) $$\frac{1}{2} m v^2$$, and potential energy (due to location or state), of which the formula depends on the situation.

## 3. Law of Conservation of Angular Momentum

The Law of Conservation of Angular Momentum is a physical law that states that the total amount of angular momentum in a closed system cannot change. Angular momentum $$\vec{L}$$ of a point mass is defined as

$$\vec{L} = m \vec{r} × \vec{v}$$

where $$m$$ is the mass, $$\vec{r}$$ is the vector from the chosen origin to the position of the point mass, $$\vec{v}$$ is the velocity of the point mass relative to the origin, and × indicates the outer (or vector) product.

## 4. The Two-Body Problem

The Two-Body Problem is the problem of determining the motion of two bodies upon which act only the mutual forces of Newtonian gravity (i.e., gravity according to Newton's Law of Gravity). This "problem" has already been solved, but the name is still used for a situation where, to determine the motion of an object, all forces can be ignored except for the mutual forces of Newtonian gravity between that body and one other body — such as between the Sun and a planet, or between a planet and a moon, or between a spaceship and a Sun, planet, or moon.

### 4.1. The Relationship Between the Motions of the two Celestial Objects

Forces that act only between the two celestial objects cannot change the location of their common center of gravity (because of the Third Law of Newton, which says that each force from A on B is balanced by an equally great but oppositely directed force from B on A). This means that the trajectories of the two objects must have the same shape and that, as seen from the common center of gravity, the two objects are always in exactly opposite directions, at distances that are inversely proportional to the mass of the objects.

We use subscript $$_1$$ for the first celestial object, and subscript $$_2$$ for the second one. Their masses are $$m_1$$ and $$m_2$$, their relative masses compared to the total are $$μ_1$$ and $$μ_2$$, their distances from the common center of gravity are $$r_1$$ and $$r_2$$, and their orbital speeds $$v_1$$ and $$v_2$$. The total mass is $$m$$, the distance between the two objects is $$r$$, and the speed of the one relative to the other is $$v$$. Then we have

\begin{eqnarray} m_1 \| = \| μ_1 m \\ m_2 \| = \| μ_2 m \\ r_1 \| = \| μ_2 r \\ r_2 \| = \| μ_1 r \\ v_1 \| = \| μ_2 v \\ v_2 \| = \| μ_1 v \end{eqnarray}

Note that $$r_1$$ and $$v_1$$ are proportional to $$μ_2$$, not to $$μ_1$$ (and likewise with $$_2$$ instead of $$_1$$). The object with the least mass moves the most (because that one is easiest to get going), so the motion of each of the two objects is proportional to the mass of the other object.

If the mass of the first object is very much greater than that of the second object, then we have approximately $$μ_1 = 1$$ and $$μ_2 = 0$$, so then $$m_1 = m$$, $$r_1 = 0$$, $$r_2 = r$$, $$v_1 = 0$$, and $$v_2 = v$$, so then the massive object is (almost) motionless, and $$r$$ and $$v$$ describe the motion of the small object.

### 4.2. The Law of Conservation of Energy for the Two-Body Problem

In the two-body problem, the total amount $$E$$ of the energy in the system that can be affected by gravity is equal to

$$E = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 − \frac{G m_1 m_2}{r} \label{eq:Etot}$$

The first two terms on the right-hand side are the kinetic energies (energy due to speed) of the two objects, and the third term is the potential gravitational energy (energy due to location in a gravitational field).

The Law of Conservation of Energy says that this total amount $$E$$ of energy stays the same (if no forces other than gravity act). With the equations from Sec. 4.1 we find

$$E = \frac{1}{2} μ_1 μ_2^2 m v^2 + \frac{1}{2} μ_1^2 μ_2 m v^2 − \frac{G μ_1 μ_2 m^2}{r} = μ_1 μ_2 \left( \frac{1}{2} m v^2 − \frac{G m^2}{r} \right)$$

$$ε ≡ \frac{E}{μ_1 μ_2 m} = \frac{1}{2} v^2 − \frac{γ}{r} \label{eq:ε}$$

$$γ ≡ G m$$

It follows from these equations that in any orbit the speed decreases with increasing distance $$r$$.

An orbit can be open or closed. An open orbit is an orbit in which the body can get arbitrarily far from the other body. In a closed orbit, the body cannot get further away from the other body than a certain distance that is fixed for that orbit.

At infinite distance $$r$$, equation \eqref{eq:ε} yields

$$\frac{1}{2} v^2_∞ = ε$$

If $$ε ≥ 0$$ then this equation has the solution $$v_∞ = \sqrt{2ε}$$, so then the body can get infinitely far from the other body, so then this orbit is open. If $$ε \lt 0$$, then the equation has no solution, so then the body cannot get infinitely far from the other body, so then the orbit is closed. Orbits in the shape of a point, circle, ellipse, or finite line are closed and have $$ε \lt 0$$. Orbits in the shape of a parabola, hyperbola, or infinite line are open and have $$ε ≥ 0$$.

There is a connection between the size of the semimajor axis and the total energy per unit mass:

$$ε = −\frac{γ}{2a} \label{eq:e-bound}$$

This formula can be used also when $$ε$$ is positive (i.e., for hyperbolic orbits): then $$a$$ is negative.

For a circular orbit, $$r = a$$ is constant. Equation \eqref{eq:ε} then yields

$$v ≡ v_\text{c}(r) = \sqrt{\frac{γ}{r}}$$

We call this speed $$v_\text{c}(r)$$ the circular orbit speed. It is the speed that a celestial object has in a circular orbit at distance $$r$$ from the Sun.

### 4.3. The Law of Conservation of Momentum for the Two-Body Problem

For the two-body problem we choose the origin to be in the common center of gravity of the two celestial objects. Then we can write the angular momentum $$L$$ as

$$L = m_1 v_{\text{t1}} r_1 + m_2 v_{\text{t2}} r_2$$

where $$v_{\text{t1}}$$ is the component of the speed $$v_1$$ perpendicular to the direction to the origin, and likewise $$v_{\text{t2}}$$ is the tangential component of $$v_2$$. According to the Law of Conservation of Angular Momentum this $$L$$ is constant.

The sign of the angular momentum $$L$$ depends on whether the motion is clockwise or counterclockwise. We choose that counterclockwise motion is positive, and clockwise motion negative. This then holds as well for the tangential speeds.

Using the equations from Sec. 4.1, we find

\begin{eqnarray} v_{\text{t1}} \| = \| v_{\text{t}} μ_2 \\ v_{\text{t2}} \| = \| v_{\text{t}} μ_1 \\ L \| = \| μ_1 μ_2 m v_{\text{t}} r \\ l \| ≡ \| \frac{L}{μ_1 μ_2 m} = v_{\text{t}} r \label{eq:l} \end{eqnarray}

The total energy and the total angular momentum depend only on the relative position $$r$$ and speeds $$v$$, $$v_\text{t}$$ of the two objects with respect to each other, but not explicitly of their position or speed with respect to the center of gravity. With the equations given before, you can calculate the position or speed of each of the objects with respect to the center of gravity from their relative position or speed with respect to each other.

### 4.4. The Solutions to the Two-Body Problem

The solution of the two-body problem is that the bodies will move along orbits that are conic sections: circles, ellipses, parabolas, hyperbolas, straight lines, or a point. Both bodies follow an orbit of the same shape (for example, for both an ellipse with the same ratio of length to width), of which the size is inversely proportional to the mass of the body. The orbits of the two bodies lie in the same plane and share a focus (focal point), and seen from that shared focus both bodies are always in exactly opposite directions.

Both celestial bodies ply their orbits around their common center of gravity. In practice, one of the two bodies is usually far more massive than the other one. If the mass of the small body can be neglected compared with the mass of the large body, then the common center of gravity is in the center of the large body and the orbit of the large body has a negligible size. We'll assume that we have such a case here, with a spacecraft orbiting around the Sun, but the results can also be used if the two objects have more comparable masses, using the equations from Sec. 4.1ff.

Fig. 1: Diagram of a Solution of the Two-Body Problem
Figure 1 shows a typical solution (an ellipse) in red. The orbit is around the Sun (shown in yellow) in the center C, which is a focus of the conic section. Point P is the perifocus, and point A the apofocus. $$q$$ is the distance PC between the perifocus and the Sun, and $$Q$$ the distance CA between the Sun and the apofocus. The true anomaly $$ν$$ (nu) is the direction to the spacecraft, as seen from the Sun and measured counterclockwise from the perifocus.

Appropriate velocities are indicated in green for two points K and L at the same distance $$r$$ from the center. The total speed $$v$$ is the speed of the spacecraft along its orbit. The tangential speed $$v_\text{t}$$ is the component of the total speed that is perpendicular to the direction to the center. This speed is positive if it is in the counterclockwise direction (as in the diagram, as seen from the center), and negative if it is in the clockwise direction. The radial speed $$v_\text{r}$$ is the component of the total speed that is along the direction to the center. This is taken to be positive if it is aiming away from the center, and negative if it is aiming towards the center. The angle $$α$$ (alfa) indicates how much you have to rotate from the tangential direction in the direction away from the Sun to end up at the direction of motion of the spacecraft, and cannot be larger than 90°.

For a circle, ellipse, parabola, and hyperbola we have

$$r = \frac{l^2}{γ(1 + e \cos ν)} \label{eq:r0}$$

where $$r$$ is the distance between the Sun and the spacecraft, $$e$$ is the eccentricity, and $$ν$$ (nu) is the true anomaly. The value of $$e$$ depends on the shape and size of the orbit.

For those same orbits except parabolic ones we have

\begin{eqnarray} r \| = \| \frac{A}{1 + e \cos ν} \label{eq:r} \\ A \| = \| a(1 − e^2) \label{eq:A} \end{eqnarray}

where $$a$$ is the length of the semimajor axis, which depends on the size of the orbit. I call $$A$$ the "characteristic size" of the orbit.

If $$ν = 0$$ then we find the perifocus distance $$q$$, and if $$ν = 180°$$ then we find the apofocus distance $$Q$$ (if $$e \lt 1$$)

\begin{eqnarray} q \| = \| a(1 − e) \label{eq:q} \\ Q \| = \| a(1 + e) \label{eq:Q} \end{eqnarray}

Combining equations \eqref{eq:Etot} and \eqref{eq:r}, we find that the speed $$v$$ of the spacecraft relative to the Sun is

\begin{eqnarray} v \| = \| v_\text{A} \sqrt{1 + 2 e \cos ν + e^2} \\ v_\text{A} \| = \| \sqrt{\frac{γ}{A}} = v_\text{c}(A) \label{eq:vae} \end{eqnarray}

The speeds in the perifocus and apofocus are

\begin{eqnarray} v(q) \| = \| v_\text{A} (1 + e) \| = \| v_\text{c}(q) \sqrt{1 + e} \| = \| \sqrt{\frac{γ}{a} \frac{1 + e}{1 − e}} \label{eq:vq} \\ v(Q) \| = \| v_\text{A} (1 − e) \| = \| v_\text{c}(Q) \sqrt{1 − e} \| = \| \sqrt{\frac{γ}{a} \frac{1 − e}{1 + e}} \label{eq:vQ} \end{eqnarray}

so for a fixed perifocus distance $$q$$ the speed is $$\sqrt{1 + e}$$ times the circular orbit speed and for a fixed apofocus distance $$Q$$ (for $$e ≤ 1$$) it is $$\sqrt{1 − e}$$ times the circular orbit speed.

In the perifocus and apofocus, $$v = v_\text{t}$$, so then equation \eqref{eq:l} leads to

$$l = v_\text{t} r = v(q)q = v(Q)Q = v_\text{A} A = \sqrt{γa(1 − e^2)} \label{eq:l2}$$

For the tangential speed, it follows that

$$v_\text{t} = \frac{l}{r} = v_\text{A} (1 + e \cos ν) \label{eq:vt}$$

Using Pythagoras' theorem, we find

$$v_\text{r}^2 = v^2 − v_\text{t}^2$$

and from that

$$v_\text{r} = v_\text{A} e \sin ν \label{eq:vr}$$

We also find the following formulas:

\begin{eqnarray} ψ \| ≡ \| \frac{v}{v_\text{c}} \\ χ \| ≡ \| \frac{v_\text{t}}{v} \\ ρ \| ≡ \| \frac{v_\text{r}}{v} \\ v_\text{A} \| = \| \frac{γ}{v_\text{t}r} = \frac{v_\text{c}^2}{v_\text{t}} = \frac{v_\text{c} χ}{ψ} \\ e \cos ν \| = \| \frac{v_\text{t}}{v_\text{A}} − 1 = \frac{v_\text{t}^2r}{γ} − 1 = \frac{v_\text{t}^2}{v_\text{c}^2} − 1 = χ^2ψ^2 − 1 \\ e \sin ν \| = \| \frac{v_\text{r}}{v_\text{A}} = \frac{v_\text{t}v_\text{r}r}{γ} = \frac{v_\text{t}v_\text{r}}{v_\text{c}^2} = χρψ^2 \\ δ_v \| = \| \frac{2γ}{r} − v^2 = 2v_\text{c}^2 − v^2 = v_\text{c}^2 (2 − ψ^2) \\ \\ e^2 \| = \| 1 − \frac{r^2 v_\text{t}^2 δ_v}{γ^2} = 1 − \frac{v_\text{t}^2δ_v}{v_\text{c}^4} = 1 − χ^2ψ^2 (2 − ψ^2) \label{eq:toe2} \\ a \| = \| \frac{γ}{δ_v} = \frac{γ}{2v_\text{c}^2 − v^2} = \frac{rv_\text{c}^2}{2v_\text{c}^2 − v_\text{t}^2 − v_\text{r}^2} = \frac{r}{2 − ψ^2} \label{eq:toa} \end{eqnarray}

$$δ_v$$ (delta v) is equal to 0 when the spaceship has exactly the local escape velocity (and hence follows a parabolic orbit). $$δ_v$$ is positive when the spaceship does not reach the local escape velocity (and hence follows a circular orbit or an elliptic orbit) and is negative when the spaceship goes faster than the local escape velocity (and hence follows a hyperbolic orbit).

For the angle $$α$$ (alpha) between the tangential direction and the total speed we find

\begin{eqnarray} \sin α \| = \| \frac{e \sin(ν)}{\sqrt{1 + 2 e \cos ν + e^2}} \\ \tan α \| = \| \frac{e \sin(ν)}{|1 + e \cos ν|} \\ v_\text{t} \| = \| v \cos α \\ v_\text{r} \| = \| v \sin α \end{eqnarray}

The angular speed $$ω$$ of the true anomaly, measured in radians, is equal to

$$ω = \frac{v_\text{t}}{r} = \frac{v_\text{A}^3}{γ} (1 + e \cos ν)^2$$

Then the orbital period $$T$$ is equal to, for $$e \lt 1$$ (i.e., for circular orbits and elliptical orbits),

$$T = \frac{2π}{\sqrt{γ}} a^{3/2} \label{eq:jaar}$$

That the square of the orbital period is proportional to the cube of the length of the semimajor axis is Kepler's Third Law.

### 4.5. Circular Orbits

For a circular orbit we have

\begin{eqnarray} e \| = \| 0 \\ r \| = \| q = Q = A = a \\ v \| = \| ±\sqrt{\frac{γ}{r}} \label{eq:cirkel} \\ v(q) \| = \| v(Q) = v_\text{t} = v \\ v_\text{r} \| = \| 0 \\ α \| = \| 0 \end{eqnarray}

### 4.6. Elliptical Orbits

Fig. 2: Elliptic Orbit
Figure 2 shows an elliptic orbit in red, around the Sun C (in a focus of the ellipse) shown in yellow. P is the perifocus and A the apofocus.

Fig. 3: Ellipses with different eccentricities
The shape of an elliptical orbit is determined by its eccentricity $$e$$, which is between 0 and 1 for an elliptical orbit. The right (second) diagram shows ellipses with semimajor axis 1 and eccentricities between 0 (circle) and 0.95 in steps of 0.05. The little square is one of the foci of each of the ellipses.

Equations \eqref{eq:q} and \eqref{eq:q} yield

\begin{eqnarray} a \| = \| \frac{Q + q}{2} \label{eq:a} \\ e \| = \| \frac{Q − q}{Q + q} \label{eq:e} \end{eqnarray}

All formulas from Sec. 4.4 hold for ellipses.

The average angular speed $$n$$ (as seen from the focus) in an elliptical orbit, measured in radians per unit time, is equal to

$$n = \sqrt{\frac{γ}{a^3}}$$

If $$t$$ is the time since the last passage through the perifocus, then the mean anomaly $$M$$ is equal to

$$M = nt = \sqrt{\frac{γ}{a^3}} t \label{eq:muitt}$$

The eccentric anomaly $$E$$ is determined by Kepler's Equation:

$$M = E − e \sin E \label{eq:kepler}$$

If you have calculated the eccentric anomaly $$E$$ from this, then the true anomaly $$ν$$ (nu) follows from

\begin{eqnarray} \cos ν \| = \| \frac{e − \cos E}{e \cos E − 1} \\ \tan\left( \frac{1}{2}ν \right) \| = \| \sqrt{\frac{1 + e}{1 − e}} \tan\left( \frac{1}{2}E \right) \label{eq:half-e} \end{eqnarray}

The other way around,

\begin{eqnarray} \cos E \| = \| \frac{e + \cos ν}{1 + e \cos ν} \\ \tan\left( \frac{1}{2}E \right) \| = \| \sqrt{\frac{1 − e}{1 + e}} \tan\left( \frac{1}{2}ν \right) \label{eq:νtoE} \end{eqnarray}

The distance $$r$$ from the Sun is then

$$r = \frac{a(1 − e^2)}{1 + e \cos ν} = \frac{A}{1 + e \cos ν} = a(1 − e \cos E) \label{eq:ruite}$$

### 4.7. Parabolic Orbits

Fig. 4: Parabolic Orbit
Figure 4 shows a parabolic orbit in red, around the Sun C (in the focus of the parabola) shown in yellow. P is the perifocus.

For a parabolic orbit we have

\begin{eqnarray} a \| = \| ∞ \\ ε \| = \| 0 \\ e \| = \| 1 \\ r \| = \| \frac{2 q}{1 + \cos ν} \label{eq:rpar} \\ A \| = \| 2q \\ v_\text{A} \| = \| \sqrt{\frac{γ}{2q}} \\ v \| = \| v_\text{A} \sqrt{2 + 2 \cos ν} = \sqrt{2} v_\text{c}(r) \\ v_\text{t} \| = \| v_\text{A} (1 + \cos ν) \\ v_\text{r} \| = \| v_\text{A} \sin ν \\ \sin α \| = \| \frac{\sin(ν)}{\sqrt{2 + 2 \cos ν}} \end{eqnarray}

which means that the speed is everywhere $$\sqrt{2}$$ times as great as the local circular orbit speed.

For a parabolic orbit Equation \eqref{eq:A} cannot be used because for such an orbit $$a$$ is infinitely large and at the same time $$e$$ is equal to 1 so that the value $$a(1 − e^2) = ∞ × 0$$ cannot be calculated. However, Equation \eqref{eq:rpar} shows that $$A = 2q$$.

The angular speed $$n$$ (as seen from the focus) in the perifocus of a parabolic orbit, measured in radians per unit time, is equal to

$$n_p = \sqrt{\frac{2γ}{q^3}}$$

If $$t$$ is the time since the last passage through the perifocus, then the parabolic anomaly $$M_p$$ is equal to

$$M_p = n_pt = \sqrt{\frac{2γ}{q^3}} t \label{eq:parabool-m}$$

The true anomaly is then determined by

$$\frac{1}{2}M_p = \tan\left( \frac{1}{2}ν \right) + \frac{1}{3} \tan^3\left( \frac{1}{2}ν \right)$$

(This is a form of the Equation of Barker.) The solution for $$ν$$ follows from

\begin{eqnarray} \tan\left( \frac{1}{2}ν \right) \| = \| u − \frac{1}{u} \label{parabool-nu} \\ u \| = \| \left( \sqrt{W^2 + 1} + W \right)^{1/3} \label{eq:u} \\ W \| = \| \frac{3}{4}M_p \end{eqnarray}

The distance $$r$$ from the Sun is then

$$r = \frac{2q}{1 + \cos ν} = q \left( 1 + \tan^2\left( \frac{1}{2}ν \right) \right)$$

### 4.8. Hyperbolic Orbits

Fig. 5: Hyperbolic Orbit
Figure 5 shows a (curved) hyperbolic orbit in red, around the Sun in focus C in yellow. P is the perifocus.

Fig. 6: Hyperbolic Orbits With Different Eccentricities
The shape of the orbit depends on the eccentricity $$e$$. Figure 6 shows hyperbolic orbits with fixed semimajor axis length $$a = −1$$ with eccentricities ranging from 1.05 to 2 in steps of 0.05. The orbit for eccentricity 1.05 is shown in blue.

The formulas from section 4.4 hold also for hyperbolic orbits, but only if $$a \lt 0$$. Usually, $$a$$ (negative) occurs in formulas multiplied by $$1 − e$$ (also negative), so that the product is positive.

At large distances from the object, the orbit gets ever closer to two lines, the asymptotes of the hyperbola, which form the big cross in the diagram, with intersection X. We call the (perpendicular) distance BC from the object to the asymptotes $$b$$, indicated in the diagram in blue. This distance is equal to

$$b = \sqrt{a^2(1 − e^2)} = |a|\sqrt{e^2 − 1} = \sqrt{|a|A}$$

The half-angle $$γ$$ (gamma) of the hyperbola is indicated in the diagram in cyan and is equal to

$$γ = \arccos\left( \frac{1}{e} \right)$$

This angle varies between 0° (for $$e ↓ 1$$) and 90° (for $$e → ∞$$).

We call the distance of the object at C to the intersection X of the asymptotes $$c$$, indicated in the diagram in green. This distance is equal to

$$c = \frac{|b|}{\sin(γ)} = |a|e = \frac{qe}{e − 1}$$

If a spacecraft moves along the hyperbola in this figure from the "top" to the "bottom", then its direction has changed over an angle $$δ$$ (delta), displayed in the diagram in yellow. This angle is equal to

$$δ = 180° − 2γ = \arccos\left( 1 − \frac{2}{e^2} \right) \label{eq:delta}$$

This angle varies between 180° (for $$e ↓ 1$$) and 0° (for $$e → ∞$$). The geometry of the situation determines whether the change of direction is clockwise or counterclockwise.

For an hyperbolic orbit we have $$ε \gt 0$$. If we call the speed at infinite distance $$v_∞$$, then we find

\begin{eqnarray} v^2 \| = \| v_∞^2 + \frac{2γ}{r} \\ v_∞^2 \| = \| −\frac{γ}{a} \label{eq:hyp-e} \end{eqnarray}

$$l = v_\text{t} r = v(q) q = v_\text{A} A = \sqrt{γA} = \sqrt{γa(1 − e^2)} = \frac{γ \sqrt{e^2 − 1}}{v_∞} = v_∞b \label{eq:hyp-L}$$

The standard angular speed $$n$$ in a hyperbolic orbit is equal to

$$n = \sqrt{\frac{γ}{|a|^3}}$$

If $$t$$ is the time since the passage through the perifocus, then the mean anomaly $$M$$ is equal to

$$M = nt = \sqrt{\frac{γ}{|a|^3}} t$$

The eccentric anomaly $$H$$ is determined by

$$M = e \sinh H − H \label{eq:kepler_h}$$

When you have calculated the eccentric anomaly $$H$$ from this, then the true anomaly $$ν$$ follows from

\begin{eqnarray} \cos ν \| = \| \frac{e − \cosh H}{e \cosh H − 1} \\ \tan\left( \frac{1}{2}ν \right) \| = \| \sqrt{\frac{e + 1}{e − 1}} \tanh\left( \frac{1}{2}H \right) \end{eqnarray}

And the other way around

\begin{eqnarray} \cosh H \| = \| \frac{e + \cos ν}{1 + e \cos ν} \\ \tanh\left( \frac{1}{2}H \right) \| = \| \sqrt{\frac{e − 1}{e + 1}} \tan\left( \frac{1}{2}ν \right) \label{eq:νtoH} \end{eqnarray}

The distance $$r$$ from the Sun is then

$$r = \frac{A}{1 + e \cos ν} = \frac{a(1 − e^2)}{1 + e \cos ν} = a(1 − e \cosh H)$$

### 4.9. Near-Parabolic Orbits

Fig. 7: Orbits with Fixed Pericenter Distances
The formulas to calculate the true anomaly $$ν$$ from the time $$t$$ since the last passage through the perifocus look rather different for ellipses ($$e \lt 1$$), parabolas ($$e = 1$$), and hyperbolas ($$e \gt 1$$), but if the eccentricity $$e$$ is very close to 1, then the three kinds of orbits are difficult to tell apart at first glance, if you can see only the part near the perifocus, such as is the case for many comets that come close to the Sun only once. Figure 7 shows orbits with fixed perifocus distance $$q$$ for eccentricities ranging from 0 (circle) via 1 (parabola) to 2 (hyperbola) in steps of 0.1. The circle and ellipses are drawn in black, the parabola in blue, and the hyperbolas in red.

The page about Kepler's Equation explains how you can solve that equation, and provides approximations for the near-parabolic case (and others).

### 4.10. Straight-Line Orbits

If the total angular momentum $$l$$ is zero, then Eq. \eqref{eq:l} says that the tangential speed $$v_\text{t}$$ is zero, too, so then only the radial speed $$v_\text{r}$$ can be non-zero, so then you get an orbit that is a straight line through the common barycenter.

If the total energy $$ε$$ is negative then you can view the orbit as an infinitely thin ellipse. In the apofocus the radial speed is zero, and so the total speed is zero, too. Eq. \eqref{eq:vQ} then says that $$e = 1$$, and Eq. \eqref{eq:q} says that the perifocus distance $$q$$ is equal to zero.

If the total energy $$ε$$ is positive then you can view the orbit as an infinitely thin hyperbola. Equation \eqref{eq:hyp-L} then implies that $$e = 1$$ and $$b = 0$$, and equation \eqref{eq:delta} that $$δ = 180°$$, and equation \eqref{eq:q} that $$q = 0$$.

If the total energy $$ε$$ is zero then you can view the orbit as an infinitely thin parabola. The perifocus distance $$q$$ is equal to 0.

In equation \eqref{eq:g}, $$g$$ is the acceleration, i.e., (minus) the second derivative $$\ddot{r}$$ (with respect to time) of the distance $$r$$. With this, we can rewrite equation \eqref{eq:g} as

$$\ddot{r} r^2 = −G M = −γ$$

The minus sign is needed because gravity pulls downward (against the direction of increasing $$r$$).

#### 4.10.1. Straight Line With Escape Velocity

For an orbit that is a straight line, we can find a formula that describes the location as a function of time for one case. If we try the simple test solution $$r = b t^n$$, then we find for the first derivative (with respect to time)

$$\dot{r} = n b t^{n−1}$$

and for the second derivative with respect to time

$$\ddot{r} = n(n−1) b t^{n−2}$$

so

$$n(n−1) b t^{n−2} (b t^n)^2 = n(n−1) b^3 t^{3n−2} = −γ$$

The right-hand side of the equation is constant, so the left-hand side must be constant as well, but that is only possible if $$3n − 2 = 0$$ which means $$n = 2/3$$. For that value of $$n$$ the last equation simplifies to $$(−2/9) b^3 = −γ$$ which means $$b = ((9/2) γ)^{1/3}$$. With this we find the following solution for the motion along a straight line (in the two-body problem):

$$r = \left( \frac{9}{2} \right)^{1/3} (t\sqrt{γ})^{2/3} ≅ 1.651 (t\sqrt{γ})^{2/3}$$

If $$t = 0$$ then we find $$r = 0$$, so $$t = 0$$ is the moment at which the spaceship passes the center of the Sun (if the spaceship could pass through the Sun without noticing, and all of the Sun's mass were concentrated in one point in its center).

The speed $$\dot{r}$$ of this solution is

$$\dot{r} = n b t^{n−1} = \left( \frac{4γ}{3t} \right)^{1/3}$$

If $$t → ±∞$$ then $$\dot{r} → 0$$, so this solution is the one for which the speed at infinity is zero, which means that the total energy $$ε$$ is exactly zero. With that, this solution is a special case of the motion along a parabola. In this solution, the spaceship everywhere has just the escape speed from that location.

#### 4.10.2. Straight Line From Zero Velocity

If we start from zero velocity, then we can regard the orbit as an infinitely thin elliptical orbit, with its eccentricity equal to 1. In that case we cannot find a formula that yields the location as a function of time, but we can find a formula that yields the time as a function of location.

We can also easily calculate how long it takes to fall all the way into the Sun, because then we've completed exactly half of a complete orbit. If $$e = 1$$ then $$q = 0$$ so $$a = Q/2$$ and then the orbital period is $$(1/2)^{3/2}$$ times as long as for a circular orbit that touches the straight line at its furthest point. Falling takes half such a period, so $$(1/2)^{5/2} ≅ 0.1768 ≅ 6/34$$ times as long as for the circular orbit. Falling from the orbit of the Earth to the center of the Sun (if all mass of the Sun were concentrated into a single point) would therefore take about 6/34 year, or about 64 days.

We use the formulas that were provided above (section 4.5) for elliptical orbits. If we start from zero velocity, then we must be in the apofocus at $$r = Q$$. With $$e = 0$$ we find

$$a = Q/2 \label{eq:aq2}$$

Equation \eqref{eq:ruite} then yields

$$\cos E = 1 − \frac{2r}{Q}$$

and then we are still free to choose whether $$E$$ should be positive or negative. Equation \eqref{eq:kepler} then yields

$$M = E − \sin E = ±\left( \arccos\left( 1 − \frac{2r}{Q} \right) − \sqrt{1 − \left( 1 − \frac{2r}{Q} \right)^2} \right)$$

and, taking into account equation \eqref{eq:muitt},

$$t = \sqrt{\frac{Q^3}{8γ}} M = ±\sqrt{\frac{Q^3}{8γ}} \left( \arccos\left( 1 − \frac{2r}{Q} \right) − \sqrt{1 − \left( 1 − \frac{2r}{Q} \right)^2} \right)$$

We can use this equation to calculate for any distance $$r$$ between 0 and $$Q$$ at what time $$t$$ that distance is reached.

For $$r = 0$$ we find $$t = 0$$ and for $$r = Q$$ we find $$t = ±π\sqrt{Q^3/8γ}$$. We select the minus sign, so we fall from zero velocity at $$r = Q$$ at $$t = −π\sqrt{Q^3/8γ}$$ and end up at $$r = 0$$ at $$t = 0$$.

If we define

$$τ = 1 + \frac{t}{π\sqrt{\frac{Q^3}{8γ}}}$$

then

$$τ = 1 − \frac{1}{π} \left( \arccos\left( 1 − \frac{2r}{Q} \right) − \sqrt{1 − \left( 1 − \frac{2r}{Q} \right)^2} \right)$$

and then the falling begins at $$τ = 0$$ and lasts until $$τ = 1$$. The period in a circular orbit at distance $$Q$$ is equal to $$P_Q = 2π\sqrt{Q^3/γ}$$, hence

$$τ = 1 − \frac{4t\sqrt{2}}{P_Q} ≅ 1 − \frac{5.657 t}{P_Q}$$

Example: How long does it take to fall for exactly half the distance? Then $$r = Q/2$$ so $$τ = 1 − (1/π) (\arccos(0) − \sqrt{1 − 0^2}) = 1 − (1/π) (π/2) = 1/2$$, so it takes exactly half the time to fall half the distance.

We now have a formula to calculate $$t$$ if we know $$r$$, but we cannot turn that formula around to calculate $$r$$ from $$t$$. This is basically the same problem that we also have with Kepler's equation.

We can try to find an approximation formula. I found

$$r = Q\sqrt{1 − 2.23 τ^2 + 4.58 τ^3 − 6.62 τ^4 + 3.27 τ^5}$$

which has a standard deviation, compared to the true solution, of 0.006 $$Q$$, and a greatest deviation of 0.014 $$Q$$.

The speed can be calculated from Equations \eqref{eq:ε}, \eqref{eq:e-bound} and \eqref{eq:aq2}, and is

$$v = \sqrt{\frac{2γ (Q − r)}{Qr}}$$

### 4.11. Summary

The following table displays some facts of the various kinds of orbits. $$ε$$ is the total energy, $$e$$ the eccentricity, and $$L$$ the angular momentum.

type $${ε}$$ $${e}$$ $${L}$$
circle $${\lt 0}$$ 0 ≠ 0
ellipse $${\lt 0}$$ $${0 \lt e \lt 1}$$ ≠ 0
parabola 0 1 ≠ 0
hyperbola $${\gt 0}$$ $${\gt 1}$$ ≠ 0
line $${\gt −∞}$$ 0
point $${−\infty}$$ 0

## 5. Multi-Body Problem or n-Body Problem

The n-body problem is the problem of determining the motion of more than two bodies upon which act only the mutual forces of Newtonian gravity (i.e., gravity according to Newton's Law of Gravity). It turns out that this problem cannot in general be solved (such that you get formules with which you can calculate the positions of all bodies at all times), and that the motion of the bodies can even be chaotic (such that very small changes to the situation at the beginning can make enormous changes in the positions after some time).

### 5.1. Orbiting Regular Polygons

The n-body problem can be solved if the bodies all have the exact same mass and form a regular polygon, and if the velocity of each of the bodies lies in the plane of the polygon and is perpendicular to the direction to the center of the polygon and has the exact right size, because then each of the bodies follows a circular orbit of the same size around the center of the polygon. It is of course very unlikely that we'll find such a case in the Universe, so this case is not very interesting in practice.

We calculate how big that speed has to be. We define a regular polygon with $$n$$ vertices, with its center at coordinates (0,0). The coordinates of the vertices (where the bodies are) are

$$x_k = r \cos\left( \frac{2πk}{n} \right)$$

$$y_k = r \sin\left( \frac{2πk}{n} \right)$$

where all angles are measured in radians, and $$k$$ runs from 0 to $$n − 1$$ and indicates the different bodies. The situation is symmetric, so the magnitude of the gravitational acceleration of all bodies is the same and points at the center of the polygon. We choose to calculate the acceleration for the body with $$k = 0$$. That body has coordinates $$(r,0)$$ and its acceleration points to the middle at (0,0), so the acceleration only has a component in the x-direction.

According to Newton, that acceleration is equal to

$$a_x = ∑_k \frac{Gm (x_k − x_0)}{|r_{k0}|^3}$$

where $$m$$ is the mass of each of the bodies, the summation is for $$k$$ from 1 to $$n − 1$$, and $$r_{k0}$$ is the distance between bodies $$k$$ and 0.

$$\begin{split} r_{k0} \| = \sqrt{(x_k − x_0)^2 + (y_k − y_0)^2} \\ \| = \sqrt{(x_k − r)^2 + y_k^2} \\ \| = \sqrt{2r\left( 1 − \cos\left( \frac{2πk}{n} \right) \right)} \end{split}$$

It then follows for the magnitude of the acceleration that

$$\begin{split} |a_x| \| = ∑_k \frac{Gm}{r^2} \frac{1 − \cos\left( \dfrac{2πk}{n} \right)}{\left( 2 − 2\cos\left( \dfrac{2πk}{n} \right) \right)^{3/2}} \\ \| = \frac{Gm}{2^{3/2}r^2} ∑_k \frac{1}{\sqrt{1 − \cos\left( \dfrac{2πk}{n} \right)}} \end{split}$$

That acceleration must be equal to the centrifugal acceleration, which is defined by

$$a_c = \frac{v^2}{r}$$

where $$v$$ is the speed perpendicular to the direction to the center of the circle. It follows that

$$v^2 = r|a_x| = \frac{Gm}{2^{3/2}r} ∑_k \frac{1}{\sqrt{1 − \cos\left( \dfrac{2πk}{n} \right)}}$$

We define the speed to be equal to $$v = c_n \sqrt{GM/r}$$ where $$M = nm$$ is the total mass of all bodies together, and $$c_n$$ depends on the number $$n$$ of bodies. The next table shows some results.

ncn
2 $\frac{1}{2\sqrt{2}}$ 0.3535
3 $\frac{1}{3}\sqrt[4]{3}$ 0.4387
4 $\frac{1}{4}\sqrt{1 + 2\sqrt{2}}$ 0.4892
5 $\frac{\sqrt[4]{2}}{10}\sqrt{\sqrt{25 − 5\sqrt{5}} + 5\sqrt{5 − \sqrt{5}} − (\sqrt{5} − 5)\sqrt{5 + \sqrt{5}}}$ 0.5247
10 0.6215
100 0.8677
1000 1.0580
104 1.2190
105 1.3610
106 1.4896

There are two columns for $$c_n$$: one that shows the formula (for $$n$$ up to 5) and one that shows the approximate value. The formula gets more complicated when the number of bodies increases, so we don't show it anymore for large numbers.

## 6. Space Journeys

We now calculate a space journey from one planet to another one. Besides the two planets and the spacecraft, the Sun is also important, so we're dealing with at least a four-body problem, for which no solutions can be found in general. We can, however, find approximate solutions.

It is very expensive to bring mass into space, so it is desirable to find a route for which you need the least amount of propellant, which corresponds to the smallest possible changes in velocity (compared to what you get for free due to gravity).

You obviously need propellant to be able to get from your home planet's surface into space. You also need propellant to synchronize your speed with that of the target planet, or else you fly right into or past that planet.

### 6.1. Hohmann Transfer Orbits

A Hohmann orbit is an elliptical orbit of which the perifocus lies on the orbit of the one planet and the apofocus on the orbit of the other planet. Such an orbit is often the cheapest ― but trajectories from the Interplanetary Transport Network discovered in 1997 can be even cheaper.

Fig. 8: Hohmann Transfer Orbit
Figure 8 shows a Hohmann transfer orbit in red, which touches the small orbit (of planet 1) at the bottom and the large orbit (of planet 2) at the top. The small blue circles show the locations of the planets when the spacecraft leaves planet 1, and the small yellow circle shows the position of planet 2 when the spacecraft arrives.

The trick of a Hohmann orbit is to use the engines of your spacecraft only twice: once to transfer from the orbit of planet 1 into the Hohmann transfer orbit, and once more to transfer from the Hohmann transfer orbit into the orbit of planet 2.

We call the radius of the orbit of planet 1 $$r_1$$ and that of planet 2 $$r_2$$. The smallest of those two is the perifocus distance $$q$$ of the Hohmann orbit, and the largest is the apofocus distance $$Q$$.

\begin{eqnarray} q \| = \| \min(r_1, r_2) \\ Q \| = \| \max(r_1, r_2) \\ Q \| ≥ \| q \end{eqnarray}

Then the semimajor axis $$a$$ and eccentricity $$e$$ of the Hohmann transfer orbit are (according to equation \eqref{eq:a}ff), and also the "characteristic size" $$A$$

\begin{eqnarray} a \| = \| \frac{r_2 + r_1}{2} \| = \| \frac{Q + q}{2} \\ e \| = \| \frac{|r_2 − r_1|}{r_2 + r_1} \| = \| \frac{Q − q}{Q + q} \\ A \| ≡ \| a(1 − e^2) \| = \| \frac{2Qq}{Q + q} \end{eqnarray}

The speeds in the Hohmann transfer orbit at the perifocus and apofocus of the journey are, according to equation \eqref{eq:vq}ff,

\begin{eqnarray} v_q \| = \| v_\text{A} (1 + e) \\ v_Q \| = \| v_\text{A} (1 − e) \\ v_\text{A} \| = \| v_\text{c}(A) = \sqrt{\frac{γ}{a(1 − e^2)}} \end{eqnarray}

The speeds in the circular orbits are, according to equation \eqref{eq:cirkel},

\begin{eqnarray} v_\text{c}(q) \| = \| \sqrt{\frac{γ}{q}} = v_\text{A} \sqrt{1 + e} \\ v_\text{c}(Q) \| = \| \sqrt{\frac{γ}{Q}} = v_\text{A} \sqrt{1 − e} \end{eqnarray}

If your spacecraft travels around the Sun in the same direction as the departure planet and the target planet, then the speed differences that your spacecraft's engines will have to produce are

\begin{eqnarray} ∆v_q \| = \| v_q − v_\text{c}(q) = v_\text{A} \left( 1 + e − \sqrt{1 + e} \right) \\ ∆v_Q \| = \| v_\text{c}(Q) − v_Q = v_\text{A} \left( \sqrt{1 − e} − (1 − e) \right) \end{eqnarray}

If your spaceship travels around the Sun in the opposite direction as the departure planet and the target planet, then the speed differences are much greater. If you can provide that much speed difference then you do not need the frugal Hohmann transfer orbit.

Fig. 9: Speed differences in a Hohmann transfer orbit, compared to the characteristic speed

In a Hohmann transfer orbit to an outer planet (further from the Sun), you begin in the perifocus of the orbit and go faster than the departure planet when you leave it, and slower than the target planet when you arrive there. In a Hohmann transfer orbit to an inner planet (closer to the Sun), it is the other way around: You begin in the apofocus and go slower than the departure planet when you leave it, and faster than the target planet when you arrive there.

The total speed difference is a measure of the cost, and is

$$∆v_\text{tot} = ∆v_q + ∆v_Q = v_\text{A} (2e + \sqrt{1 − e} − \sqrt{1 + e})$$

Figure 9 shows how these speeds vary with eccentricity $$e$$. The greatest value of $$∆v_\text{tot}/v_\text{A}$$ is reached for

$e = \frac{\sqrt{111 − \sqrt{33}}}{8\sqrt{2}} ≈ 0.9068$

Then

$∆v_\text{tot} = \frac{\sqrt{207 − 33\sqrt{33}}}{4\sqrt{2}} v_\text{A} ≈ 0.7380 v_\text{A}$

But because $$v_\text{A}$$ itself depends on $$e$$, this highest value is not very useful. It is more useful to compare $$∆v_\text{tot}$$ with $$v_\text{c}(q)$$ if we travel to an outer planet or with $$v_\text{c}(Q)$$ if we travel to an inner planet, because those speeds are constant for the departure planet.

Fig. 10: Speed differences in a Hohmann transfer orbit, compared to the circular orbit speed at perifocus

The comparison with $$v_\text{c}(q)$$ is displayed in Figure 10. Here

\begin{eqnarray} ∆v_q \| = \| v_\text{c}(q) \frac{1 + e − \sqrt{1 + e}}{\sqrt{1 + e}} = v_\text{c}(q) \left( \sqrt{1 + e} − 1 \right) \\ ∆v_Q \| = \| v_\text{c}(q) \frac{\sqrt{1 − e} − (1 − e)}{\sqrt{1 + e}} \\ ∆v_\text{tot} \| = \| ∆v_q + ∆v_Q \end{eqnarray}

The highest value of $$∆v_\text{tot}$$ is achieved when

$e = 2 \cos(20°) − 1 ≈ 0.8794$

and is approximately equal to $$∆v_\text{tot} ≈ 0.5363 v_q$$. At that $$e$$ we have $$Q/q ≈ 15.5817$$, so $$∆v_\text{tot}$$ is greatest for a Hohmann transfer orbit to a planet that is that many times as far from the Sun as the departure planet.

Fig. 11: Speed differences in a Hohmann transfer orbit, compared to the circular orbit speed at apofocus

The comparison with $$v_\text{c}(Q)$$ is displayed in Figure 11. Here,

\begin{eqnarray} ∆v_q \| = \| v_\text{c}(Q) \frac{1 + e − \sqrt{1 + e}}{\sqrt{1 − e}} \\ ∆v_Q \| = \| v_\text{c}(Q) \frac{\sqrt{1 − e} − (1 − e)}{\sqrt{1 − e}} = v_\text{c}(Q) \left( 1 − \sqrt{1 − e} \right) \\ ∆v_\text{tot} \| = \| ∆v_q + ∆v_Q \end{eqnarray}

Now $$∆v_\text{tot}$$ rises without limit when $$e$$ grows from 0 to 1, i.e., when the perifocus gets ever closer to the zero point. For $$q/Q ≈ 0.409$$, $$v/v_Q$$ reaches the same value as the highest value of $$v/v_q$$, so for smaller $$q/Q$$ than that it is more expensive (in terms of $$∆v_\text{tot}$$) to travel to an inner planet than to any outer planet.

For a Hohmann transfer orbit from the Earth to Mars (where we pretend that those planets follow circular orbits) we have $$q = 1 \text{ AU}$$ and $$Q = 1.5237 \text{ AU}$$, so

\begin{eqnarray*} a \| = \frac{1 + 1.5237}{2} = 1.2619 \text{ AU} \\ e \| = \frac{1.5237 − 1}{1.5237 + 1} = 0.2075 \\ A \| = a(1 − e^2) = 1.2075 \\ v_\text{A} \| = \sqrt{\frac{γ}{a(1 − e^2)}} = 27.0 \text{ km/s} \end{eqnarray*}

(with $$\sqrt{γ}$$ equal to 29.8 km/s for all orbits around the Sun if $$a$$ is measured in AU). It follows that

\begin{eqnarray*} v_q \| = \| v_\text{A} (1 + 0.2075) = 1.208 v_\text{A} = 32.4 \text{ km/s} \\ v_Q \| = \| v_\text{A} (1 − 0.2075) = 0.792 v_\text{A} = 21.4 \text{ km/s} \\ v_\text{c}(q) \| = \| v_\text{A} \sqrt{1 + 0.2075} = 1.099 v_\text{A} = 29.8 \text{ km/s} \\ v_\text{c}(Q) \| = \| v_\text{A} \sqrt{1 − 0.2075} = 0.890 v_\text{A} = 24.1 \text{ km/s} \\ ∆v_q \| = \| 32.4 − 29.8 = 2.9 \text{ km/s} \\ ∆v_Q \| = \| 24.1 − 21.4 = 2.6 \text{ km/s} \\ ∆v_\text{tot} \| = \| 5.6 \text{ km/s} \end{eqnarray*}

The below table shows corresponding results for Hohmann transfers from Earth to all other planets.

Mercury Venus Mars Jupiter Saturn Uranus Neptune
$${a}$$ 0.6935 0.8617 1.262 3.101 5.277 10.11 15.56 AU
$${e}$$ 0.4419 0.1605 0.2075 0.6776 0.8105 0.9011 0.9357
$${v_\text{A}}$$ 39.75 32.42 27.03 22.93 22.07 21.54 21.35 km/s
$${v_q}$$ 57.32 37.62 32.64 38.47 39.96 40.95 41.32 km/s
$${v_Q}$$ 22.19 27.21 21.42 7.394 4.182 2.131 1.372 km/s
$${v_\text{c}(Q)}$$ 47.74 34.92 24.06 13.02 9.608 6.775 5.413 km/s
$${∆v_q}$$ 27.62 7.920 2.936 8.768 10.26 11.25 11.62 km/s
$${∆v_Q}$$ 25.55 7.709 2.641 5.627 5.426 4.644 4.040 km/s
$${∆v_\text{tot}}$$ 53.17 15.63 5.578 14.39 15.69 15.89 15.66 km/s

Of Hohmann journeys to all planets, the one to Mars is by far the cheapest, with the smallest $$∆v_\text{tot}$$.

It is simple to calculate the travel time $$t$$ in a Hohmann transfer orbit between two planets, because that is exactly one half of the orbital period of the whole ellipse, and that period follows from equation \eqref{eq:jaar}.

$$t_\text{H} = \frac{1}{2} \sqrt{\frac{4π^2}{γ}} a^{3/2} = \sqrt{\frac{π^2}{γ}} a^{3/2}\label{eq:th}$$

For Hohmann transfer orbits around the Sun, $$\sqrt{4π^2/γ}$$ equals 1 year/AU3/2, so then $$t_\text{H}/[\text{year}] = (a/[\text{AU}])^{3/2}$$.

The Hohmann transfer orbit from the Earth to Mars takes $$\frac{1}{2} × 1.2619^{3/2} = 0.7088$$ years, or about 259 days.

Mercury Venus Mars Jupiter Saturn Uranus Neptune
$${t_\text{H}}$$ 0.2888 0.3999 0.7087 2.731 6.062 16.07 30.67 year

A journey to another planet via a Hohmann transfer orbit cannot start at just any time you want, because the travel time is fixed (for planets in circular orbits) and you would probably like for the target planet to be at the end of the Hohmann transfer orbit just when your spacecraft is there, too, so you must leave the home planet at just the right time. When is that?

First we look at the relative position of the two planets. We measure positions along the orbits in degrees, and the two planets are in the same relative position if the line through them also goes through the center of the Sun. We define the position of the first planet at time $$t = 0$$ to be zero. Then the positions $$φ_1, φ_2$$ of the planets as a function of time are

\begin{eqnarray} φ_1(t) \| = \| n_1 t \pmod{360°} \\ φ_2(t) \| = \| n_2 t + φ_0 \pmod{360°} \\ n_1 \| = \| \sqrt{\frac{γ}{2π}} r_1^{−3/2} \\ n_2 \| = \| \sqrt{\frac{γ}{2π}} r_2^{−3/2} \end{eqnarray}

where $$t$$ is the time, $$φ_0$$ is the position of planet 2 at $$t = 0$$ (see the diagram above), and $$n_1, n_2$$ are the angular speeds of the planets (measured in degrees per unit time). The "mod 360°" notation means that complete orbits can be disregarded: when you've gone around 2.4 times (i.e., 2.4 times 360°) then you're in the same place as when you've gone around 0.4 times (i.e., 0.4 times 360°).

If distances are measured in AU and times in years, and if we look at Hohmann transfer orbits around the Sun, then $$\sqrt{γ/2π}$$ is equal to 360.

A conjunction takes place whenever $$φ_1$$ and $$φ_2$$ are equal, disregarding complete orbits. This means that the faster planet must have completed one more orbit than the slower planet, or that

$$φ_1 = φ_2 ⇔ n_1 t = n_2 t + φ_0 \pmod{360°} ⇔ t = \frac{φ_0}{n_1 − n_2} \pmod{\frac{360°}{n_1 − n_2}} \label{eq:conj}$$

so the time between two consecutive conjunctions is

$$t_\text{conj} = \frac{360°}{|n_1 − n_2|}$$

We set time $$t$$ to zero when our spaceship leaves planet 1. Then, after a time $$t_\text{H}$$ (see equation \eqref{eq:th}) has passed, the spaceship has reached the orbit of planet 2, halfway around the Sun. This means that at $$t = t_\text{H}$$, the position of the second planet should be equal to $$φ_2 = 180° \pmod{360°}$$. From this and equation \eqref{eq:conj} we deduce the value of $$φ_0$$ (i.e., where planet 2 must be when we start the journey from planet 1):

$$φ_0 = 180° − n_2 t_\text{H} \pmod{360°}$$

Suppose we leave the second planet on our return trip at time $$t = t_2$$. Then we want the first planet to be halfway around the Sun when our spaceship arrives there, so

$$\begin{split} φ_1(t_2 + t_\text{H}) \| = φ_2(t_2) + 180° \pmod{360°} \\ \| ⇒ n_1 t_2 + n_1 t_\text{H} = n_2 t_2 + φ_0 + 180° \pmod{360°} \\ \| ⇒ (n_1 − n_2) t_2 = −n_1 t_\text{H} + φ_0 + 180° = −(n_1 + n_2) t_\text{H} \pmod{360°} \\ \| ⇒ t_2 = −\frac{n_1 + n_2}{n_1 − n_2} t_\text{H} \pmod{\frac{360°}{n_1 − n_2} = t_\text{conj}} \end{split}$$

so a new opportunity to return comes after every period $$tcircle$$, as expected. The first opportunity to return after time $$t$$ is at

$$t_2 = \left(\left( −\frac{n_1 + n_2}{n_1 − n_2} t_\text{H} − t \right) \bmod{t_\text{conj}}\right) + t$$

The return journey takes $$t_\text{H}$$ again so the earliest time at which you can be back on your starting planet is

$$t_3 = t_2 + t_\text{H}$$

In the case of a return trip from the Earth to Mars, we found earlier that $$t = 0.7087$$ years (259 days), and we find now that

\begin{align*} n_1 \| = 360 \text{ degrees/year} \\ n_2 \| = 360 × 1.5237^{−3/2} = 191,4 \text{ degrees/year} \\ φ_0 \| = 180° − 191.4° × 0.7087 = 44.3° \pmod{360°} \\ t_\text{conj} \| = \frac{360}{360 − 191.4} = 2.1353 \text{ year} = 780 \text{ days} \\ t_2 \| = −\frac{360 + 191.4}{360 − 191.4} × 0.7087 = −2.3180 \pmod{2.1353 \text{ year}} \end{align*}

The first $$t_2$$ after $$t = t_\text{H}$$ is at $$t_2 = −2.3180 + 2×2.1353 = 1.9526$$ years (713 days). Add $$t_\text{H}$$ for the return journey, and the first opportunity to be back on Earth is at

$t_3 = 1.9526 + 0.7087 = 2.6613 \text{ year}$

A return trip to Mars using Hohmann transfer trajectories would take at least about 972 days in total.

Mercury Venus Mars Jupiter Saturn Uranus Neptune
$${t_\text{conj}}$$ 0.317 1.599 2.135 1.092 1.035 1.012 1.006 year
$${φ_0}$$ 108.3 306.0 44.34 97.16 106.1 111.3 113.2
°
$${t_2}$$ 0.4720 1.679 1.953 3.319 6.969 16.94 31.33 year
$${t_3}$$ 0.7608 2.079 2.661 6.050 13.03 33.01 62.00 year

Where is planet 2 when the spacecraft leaves planet 1? The distance $$Δ_{12}$$ from planet 2 to planet 1 is then

$$Δ_{12} = \sqrt{r_1^2 + r_2^2 − 2 r_1 r_2 \cos φ_0}$$

The elongation $$ψ$$ of planet 2 as seen from planet 1 (see the preceding diagram) is then determined by

$$\sin ψ = \frac{r_2}{Δ_{12}} \sin φ_0$$

We find for the journey from Earth to Mars that

\begin{eqnarray*} Δ_{12} \| = \| \sqrt{1^2 + 1.5237^2 − 2×1×1.5237×\cos(44.3°)} = 1.0688 \text{ AU} \\ ψ \| = \| \arcsin((1.5237/1.0688) × \sin(44.3°)) = 85.2° \end{eqnarray*}

so we can start a journey to Mars along a Hohmann transfer orbit when Mars is about 85 degrees from the Sun, as seen from Earth. Mars has to be east of the Sun then, so then it is visible in the evening.

Mercury Venus Mars Jupiter Saturn Uranus Neptune
$${∆_{12}}$$ 1.180 0.8207 1.069 5.419 9.879 19.60 30.52 AU
$${ψ}$$ 18.14 −45.51 85.19 72.29 68.31 65.95 65.11
°

### 6.2. Gravity Assists

You can save a lot of propellant if on your way to some far-away planet you first swing by one or more other planets, because those other planets can give you a so-called gravity assist.

#### 6.2.1. A Spot of Tennis

This works in about the same way as for a ball that is hit by a tennis racket. In the ideal case (when the collision is fully elastic and we turn off other effects − such as gravity − and if the ball does not permanently influence the racket) the speed at which the ball leaves the racket is the same as the speed at which it hit the racket − if those speeds are measured relative to the racket. (If the collision is not fully elastic, then the outgoing speed will be less than the incoming speed.)

Suppose that the racket has horizontal speed $$v_\text{r}$$ to the right relative to the ground, and that the ball has speed $$v_1$$ to the right. (We count speeds to the right as positive, and those to the left as negative.) Then the incoming speed $$v_2$$ of the ball (compared to the racket) is equal to

$$v_2 = v_1 − v_\text{r}$$

The outgoing speed $$v_3$$ (compared to the racket) is equally great but in the opposite direction, so

$$v_3 = −v_2 = v_\text{r} − v_1$$

The velocity relative to the ground is $$v_\text{r}$$ greater than the velocity relative to the racket, so the return velocity $$v_4$$ compared to the ground is

$$v_4 = v_3 + v_\text{r} = 2 v_\text{r} − v_1$$

If the racket is stationary ($$v_\text{r} = 0$$), then $$v_4 = −v_1$$ so then the return speed is equally large (but in the opposite direction) as the initial speed. If the racket itself moves, then the return speed will be different from the initial speed. You can make the ball go faster after the hit (compared to the ground) by moving the racket towards the ball. If instead you move the racket in the same direction in which the ball is going, then the ball will be slower after the hit (if it can catch up to the racket at all).

We learn from this spot of tennis that in a collision between a light and a heavy object the final speed of the light object depends also on the speed of the heavy object. If you use the right strategy, then the light object can gain speed.

#### 6.2.2. And Now With a Planet: The Speed Relative to the Planet

If a spacecraft flies by a planet (without using its engines), then this is comparable to an elastic collision. When the spacecraft has entered the domain of the planet (where the planet's gravity has a lot more influence than the Sun's gravity), then as seen from the planet it follows a hyperbolic orbit around the planet. The Law of Conservation of Energy holds to a good approximation for the planet and the spacecraft together, and the absence of drag forces in space means that the spacecraft eventually leaves the planet at the same speed (but in a different direction) relative to the planet.

Suppose that the spacecraft flies around the Sun in an orbit with semimajor axis $$a$$ and eccentricity $$e$$ and that it approaches the planet that moves in a circular orbit at distance $$r$$ from the Sun. How fast does the spacecraft go relative to the planet?

In what follows we assume that the spacecraft and the planet move around the Sun in the same plane.

The formulas from section 4.4 yield, for the spacecraft,

\begin{eqnarray} v \| = \| v_\text{A} \sqrt{1 + 2 e \cos ν + e^2} \\ v_\text{t} \| = \| v \cos α \\ v_\text{r} \| = \| v \sin α \end{eqnarray}

The speed $$v_0$$ of the planet relative to the Sun follows from equation \eqref{eq:cirkel}:

$$v_0 = \sqrt{\frac{γ}{r}}$$

Then the tangential speed $$v_\text{tp}$$ of the spacecraft relative to the planet is

$$v_\text{tp} = v_\text{t} − v_0 = v \cos(α) − v_0$$

The radial speed $$v_\text{rp}$$ of the spacecraft compared to the planet is equal to

$$v_\text{rp} = v_\text{r} = v \sin α$$

because the speed of the planet has no radial component. The total speed $$v_\text{p}$$ relative to the planet is then (again according to Pythagoras)

$$\begin{split} v_\text{p} \| = \sqrt{v_\text{rp}^2 + v_\text{tp}^2} \\ \| = \sqrt{v_\text{r}^2 + (v_\text{t} − v_0)^2} \\ \| = \sqrt{v^2 + v_0^2 − 2 v_\text{t} v_0} \\ \| = \sqrt{v^2 + v_0^2 − 2 v v_0 \cos α} \end{split}$$

#### 6.2.3. The Transition From the Solar Domain to the Planetary Domain

We have now calculated how fast the spacecraft would go relative to the planet at the planet's orbit if the planet had no mass and hence no gravity, but the planet does have mass and gravity (or else we could not get a gravity assist from that planet, so then it would not be interesting for the current problem). It is not possible in general to find a formula that says where the spacecraft is at any given moment, under influence of the gravity of both the planet and the Sun. So, we'll make some simplifying assumptions that are not entirely correct but do make the calculations a lot simpler.

We use the following assumptions:

1. The mass of the spacecraft can be ignored compared to the mass of the planets and the Sun.
2. If the spacecraft is "far" from the planets, then only the gravity of the Sun counts. The spacecraft is then in the domain of the Sun.
3. If the spacecraft is "near" one of the planets, then only the gravity of that planet counts. The spacecraft is then in the domain of that planet.
4. The solar domain trajectory of the spacecraft touches (near the planet's orbit) the asymptote of the planet domain trajectory of the spacecraft. See below.
5. The speed of the spacecraft at infinite distance from the planet if the Sun had no influence is equal in size and direction to the speed of the spacecraft at the location of the planet if the planet had no influence.

and from this we get a method for transferring from the domain of the Sun to the domain of the planet:

1. Calculate the speed (in magnitude and direction) that the spacecraft would have near the planet relative to the planet if the planet had no influence (i.e., still in the domain of the Sun).
2. Declare this speed (in magnitude and direction) equal to the speed at infinite distance in a hyperbolic orbit around the planet if the Sun had no influence (i.e., in the domain of the planet).

Fig. 12: Transition Orbit
Figure 12 illustrates assumption 4. The yellow trajectory is the one of the planet (which is now at point C). The blue trajectory shows how the spacecraft would move if there were no planet but the gravity of the Sun did act. The red trajectory shows how the spacecraft could move if the Sun were not there (but the gravity from the planet did act). The red orbit was chosen such that the appropriate asymptote (the black line) of that orbit touches the blue orbit in the intersection D of the asymptote (blue) and the planet's orbit (yellow).

The real trajectory of the spacecraft (under the simultaneous influence of the gravity of both the Sun and the planet) will be something that looks like the green orbit from the diagram. When the spacecraft is still far from the planet (in the domain of the Sun), then that orbit follows the blue ("Sun only") orbit. When the spacecraft has gotten very close to the planet (in the domain of the planet), then the green orbit looks at lot like the red ("planet only") orbit instead, as if the spacecraft has switched from the blue to the red orbit.

To show that assumption 4 is an approximation, I've kept the green and red orbits slightly apart.

#### 6.2.4. The Orbit in the Planetary Domain

According to assumption 5, we set the speed $$v_\text{p}$$ relative to the planet equal to the speed $$v_\text{h}$$ of the hyperbolic orbit in the planetary domain. Then we find (also using equation \eqref{eq:hyp-e})

$$v^2 + v_0^2 − 2 v v_0 \cos α = v_\text{h}^2 = −\frac{γ_\text{p}}{a_\text{p}} \label{eq:e-planet}$$

where $$γ_\text{p}$$ is the gravity parameter of the planet (like $$γ$$ is of the Sun) and $$a_\text{p}$$ describes the hyperbolic orbit around the planet (like $$a$$ describes the orbit around the Sun). We assume that $$a_\text{p}$$ is negative and the orbit around the planet is therefore a hyperbola. If $$a_\text{p}$$ were positive, then the orbit around the planet would be an ellipse, which violates assumption 4. In such a case, the current set of assumptions don't work.

For the spacecraft travelling from Earth to Mars via a Hohmann orbit we found $$v_\text{h}$$ equal to 2.641 km/s. The mass of Mars is 0.107 times that of Earth, so the gravity parameter of Mars is 0.107 times that of the Earth as well:

$γ_\text{p} = 0.107 × 3.987×10^{14} = 4.266×10^{13} \text{ m}^3/\text{s}^2$

Then

$a_\text{p} = −\frac{γ_\text{p}}{v^2_\text{h}} = −\frac{4.266×10^{13}}{2641^2} = −6.115×10^6 \text{ m} = −6115 \text{ km}$

Mercury Venus Mars Jupiter Saturn Uranus Neptune
$${a_\text{p}}$$ −238.7 −44.61 × 103 −6115 −4.004 × 106 −1.287 × 106 −268.1 × 103 −417.7 × 103 km

We have not yet found the eccentricity $$e_\text{p}$$ of the hyperbola. This eccentricity depends on how close the spacecraft gets to the planet. We express this in terms of the so-called collision parameter $$b$$, which indicates how close the spacecraft would get to the planet if the planet did not attract the spacecraft.

It follows from equations \eqref{eq:hyp-e} and \eqref{eq:hyp-L} that

$$e_\text{p}^2 = 1 + \frac{1}{P^2}$$

where

$$P ≡ \frac{γ_\text{p}}{v_\text{h}^2b} = \frac{|a_\text{p}|}{b} \label{eq:p}$$

The greater $$P$$ is, the more the planet changes the direction of the spacecraft.

With equation \eqref{eq:delta} we then find the angle over which the direction of the spacecraft is changed (where the geometry determines whether the rotation is clockwise or counterclockwise):

$$δ = \arccos\left( 1 − \frac{2}{e_\text{p}^2} \right) = \arccos\left( \frac{1 − P^2}{1 + P^2} \right) = 2 \arctan(P) \label{eq:delta-p}$$

The limiting cases are

\begin{eqnarray} δ \| → \| \frac{360°}{π} P \| \qquad (P ↓ 0) \\ δ \| → \| 180° − \frac{360°}{Pπ} \| \qquad (P → ∞) \end{eqnarray}

The perifocus distance follows from equations \eqref{eq:e-planet} and \eqref{eq:q}:

$$q = a_\text{p}(1 − e_\text{p}) = b\left(\sqrt{P^2 + 1} − P\right)$$

The limiting cases are

\begin{eqnarray} q \| → \| b (1 − P) = b − |a_\text{p}| \| \qquad (P ↓ 0) \\ q \| → \| \frac{b}{2P} = \frac{b^2}{2|a_\text{p}|} \| \qquad (P → ∞) \end{eqnarray}

The speed at the perifocus then follows from equation \eqref{eq:hyp-L}:

$$v_\text{p}(q) = \frac{v_\text{h}}{\sqrt{P^2 + 1} − P}$$

The limiting cases are

\begin{eqnarray} v_\text{p}(q) \| → \| v_\text{h} (1 + P) \| \qquad (P ↓ 0) \\ v_\text{p}(q) \| → \| 2v_\text{h} P \| \qquad (P → ∞) \end{eqnarray}

#### 6.2.5. Don't Hit the Planet

The spacecraft must not hit the planet. If $$R$$ is the radius of the planet, then we want $$q \gt R$$, from which follow

\begin{eqnarray} \sqrt{P^2 + 1} − P \| \gt \| \frac{R}{b} \\ P \| \lt \| \frac{b}{2R} − \frac{R}{2b} ≡ P_\text{max} \end{eqnarray}

Fig. 13: δ_max, P_max, b_min

With equation \eqref{eq:p} we then find

$$b \gt R\sqrt{1 + \dfrac{1}{f^2}} = R \dfrac{\sqrt{v_\text{h}^2 + v_\text{esc}^2}}{v_\text{h}} ≡ b_\text{min}$$

with the escape speed $$v_\text{esc}$$ from the surface of the planet and $$f$$ given by

\begin{eqnarray} v^2_\text{esc} \| ≡ \| \frac{2γ_\text{p}}{R} \label{eq:vesc} \\ f \| = \| \frac{v_\text{h}}{v_\text{esc}} \label{eq:f} \end{eqnarray}

$$P_\text{max}$$ leads to

$$\begin{split} δ_\text{max} \| = \| 2 \arctan(P_\text{max}) \\ \| = \| 2 \arctan\left( \dfrac{1}{2f \sqrt{f^2 + 1}} \right) \\ \| = \| 2 \arctan\left( \dfrac{v_\text{esc}^2}{2v_\text{h} \sqrt{v_\text{h}^2 + v_\text{esc}^2}} \right) \end{split} \label{eq:δmax}$$

$$δ_\text{max}$$ increases if $$v_\text{h}$$ decreases or $$v_\text{esc}$$ increases. The variation of $$δ_\text{max}$$, $$P_\text{max}$$, and $$b_\text{min}$$ is displayed in Figure 13.

The radius of Mars is 3396 km, so the escape speed at the surface of Mars is

$v_\text{esc} = \sqrt{\frac{2×4.266×10^{13}}{3.396×10^6}} = 5013 \text{ m/s}$

If our spacecraft must not hit Mars, then we must have

\begin{eqnarray*} b \| \gt \| 3396 × \sqrt{1 + \left( \frac{5013}{2641} \right)^2} = 7285 \text{ km} \\ P \| \lt \| \dfrac{6115}{7285} = 0.8394 \\ δ \| \lt \| 2\arctan(0.8394) = 80.0° \end{eqnarray*}

Mercury Venus Mars Jupiter Saturn Uranus Neptune
$${v_\text{esc}}$$ 4.241 10.36 5.013 59.83 35.44 21.34 23.69 km/s
$${b_\text{min}}$$ 2667 24.01 × 103 7285 756.6 × 103 398.6 × 103 119.4 × 103 144.5 × 103 km
$${P_\text{max}}$$ 0.08951 1.858 0.8394 5.293 3.228 2.245 2.890
$${δ_\text{max}}$$ 10.23 123.4 80.02 158.6 145.6 132.0 141.8
°

#### 6.2.6. Back to the Domain of the Sun

The spacecraft flies past the planet, its direction changes over an angle $$δ$$, and then the spacecraft moves away from the planet again, back to the domain of the Sun. We handle the transition from the domain of the planet to the domain of the Sun the same as the earlier transition from the domain of the Sun to the domain of the planet.

The radial and tangential speeds of the spacecraft relative to the planet after the encounter with the planet are rotated over an angle $$δ$$ compared to before the encounter (where the sign of $$δ$$ is determined in the same way as that of $$α$$):

\begin{eqnarray} v_\text{tpx} \| = \| v_\text{tp} \cos δ − v_\text{rp} \sin δ \| = \| v \cos(α + δ) − v_0 \cos δ \\ v_\text{rpx} \| = \| v_\text{tp} \sin δ + v_\text{rp} \cos δ \| = \| v \sin(α + δ) − v_0 \sin δ \end{eqnarray}

The corresponding speeds relative to the Sun are

\begin{eqnarray} v_\text{tx} \| = \| v_\text{tpx} + v_0 \| = \| v \cos(α + δ) + v_0 (1 − \cos δ) \\ v_\text{rx} \| = \| v_\text{rpx} \| = \| v \sin(α + δ) − v_0 \sin δ \end{eqnarray}

The total speed $$v_\text{x}$$ relative to the Sun is then determined by

\begin{eqnarray} v_\text{x}^2 \| = \| v_\text{t2}^2 + v_\text{r2}^2 = v^2 + 2v_0^2(1 − \cos δ) + 2 v v_0 (\cos(α + δ) − \cos α) \\ \| = \| v^2 + 4v_0^2 \sin^2(δ/2) − 4vv_0 \sin(α + δ/2) \sin(δ/2) \label{eq:vx2} \end{eqnarray}

The length of the semimajor axis $$a_\text{x}$$ after the encounter follows from equations \eqref{eq:Etot} and \eqref{eq:e-bound}

$$a_\text{x} = \left( \frac{2}{r} − \frac{v_\text{x}^2}{γ} \right)^{−1} = \frac{γr}{2γ − v_\text{x}^2r} = \frac{v_0^2r}{2v_0^2 − v_\text{x}^2}$$

and the eccentricity $$e_\text{x}$$ follows from equations \eqref{eq:l} and \eqref{eq:l2}

$$\begin{split} e_\text{x} \| = \sqrt{1 − \frac{v_\text{tx}^2 r^2}{γa_\text{x}}} \\ \| = \sqrt{1 − \frac{v_\text{tx}^2 r^3}{2γ − v_\text{x}^2 r}} \\ \| = \sqrt{1 − \frac{v_\text{tx}^2}{v_0^4} (2v_0^2 − v_\text{x}^2)} \\ \| = \sqrt{1 − \frac{v_\text{tx}^2}{v_0^2} \frac{r}{a_\text{x}}} \end{split}$$

#### 6.2.7. Greatest and Smallest End Speeds

What are the greatest and least possible values of $$v_\text{x}$$?

If $$v$$ and $$v_0$$ and $$α$$ are fixed and we vary $$δ$$, then we find a greatest or least value for $$v_\text{x}^2$$ when

$$\tan(δ) = −\dfrac{v \sin(α)}{v \cos(α) − v_0}$$

The maximum value is

$$\max_δ v_\text{x} = v_0 + \sqrt{v^2 − 2 v v_0 \cos(α) + v_0^2}$$

which is reached when

$$δ = \arctan( −v \sin(α), v \cos(α) − v_0 )$$

The minimum value is

$$\min_δ v_\text{x} = \left| v_0 − \sqrt{v^2 − 2 v v_0 \cos(α) + v_0^2} \right|$$

which is reached when

$$δ = \arctan( −v \sin(α), v \cos(α) − v_0 ) + 180° = \arctan( v \sin(α), v_0 − v \cos(α))$$

If we vary both $$α$$ and $$δ$$ and keep $$v$$ and $$v_0$$ fixed, then $$v_\text{x}$$ reaches its greatest value when $$α = 180°$$ and $$δ = 180°$$, and that value is

$$\max_{α,δ} v_\text{x} = v + 2 v_0$$

This case cannot happen in practice, because $$δ = 180°$$ means that the spacecraft hits the planet.

$$v_\text{x}$$ reaches its smallest possible value when $$\cos(α) = v/(2v_0)$$, but that is only possible if $$v ≤ 2v_0$$. That smallest value is

$$\min_{α,δ} v_\text{x} = 0$$

so then the spacecraft falls into the Sun.

If $$v \gt 2v_0$$ then $$v_\text{x}$$ reaches its smallest value if $$α = 0°$$ and $$δ = 180°$$, and that value is

$$\min_{α,δ} v_\text{x} = v − 2 v_0$$

But $$δ = 180°$$ means that the spacecraft hits the planet.

#### 6.2.8. Gravity Assist After a Hohmann Trajectory

For a Hohmann trajectory, $$α = 0$$ at the target planet. Then

\begin{eqnarray} v_\text{x}^2 \| = \| v^2 + 2 v_0 (v_0 − v) (1 − \cos(δ)) \\ \| = \| v^2 + 4 v_0 (v_0 − v) \sin^2\left( \frac{1}{2} δ \right) \end{eqnarray}

If $$δ = 0$$ then $$v_\text{x} = v$$, so then the spacecraft doesn't notice the planet. The other extreme value is reached when $$|δ| = 180°$$. Then $$v_\text{x} = |2v_0 − v|$$, but for that $$δ$$ the spacecraft would hit the planet.

The relative speed of the spacecraft with respect to the target planet is

$$v_\text{h} = |v_0 − v|$$

Equation \eqref{eq:δmax} then yields, for the case in which $$δ$$ is as large as possible (equal to $$δ_\text{max}$$), where the spacecraft flies very close to the planet,

$$v_\text{x}^2 = v_\text{x,extr}^2 ≡ v^2 + \dfrac{4 v_0 (v_0 − v)}{(2f^2 + 1)^2} \label{eq:vx2h}$$

with $$f = v_\text{h}/v_\text{esc}$$ from equation \eqref{eq:f}.

With

\begin{eqnarray} ξ \| ≡ \| \dfrac{v}{v_0} \label{eq:ξ} \\ ξ_\text{x,extr} \| ≡ \| \dfrac{v_\text{x,extr}}{v_0} \\ ξ_\text{esc} \| ≡ \| \dfrac{v_\text{esc}}{v_0} \end{eqnarray}

we find

\begin{eqnarray} f \| = \| \dfrac{|1 − ξ|}{ξ_\text{esc}} \\ ξ_\text{x,extr} \| = \| \sqrt{ξ^2 + \dfrac{4(1 − ξ)}{(2f^2 + 1)^2}} \end{eqnarray}

For the journey to Mars, we found earlier that $$v = 21.42$$ km/s, $$v_0 = 24.06$$ km/s, $$v_\text{esc} = 5.013$$ km/s, $$δ_\text{max} = 80.02$$°. Then

$v_\text{x,extr} = \sqrt{21.42^2 + 4×24.06×(24.06 − 21.42)×\sin^2(80.02°/2)} = 23.75 \text{ km/s}$

and also

\begin{eqnarray*} f \| = \| \dfrac{24.06 − 21.42}{5.013} = 0.5266 \\ v_\text{x,extr} \| = \| \sqrt{21.42^2 + \dfrac{4×24.06×(24.06 − 21.42)}{(2×0.5266^2 + 1)^2}} = 23.75 \text{ km/s} \end{eqnarray*}

For a gravity assist after a Hohmann trajectory from Earth to a planet we find, for all planets:

Mercury Venus Mars Jupiter Saturn Uranus Neptune
$${v_\text{x,extr}}$$ 57.19 33.51 23.75 18.38 14.41 10.47 8.944 km/s
$${ξ}$$ 1.201 1.077 0.8902 0.5678 0.4353 0.3145 0.2535
$${ξ_\text{esc}}$$ 0.08883 0.2968 0.2083 4.595 3.688 3.149 4.377
$${ξ_\text{x,extr}}$$ 1.198 0.9596 0.9869 1.411 1.500 1.545 1.653

To escape from the Solar System, the final speed $$v_\text{x}$$ relative to the Sun must be at least $$\sqrt{2} v_0$$, because that is the escape speed at the distance of the target planet's orbit. Equation \eqref{eq:vx2h} says that $$v_\text{x} ≥ \sqrt{2} v_0$$ if

$$f ≤ \sqrt{\sqrt{\dfrac{v_0 (v_0 − v)}{2v_0^2 − v^2}} − \dfrac{1}{2}} = \sqrt{\sqrt{\dfrac{1 − ξ}{2 − ξ²}} − \dfrac{1}{2}} ≡ θ \label{eq:ff}$$

The right-hand side of equation \eqref{eq:ff} only exists (as a real number) if $$ξ ≤ 2 − \sqrt{2}$$ or $$2 ≤ ξ ≤ 2 + \sqrt{2}$$. The values greater than or equal to 2 are discarded because a Hohmann transfer orbit is bound, because for a bound orbit $$ξ \lt \sqrt{2}$$. So $$ξ ≤ 2 − \sqrt{2} ≈ 0.5858$$ remains.

Equations \eqref{eq:vq}, \eqref{eq:vQ}, and \eqref{eq:e} yield

$$\frac{r_2}{r_1} = \frac{2}{ξ²} − 1$$

where $$r_1$$ is the radius of the orbit of the departure planet, and $$r_2$$ is the radius of the orbit of the target planet. To $$ξ ≤ 2 − \sqrt{2}$$ corresponds $$r_2/r_1 ≥ (2\sqrt{2} − 2)/(3 − 2\sqrt{2}) ≈ 4.8284$$.

Equations \eqref{eq:f} and \eqref{eq:ξ} yield

$$\dfrac{v_\text{esc}}{v_0} ≥ \dfrac{1 − ξ}{θ}$$

$$v_\text{esc}/v_0$$ reaches a minimum of about 2.1451 for $$r_2/r_1 ≈ 36.3507$$. For larger $$r_2/r_1$$, $$v_\text{esc}/v_0$$ rises ever more slowly to $$\sqrt[4]{2} \sqrt{2 − \sqrt{2}} (\sqrt{2} + 1) ≈ 2.1974$$. For smaller $$r_2/r_1$$ down to $$(2\sqrt{2} − 2)/(3 − 2\sqrt{2}) ≈ 4.8284$$, $$v_\text{esc}/v_0$$ quickly grows to infinity. Here are some values:

$${r_2/r_1}$$ $${v_\text{esc}/v_0}$$
4.91 10
4.93 9
4.96 8
5.00 7
5.07 6
5.20 5
5.46 4
6.28 3

So to be able to escape from the Solar System through a gravity assist after a Hohmann transfer orbit, the orbit of the target planet must be at least 4.8284 times further from the Sun than the departure planet's, and additionally the escape speed from the target planet's surface must be at least 2.1451 times greater than the orbital speed of the target planet. If either of those conditions is not met, then escape from the Solar System through a gravity assist after a Hohmann orbit is not possible. If both conditions are met, then such an escape may be possible.

### 6.3. Cheapest Escape

Suppose that you have a space ship and a limited amount of propellant. If you can raise your speed above the local escape velocity, then you can escape from the Sun and get to arbitrarily great distances.

Equations \eqref{eq:ε} and \eqref{eq:e-bound} yield for the speed $$v$$ in a closed orbit

$$v^2 = \frac{2γ}{r} − \frac{γ}{a} \label{eq:vv}$$

if $$r ≤ 2a$$ (but that is always true for a closed orbit). An orbit with $$ε = −γ/2a = 0$$ is one in which the space ship always travels at the local escape velocity, so we find for the local escape velocity $$v_\text{escape}$$

$$v_\text{escape}^2 = \frac{2γ}{r}$$

For the difference $$∆v = v_\text{escape} − v$$ we then find

$$∆v = \sqrt{\frac{γ}{a}} (\sqrt{x} − \sqrt{x − 1})$$

$$x = \frac{2a}{r}$$

This $$∆v$$ increases monotonically with $$r$$ and is always greater than zero (for closed orbits), so the extreme values occur for the greatest and smallest possible values of $$r$$ for that orbit, i.e., in the perihelion and aphelion of the orbit. $$∆v$$ is least in the perihelion and greatest in the aphelion.

So, if you want to escape the Solar System from a closed orbit, using the least amount of effort and no gravity assists, then you should run your engines when you're in the perihelion of your orbit, and run them in such a way that you are accelerated forward (in the direction in which you were traveling already). You must, of course, accelerate sufficiently, or else you won't escape. To escape the Solar System from the Earth's orbit you need to increase your forward speed by at least about 12 km/s = 44,000 km/h. That is no small increase!

### 6.4. How Far Away Without Gravity Assists?

If you have too little propellant to escape from the Solar System (without gravity assists), then you far can you get?

#### 6.4.1. Look at the Energy

We'll first try to find an upper boundary by looking at the energy. Equation \eqref{eq:vv} shows that for a closed orbit always $$r \lt 2a$$, so the aphelion cannot be further from the Sun than at distance $$2a$$. We rewrite equation \eqref{eq:toa} to

$$\frac{a}{r} = \frac{1}{2 − ψ^2}$$

If we want to make $$a$$ as great as possible while keeping $$r$$ and $$γ$$ fixed, then we must make $$ψ$$ as great as possible, and hence $$v$$ as great as possible. With a limited amount of propellant, we get $$v$$ as great as possible by accelerating in the direction in which we were already traveling. If our propellant can change our velocity by $$v_\text{s} ≡ σvcircle$$, then we find for the new semimajor axis $$a_\text{x}$$

$$\frac{a_\text{x}}{r} = \frac{1}{2 − (ψ + σ)^2}$$

and hence the following upper limit $$Q_\text{x}$$ for the aphelion

$$\frac{Q_\text{x}}{r} \lt \frac{2}{2 − (ψ + σ)^2}$$

For example, if we start from a circular orbit at $$r = 1$$ AU from the Sun (roughly the Earth's orbit) and apply a velocity change of 0.1 times the orbital velocity (i.e., about 3 km/s), then $$v = vcircle$$, $$v_\text{s} = 0.1 vcircle$$, $$ψ = 1$$, and $$σ = 0.1$$, so $$Q_\text{x} \lt 2/(2 − 1.1^2) = 2.53$$ AU. We certainly won't get further from the Sun than 2.53 AU. However, this is an upper boundary, so the real aphelion might be closer to the Sun.

#### 6.4.2. In General

Can we find the aphelion more precisely? Equation \eqref{eq:r} yields

$$a (1 − e^2) = r (1 + e \cos ν)$$

and $$a (1 − e^2)$$ is constant for the whole orbit, and so also for the aphelion. We find

$$Q (1 − e) = r (1 + e \cos ν) ⇒ \frac{Q}{r} = \frac{1 + e \cos ν}{1 − e}$$

You find the furthest possible aphelion if (for fixed eccentricity $$e$$) you accelerate in such a way that you end up in the perihelion of the new orbit (where $$ν = 0$$ hence $$\cos ν = 1$$), or if (for fixed true anomaly $$ν$$) you arrange to end up in the orbit with the greatest possible eccentricity. However, $$e$$ and $$ν$$ both depend on the direction and magnitude of the velocity change, so we cannot pick their values at will, so we're not done yet.

In general, the aphelion distance $$Q$$ is equal to

$$\frac{Q}{r} = \frac{1 + \sqrt{1 − χ^2ψ^2 (2 − ψ^2)}}{2 − ψ^2} \label{eq:Qalg}$$

Here, again, we have two quantities ($$χ$$ and $$ψ$$) that each depend on the direction and magnitude of the velocity change, so we cannot deduce from this equation what the best direction is in general to apply the velocity change in to get the furthest aphelion. The best way to find that direction is to try equation \eqref{eq:Qalg} for different directions and to see which one yields the furthest aphelion.

I calculated a number of random cases and found that the direction in which you should apply the velocity change to get the furthest aphelion is not always the same and is often different from the direction in which your space ship is traveling just before you start the engines. If the velocity increase gets closer to the amount that is needed to make the space ship escape from the Solar System, then the best direction does tend to the direction of motion. That fits with what we saw before (in Sec. 6.3).

#### 6.4.3. Leaving from a Circular Orbit

The general case does not provide easy answers. The simpler case in which we start from a circular orbit does: in that case, your best choice is to accelerate in the direction of motion. Then $$χ = 1$$ and hence (from Equation \eqref{eq:Qalg})

$$\frac{Q}{r} = \frac{1 + \sqrt{1 − ψ^2 (2 − ψ^2)}}{2 − ψ^2} = \frac{1 + |ψ^2 − 1|}{2 − ψ^2}$$

If $$ψ \lt 1$$ then we find $$Q/r = 1$$: in that case you decelerated rather than accelerated, and then you don't get further from the Sun but rather closer to it, and then your point of departure at distance $$r$$ becomes the aphelion of your new orbit. That case does not interest us here, so we assume that $$ψ ≥ 1$$. Then we find

$$\frac{Q}{r} = \frac{ψ^2}{2 − ψ^2}$$

Before the acceleration we have $$ψ = 1$$ and $$Q = r$$. Now we apply a velocity change of $$v_\text{s} = σvcircle$$ so $$ψ = 1 + σ$$. We find

$$e = 2σ + σ^2$$

$$Q = r \frac{1 + e}{1 − e}$$

The other way around,

$$e = \frac{Q − r}{Q + r}$$

$$σ = \sqrt{1 + e} − 1$$

For example, when we start from a circular orbit at 1 AU from the Sun (roughly the Earth's orbit) and apply a velocity change of 0.1 times the orbital velocity (i.e., about 3 km/s extra), then we end up in an orbit with $$e = 2×0.1 + 0.1^2 = 0.21$$ and $$Q = 1×(1 + 0.21)/(1 − 0.21) = 1.53$$ AU.

This is the same case as in the previous example, where we found an upper limit of 2.54 AU. Here we find the "real" aphelion, which is indeed below the upper limit, but is in fact well below that upper limit. However, the formulas for the upper limit can be used for departure from any closed orbit, whereas the formulas right before this are only valid for departure from circular orbits.

If we want to move from a circular orbit at $$r = 1$$ AU to an orbit with $$Q = 1.53$$ AU, then we need $$e = (1.53 − 1)/(1.53 + 1) = 0.21$$ and $$σ = (\sqrt{1 + 0.21} − 1) = 0.1$$ so $$v_\text{s} = 0.1 vcircle$$.

### 6.5. All Orbits Through Two Points

Fig. 14: Orbits Through 2 Points
Suppose that planets 1 and 2 are at distances $$r_1$$ and $$r_2$$ from the Sun, and are at an angular distance $$φ$$ from each other, as seen from the Sun. Which orbits connect those two planets? $$a(1 − e^2)$$ is constant along any such orbit, so we deduce from Equation \eqref{eq:r} that $$r (1 + e \cos ν)$$ is also constant, so

$$r_2 − r_1 = e (r_1 \cos(ν_1) − r_2 \cos(ν_2))$$

where $$ν_1$$ and $$ν_2$$ are the true anomalies of the spacecraft in its orbit when it is near planets 1 and 2, for which we must then have

$$ν_2 − ν_1 = φ$$

With further definitions

\begin{align} r_− \| ≡ \frac{1}{2}(r_2 − r_1) \label{eq:rm} \\ r_+ \| ≡ \frac{1}{2}(r_2 + r_1) \\ ν_− \| ≡ \frac{1}{2}(ν_2 − ν_1) = \frac{1}{2}φ \\ ν_+ \| ≡ \frac{1}{2}(ν_2 + ν_1) \end{align}

we can rewrite the above equation to

$$2r_− = e (2r_+ \sin(ν_+) \sin(ν_−) − 2r_− \cos(ν_+) \cos(ν_−))$$

or

$$e = \frac{r_−}{r_+ \sin(ν_+) \sin(ν_−) − r_− \cos(ν_+) \cos(ν_−)} \label{eq:e2}$$

We assumed that we already knew the values of $$r_1$$, $$r_2$$, and $$ν_−$$, so $$r_+$$ and $$r_−$$ are also known, and only the values of the eccentricity $$e$$ and average true anomaly $$ν_+$$ are still unknown in the above equation. That average true anomaly says where the perihelion of the connecting orbit is, relative to the two planets.

If we now assume a particular value for $$ν_+$$ then Equation \eqref{eq:e2} says which eccentricity $$e$$ goes with that.

For Equation \eqref{eq:r} we've assumed that the eccentricity $$e$$ was not negative, so we must discard negative values of $$e$$. To find an orbit with equally large but positive $$e$$, you can add 180° to $$ν_+$$.

We moeten hebben dat $$e ≥ 0$$. $$e$$ verandert van teken als de noemer van vergelijking \eqref{eq:e2} gelijk is aan 0, dus als

$$\tan(ν_+) = \dfrac{r_− \cos(ν_−)}{r_+ \sin(ν_−)}$$

De bruikbare $$ν_+$$ waarvoor $$e ≥ 0$$ zijn

\begin{eqnarray} ν_e \| ≤ \| ν_+ ≤ ν_e + 180° \\ ν_e \| = \| \arctan(r_− \cos(ν_−), r_+ \sin(ν_−)) \end{eqnarray}

What is the smallest eccentricity $$e$$ that an orbit between the two planets can have? Then $$1/e$$ must be at its greatest. Equation \eqref{eq:e2} yields

\begin{eqnarray} \frac{1}{e} \| = \| \frac{\sin(ν_+) \sin(ν_−)}{e_\text{H}} − \cos(ν_+) \cos( ν_−) \\ e_\text{H} \| ≡ \| \dfrac{r_−}{r_+} \end{eqnarray}

waarin $$e_\text{H}$$ de excentriciteit van de Hohmannbaan tussen de twee planeten is.

If we vary only $$ν_+$$ then we find an extreme value if $$\tan ν_− = −e_\text{H} \tan ν_+$$ and that extreme value is

$$\frac{1}{|e|_\text{min}} = \sqrt{\dfrac{1}{e_\text{H}^2} \sin^2(ν_−) + \cos^2(ν_−)} \label{eq:emin}$$

The right-hand side of Equation \eqref{eq:emin} reaches its greatest value ― hence $$e$$ reaches its least possible value ― when $$ν_− = 90°$$ so $$φ = 180°$$. Then $$e = e_\text{H}$$. This is the Hohmann trajectory. Of all orbits that can connect two locations, the Hohmann trajectory is the one with the least possible eccentricity.

The right-hand side of Equation \eqref{eq:emin} reaches its smallest value when $$ν_− = φ = 0$$, and that value is 1, so the smallest $$e$$ rises to 1 when $$φ$$ goes to $$0$$.

For every $$ν_−$$ there are $$ν_+$$ for which $$1/e = 0$$ hence $$e = ∞$$, namely those for which

$$\tan(ν_+) = \frac{e_\text{H}}{\tan(ν_−)}$$

Al met al varieert $$e$$ dus tussen $$e_\text{H} \lt 1$$ en $$∞$$, afhankelijk van $$ν_+$$ en $$ν_−$$.

Wat zijn de grootste en kleinste waarden die $$a$$ kan hebben? Door weer vergelijking \eqref{eq:r} te gebruiken vinden we

By using Equation \eqref{eq:r} again we find

$$a = r_2 \frac{1 + e \cos ν_2}{1 − e^2}$$

We substitueren de $$e$$ uit vergelijking \eqref{eq:e2} en herschrijven $$r_2$$ en $$ν_2$$ in termen van $$r_−, r_+, ν_−, ν_+$$ met hulp van vergelijkingen \eqref{eq:rm}ff en vinden dan $$a$$ als een functie van $$r_−, r_+, ν_−, ν_+$$:

\begin{eqnarray} a \| = \| \dfrac{A r_+^3 − B r_− r_+^2 − A r_−^2 r_+ + B r_−^3}{A r_+^2 − 2 B r_− r_+ + C r_−^2} \\ A \| = \| \sin^2(ν_−) \sin^2(ν_+) \\ B \| = \| \cos(ν_−) \sin(ν_−) \cos(ν_+) \sin(ν_+) \\ C \| = \| \cos^2(ν_−) \cos^2(ν_+) − 1 \end{eqnarray}

We zoeken uiterste waarden van $$a$$ door de afgeleide naar $$ν_+$$ gelijk te zetten aan 0. We vinden dat $$a$$ een uiterste waarde heeft als

Then we've found $$a$$ and $$e$$ and we know where the perihelion is (at an angle $$ν_+ + \frac{1}{2}φ$$ before planet 2, and $$ν_+ − \frac{1}{2}φ$$ before planet 1), so then the whole orbit is known. By varying $$ν_+$$ we get all acceptable orbits that pass through both planets.

For example, on November 1, 2007 (0h UTC) the Earth and Mars were at $$r_1 = 0.9927$$ AU and $$r_2 = 1.5079$$ AU from the Sun, and at $$φ = 27.0°$$ from each other (as seen from the Sun). Then we find $$r_− = (1.5079 − 0.9927)/2 = 0.2576$$ and $$r_+ = (1.5079 + 0.9927)/2 = 1.2503$$, and from those we find $$e = 0.2576/(0.2919 \sin ν_+ − 0.2505 \cos ν_+)$$ and $$a = 1.5079 (1 + e \cos(ν_+ + 13.5°))/(1 − e^2)$$.

We also find $$ν_0 = \arctan(0.2919, −0.2505) = 130.6°$$, so we only need to look at $$ν_+$$ between 130.6° − 90° = 40.6° and 180°. A few cases are displayed in Figure 14, where small square A indicates the Sun, B the Earth, and C Mars. Black orbits are closed ($$e \lt 1$$) and blue orbits are open ($$e \gt 1$$).

We find the travel times between the two planets via those orbits as follows. If $$e \lt 1$$, then calculate (from Equations \eqref{eq:muitt}, \eqref{eq:kepler} and \eqref{eq:νtoE})

\begin{align} E_1 \| = 2 \arctan\left( \sqrt{\frac{1 − e}{1 + e}} \tan\left( \frac{1}{2}ν_1 \right) \right) \\ M_1 \| = E_1 − e \sin E_1 \\ t_1 \| = \frac{P M_1}{2π} = \frac{a^{3/2} M_1}{\sqrt{γ}} \end{align}

Beware: $$E_1$$ and $$M_1$$ must be measured here in radians instead of degrees, or else you get wrong results. 1 radian equals 180°/π degrees.

If $$e \gt 1$$, then calculate (from Equations \eqref{eq:kepler_h} and \eqref{eq:νtoH})

\begin{align} H_1 \| = 2 \artanh\left( \sqrt{\frac{e − 1}{e + 1}} \tan\left( \frac{1}{2}ν_1 \right) \right) \\ M_1 \| = e \sinh H_1 − H_1 \\ t_1 \| = \frac{P M_1}{2π} = \frac{|a|^{3/2} M_1}{\sqrt{γ}} \end{align}

Beware: we don't use just the "ordinary" $$\tan$$ function, but also the hyperbolic $$\artanh$$ and $$\sinh$$ functions.

Because $$a$$, $$e$$, and $$ν_1$$ depend on $$ν_+$$, so does $$t_1$$.

Make similar calculations for planet 2. Then the travel time is equal to

$$∆t = t_2 − t_1$$

If for the case described in the previous example we select $$ν_+ = 126.0°$$, the we find $$e = 0.2576/(0.2919 \sin(126°) − 0.2505 \cos(126°)) = 0.6719$$ and $$a = 1.5079 (1 + 0.6719 × \cos(126° + 13.5°))/(1 − 0.6719) = 1.3444$$ AU. $$0 \lt e \lt 1$$, so this is an elliptic orbit.

For the space ship at planet 1 we find $$ν_1 = v_+ − ν_− = 126.0° − 13.5° = 112.5°$$, $$E_1 = 2 \arctan(\sqrt{(1 − 0.6719)/(1 + 0.6719)} \tan(\frac{1}{2} × 112.5°)) = 1.1708$$ radians, $$M_1 = 1.1708 − 0.6719 × \sin(1.1708) = 0.5519$$ radians, $$t_1 = 1.3444^{3/2} × 0.5519 / 2π = 0.1369$$ year. (If we measure $$a$$ in AU and $$t$$ in years, then $$P = a^{3/2}$$ in our Solar System.)

For the space ship at planet 2 we find $$ν_2 = v_+ + ν_− = 126.0° + 13.5° = 139.5°$$, $$E_2 = 2 \arctan(\sqrt{(1 − 0.6719)/(1 + 0.6719)} \tan(\frac{1}{2} × 139.5°)) = 1.7527$$ radians, $$M_2 = 1.7527 − 0.6719 × \sin(1.7527) = 1.0918$$ radians, $$t_2 = 1.3444^{3/2} × 1.0918 / 2π = 0.2709$$ year.

And for the travel time we find $$∆t = 0.2709 − 0.1369 = 0.1340$$ year.

If instead we select $$ν_+ = 72.0°$$, then we find $$e = 0.2576/(0.2919 \sin(72°) − 0.2505 \cos(72°)) = 1.2868$$ and $$a = 1.5079 (1 + 1.2868 × \cos(72° + 13.5°))/(1 − 1.2868) = −2.5314$$ AU. $$e \gt 1$$ and $$a \lt 0$$, so this is a hyperbolic orbit.

For the space ship at planet 1 we find $$ν_1 = v_+ − ν_− = 72.0° − 13.5° = 58.5°$$, $$H_1 = 2 \artanh(\sqrt{(1.2868 − 1)/(1.2868 + 1)} \tan(\frac{1}{2} × 58.5°)) = 0.4020$$, $$M_1 = 1.2868 × \sinh(0.4020) − 0.4020 = 0.1293$$ radians, $$t_1 = 2.5314^{3/2} × 0.1293 / 2π = 0.0829$$ year.

For the space ship at planet 2 we find $$ν_2 = v_+ + ν_− = 72.0° + 13.5° = 85.5°$$, $$H_2 = 2 \artanh(\sqrt{(1.2868 − 1)/(1.2868 + 1)} \tan(\frac{1}{2} × 85.5°)) = 0.6797$$ radians, $$M_2 = 1.2868 × \sin(0.6797) − 0.6797 = 0.2639$$ radians, $$t_2 = 2.5314^{3/2} × 0.2639 / 2π = 0.1691$$ year.

And for the travel time we find $$∆t = 0.0829 − 0.1691 = 0.0862$$ year.

### 6.6. Aiming for a Planet

It is nice that we can find all orbits passing through two points, but the time it takes to travel between those two points is different for each different orbit, so if you make space ships depart the first planet for the second one along all of those orbits at the same time, then they won't all arrive at the second position at the same time. The second planet moves, so it will be in that second position only for an instant. Only the space ship that happens to arrive at that second position at exactly that instant will reach the second planet. In the last few examples given above we found travel times greater than zero, but the positions of the two planets were calculated for the same instant, so if the space ship arrives at the second position after the travel time that we calculated, then the second planet won't be there anymore, so the space ship misses the planet.

You have to choose some travel time before you can calculate where the target planet will be after that travel time. You then have two points that your orbit must pass through: your current position, and the position of the target planet after the chosen travel time. There is then a wide range of orbits that passes through those two points, but each of those orbits has a different travel time between the two points. You can calculate that travel time only after you have chosen an orbit, and only then can you tell if the travel time along that orbit is the same as the desired travel time that you chose at the very beginning. If the travel time in the chosen orbit is not equal to the desired travel time, then you'll miss the target planet if you use that orbit. All in all, you need a big search operation to find an orbit that gets you to the target planet.

The procedure is as follows:

1. Choose a travel time $$∆t$$.
2. Calculate where planet 1 is at the desired departure time $$t_0$$ and where planet 2 is at time $$t_0 + ∆t$$. We need: the distance $$r_1$$ between the Sun and planet 1 (at time $$t_0$$), the distance $$r_2$$ between the Zon and planet 2 (at time $$t_0 + ∆t$$), and the angular distance $$φ$$ between planet 2 (at time $$t_0 + ∆t$$) and planet 1 (at time $$t_0$$) as seen from the Sun.
3. Now calculate for all possible orbits through those two points (by varying $$ν_+$$, see Section 6.5) what the travel time between the planets is along that orbit.
4. Find the $$ν_+$$ for which the corresponding orbit has a travel time equal to $$∆t$$.

Instead of calculating the travel time for "all" orbits, you can also try to do a more directed search for the solution, for example using one of the search methods described on the Calculation Page about Horizontal Phenomena.

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Last updated: 2020-07-18