This extended essay focuses on orbital mechanics, which is the study of the motion and trajectories of artificial satellites and spacecraft under the influence of external forces such as gravitational attraction due to celestial bodies, atmospheric drag or thrust. Specifically, I will analyse a fairly simple manoeuvre: the hyperbolic flyby, visualising it with vectors and calculating its impact on the craft through the use of trigonometry and geometry. My research question is: “What are the mathematical methods involved in defining the trajectory and speed of a spacecraft in a hyperbolic flyby manoeuvre?” This manoeuvre is defined as the hyperbola-shaped trajectory that a spacecraft with enough velocity to escape its gravitational attraction takes around a central body. By the end of this investigation, my aim is to be able to calculate the values of speed and direction of a spacecraft after a hyperbolic flyby, having set specific initial conditions. This investigation also aims to define what are the basic initial parameters that need to be defined to calculate the exit speed of a spacecraft after a hyperbolic flyby.
The choice of topic for this essay stems from my passion in aerospace, a subject I would like to study at university. I have always been interested in spacecraft construction and launches, and the idea of analysing hyperbolic flybys came while I was reading about BepiColombo, a scientific probe, named after a scientist and engineer who lived near my home town. The spacecraft, sent to study Mercury, had just performed a gravity assist on the planet, used to increase the speed and change the direction of the probe in relation to Mercury. I already knew about this manoeuvre, because it is commonly used, as it is impractical and impossible to load all the fuel needed to reach high speeds once in orbit. This was the case with the Voyager 2 probe, launched in 1977. Its purpose was to study the outer planets of the solar system and interstellar space. The spacecraft used gravity assists from Jupiter, Saturn, Uranus and Neptune, as well as these planet’s moons, to accelerate and escape from the solar system. Voyager two is currently travelling at about 15km/s, but the highest speed reached (relative to the sun), was of about 35 km/s. These speeds have been obtained respectively from two articles by NASA - “Voyager - Mission Status.”2 and “Voyager 2 - In Depth.”
These high speeds cannot be reached even with today’s launch systems, showcasing how flybys are fundamental in interplanetary missions. My initial aim was to focus on these manoeuvres and how they could be used multiple times to increase a spacecraft’s velocity and direct it towards a destination, however, after some research, I discovered that the topic was far too difficult. This is because it would require extensive knowledge on the position of planets in their orbits and a deep understanding of planet-probe interactions. For this reason, I decided to focus only on the manoeuvre itself, as they are the basis of planet-probe interactions and interplanetary travel.
The study of these flybys included vectors, a topic which I have had to study on my own, and trigonometry, applying the knowledge I had already acquired in a totally different context, proving how maths can be of great help in modelling and understanding the physical world. This essay will be structured as follows: firstly, an overview of the topic will be given, defining the terms used, the variables and the assumptions made, then the problem will be described and analysed from a mathematical perspective, yielding a set of formulae that will be later put into use in a real-life scenario, obtaining a value for the exit velocity, which will be compared with data taken from the Voyager 2 mission. Finally, conclusions will be made to evaluate the possible errors and improvements to this investigation.
\(\vec{V}_{\infty}\,_{in}\) - Spacecraft’s inbound hyperbolic excess velocity
\(\vec{V}_{\infty}\,_{out}\) - Spacecraft’s outbound hyperbolic excess velocity
\(\vec{V}_p\) - Planet’s heliocentric velocityPlanet’s heliocentric velocity
\(\vec{V}\,_{sun \,in}\) - Spacecraft’s inbound heliocentric velocity
\(\vec{V}\,_{sun\,out}\) - Spacecraft’s outbound heliocentric velocity
î - base vector in the horizontal direction (⟂ to the direction towards the sun)
ĵ - base vector in the vertical direction (direction towards the sun)
θin - Angle between\( \vec{V} _{∞\, in}\) and \(\vec{V}\,_{ p}\) or î(the horizontal direction)
θout - Angle between\(\vec {V} _{∞\,out}\) out and \(\vec{V}\,_{ p}\) or î (the horizontal direction)
αin - Angle between\( \vec{V}\, _{sun\, in }\)and \(\vec{V}\,_{ p} \)or î (the horizontal direction)
αout - Angle between V⃗ sun out and V⃗ p or î(the horizontal direction)
γ - Angle between the inbound hyperbolic excess velocity \(\vec{V}_{ ∞\,in}\) in and \(\vec{V}_{∞\,out}\)
All of the angles above are measured counterclockwise.
mp - Planet’s mass
ms - Spacecraft’s mass
In order not to complicate the maths and render the calculations extremely complex, the following are the assumptions and simplifications that have to be made for the model to be feasible.
By assuming the heliocentric velocity of the planet to be constant, it is also assumed that the planet’s orbit is perfectly circular, although this is very unlikely to happen naturally. This consequence derives from Kepler’s second law of planetary motion.
As a consequence of the previous assumption, the instantaneous heliocentric velocity of the planet is considered to be perpendicular to the direction to the sun. The velocity is tangent to the orbit, perpendicular to the radius, the direction to the sun.
This model does not account for the air resistance that may result from the probe entering the planet’s exosphere (upper layer of the atmosphere), slowing the spacecraft down.
The model does not account for differences in the gravitational attraction of the planet, which in nature occur because of the non-spherical shape of planets and their differences in density. For this reason, it is assumed that the planet is perfectly spherical and its mass is evenly distributed.
In a real scenario, there might be attraction from the planet’s moons or other planets, affecting the spacecraft’s speed and trajectory. This is not accounted for in the model.
The model does not account for this, as it has a minimal impact on the manoeuvre. For this reason, the spacecraft is considered to be point-like
A hyperbolic flyby is a manoeuvre performed by spacecraft that is travelling faster than the escape velocity of the planet it is approaching. The spacecraft’s trajectory can be modelled mathematically as an hyperbola, or a conic section with eccentricity greater than one. Planet X occupies one of the two foci of the hyperbola, while the other is left empty. In orbital mechanics, the trajectory the spacecraft follows is the one nearer to the planet. The other branch would represent the trajectory of a particle if the force exerted by the object on the opposing focus were repulsive. For example, if the object on the focus were to be a positively or negatively charged particle, the opposing branch of the hyperbola would represent the trajectory of a particle of the same charge.
During the inbound crossing, the spacecraft accelerates as a consequence of the gravitational attraction caused by the planet’s gravity well. At periapsis, the spacecraft’s velocity is maximum, and during the outbound crossing the spacecraft decelerates once more. Thus, after the manoeuvre, the spacecraft doesn’t gain speed if observed only in the planet’s frame of reference.
The problem will now be analysed through the use of vectors, both in the planet’s frame of reference and in the sun’s frame of reference, showing how the manoeuvre affects the velocity of the spacecraft, enabling a probe to increase its velocity relative to the sun without the use of propellant.
The inbound \((\vec{V}\,_{ ∞ in} )\) and outbound \((\vec{V} ∞\,_{ out} )\) hyperbolic excess velocities of a spacecraft approaching a planet can be visualised as follows. The two velocity vectors lie on the asymptotes of the hyperbola. The analysis will only be focused on the time period in which the spacecraft is in the planet’s sphere of influence.
As seen in the illustration, the velocity vector changes direction as a consequence of the gravitational field of planet X, but its magnitude remains the same.
However, this vector only represents the velocity of the spacecraft relative to the planet, which in turn moves relative to the sun. This movement is represented by the planet’s heliocentric velocity\( \vec{V}_{ p} .\)
Adding these two visualisations, moving to the sun’s frame of reference, the spacecraft’s hyperbolic excess velocities\(\vec{ V}_{∞}\,_ {in}\) and \(\vec{V} _{∞ }\,_{out}\) out are summed with the planet’s heliocentric velocity\( \vec{V}\,_{ p} \), giving the spacecraft’s heliocentric velocities \(\vec{V}_{sun\, in}\) and\( \vec{V}_{sun \,out}\)
\(\vec{V}_{ sun\, in }=\vec{ V}_ {p }+ \vec{V} _{∞ \,in}\)
\(\vec {V} _{sun \,out} =\vec{ V }_{p }+ \vec{V} _{∞ out}\)
Thanks to the optimal angle, the spacecraft’s resultant outbound heliocentric velocity \(\vec{V}_{ sun\,out }\) is greater in magnitude than the inbound heliocentric velocity \(\vec{V}_{ sun\, in}\), as seen in figure 7
Having calculated the spacecraft’s inbound and outbound heliocentric velocities\( \vec{V}_{sun\, in}\) and\(\vec{ V}_{ sun\, out}\) , the change in the spacecraft’s heliocentric velocity can be found -
\(Δ\vec{V}_ {sun} = \vec{V}_{ sun out} − \vec{V}_{ sun in}\)
Substituting for the inbound and outbound heliocentric velocities -
\(Δ\vec{V}_{ sun }= (\vec{V}_{ p} +\vec{ V}_{ ∞ \,out} ) − (\vec{V}_ {p }+ \vec{V}_{ ∞\, in} )\)
Since \(\vec{V}_{ p}\) is constant (see assumption 1), it can be cancelled out from the equation, giving -
\(Δ\vec{V}_ {sun }= \vec{V} _{∞ out} − \vec{V}_{∞ in}\)
\(Δ\vec{V}_{ sun} = Δ\vec{V}_{ ∞}\)
It has been previously shown that the hyperbolic excess velocities are equal in magnitude, because only the direction of these velocities is changed by the gravity of the planet. For this reason, the change in the spacecraft’s heliocentric velocity is given by the change in direction of the probe’s hyperbolic excess velocity The angle that determines the amount of change in heliocentric velocity is the angle between \(\vec{V}_{∞}\) in and \(\vec{V}_{∞}\) out, as shown in the following diagram.
As seen in figure 8, if angle θout is smaller than angle θin ,\( \vec{V}_{ sun \,out}\) is larger than \(\vec{V}_{ sun\, in }\) in magnitude, showing how a hyperbolic flyby impacts the heliocentric velocity of a spacecraft. Through this manoeuvre, a probe can gain or lose velocity, according to the direction it approaches the planet, and consequently how the vector sum is arranged. In a leading side flyby, the spacecraft’s velocity is reduced, while in a leading side flyby, as illustrated in the previous diagrams, the velocity is increased.
Here follows a more thorough mathematical analysis of the problem, utilizing base vectors and trigonometry to define angles and magnitudes of the previously defined velocities. The magnitude of the inbound hyperbolic excess velocity will be found, as it is an unknown that is fundamental to solving the problem.
Recalling a previously expressed equation for the spacecraft’s inbound heliocentric velocity -
\(\vec{V}_{ sun in} = \vec{V }_{ p }+ \vec{V}_{ ∞ in}\)
And rearranging for\( \vec{V}_{ ∞ in }\), which is the velocity that causes the change in the overall heliocentric velocity, gives -
\(\vec{V}_{ ∞ in} = \vec{V}_{sun in} −\vec{ V}_{ p}\)
Two base vectors will now be introduced, in order to carry on the analysis considering the vectors expressed in the two components of the plane: horizontal and vertical. The horizontal component will be directed to the left, so to be in the same direction as the planet’s heliocentric velocity\( \vec{V}_{ p}\) The base vectors are defined as -
\( \hat{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \)
\(\hat{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\)
To complete the rearranged formula for the inbound hyperbolic excess velocity, both the spacecraft’s inbound heliocentric velocity and the planet’s heliocentric velocity have to be written in base vectors.
Expressing\(\vec{ V}_{ sun\, in}\) and \(\vec{V}_{ p} \)in their components
The spacecraft’s inbound heliocentric velocity can be expressed as -
\(\vec{V}_{\text{sun in}} = \left[\vec{V}_{\text{sun in}}\right]_i + \left[\vec{V}_{\text{sun in}}\right]_j \)
\(\begin{equation} \left[ \vec{V}_{\text{sun in}} \right]_i = \left| \vec{V}_{\text{sun in}} \right| \cos(\alpha_{\text{in}}) \hat{i} \end{equation} \)
\(\begin{equation} \left[ \vec{V}_{\text{sun in}} \right]_j = \left| \vec{V}_{\text{sun in}} \right| \sin(\alpha_{\text{in}}) \hat{j} \end{equation}\)
Thus,\(\vec{ V}_{ sun\, in}\) can be expressed in base vectors as -
\( \vec{V}_{\text{sun in}} = \left| \vec{V}_{\text{sun in}} \right| \cos(\alpha_{\text{in}}) \hat{i} + \left| \vec{V}_{\text{sun in}} \right| \sin(\alpha_{\text{in}}) \hat{j} \)
Similarly, \(\vec{V}_{ sun\, out}\) can be expressed in base vectors as -
\( \vec{V}_{\text{sun out}} = \left| \vec{V}_{\text{sun out}} \right| \cos(\alpha_{\text{out}}) \hat{i} + \left| \vec{V}_{\text{sun out}} \right| \sin(\alpha_{\text{out}}) \hat{j} \)
The inbound hyperbolic excess \(\vec{V}_{ ∞\, in}\) can be expressed in base vectors as
\( \left[ \vec{V}_{\infty \text{in}} \right]_i = \left| \vec{V}_{\infty \text{in}} \right| \cos(\alpha_{\text{in}}) \hat{i} \)
\( \left[ \vec{V}_{\infty \text{in}} \right]_i = \left| \vec{V}_{\infty \text{in}} \right| \sin(\alpha_{\text{in}}) \hat{i} \)
Thus, \(\vec{V}_{sun\, in}\) can be expressed in base vectors as -
\( \vec{V}_{\infty\, \text{in}} = \left| \vec{V}_{\infty\, \text{in}} \right| \cos(\alpha_{\text{in}}) \hat{i} + \left| \vec{V}_{\infty\, \text{in}} \right| \sin(\alpha_{\text{in}}) \hat{j} \)
The planet’s velocity \(\vec{V}_{ p}\) can be expressed in its components as -
\( \left[ \vec{V}_p \right]_i = \left| V_p \right| \hat{i} \)
\( \left[ \vec{V}_p \right]_j = 0 \)
Note that, as the planet’s heliocentric velocity\(\vec{ V}_{ p}\) lays on the horizontal axis (assumption 2), its vertical component \([\vec{V}_{ p}]\) j is zero. For this reason, \(\vec{V}_{ p} \)can be expressed in base vectors as -
\(\vec {V}_{ p} = |\vec{V}_{ p}|î\)
All these expressions will be used in the following subsection, in order to find \(\vec{V}_{ ∞ \,in}\)
Solving for \(\vec{ V}_{ ∞\, in}\)
Substituting these expressions in the previous equation for the inbound hyperbolic excess velocity \(\vec{V}_{∞ \,in}\)
\( \vec{V}_{\infty \text{in}} = |\vec{V}_{\text{sun in}}| \cos(\alpha_{\text{in}}) \hat{i} - |\vec{V}_{p}| \hat{i} + |\vec{V}_{\text{sun in}}| \sin(\alpha_{\text{in}}) \hat{j} \)
Thus, re-writing the equation completely in base vectors -
\( |\vec{V}_{\infty \text{in}}| \cos(\alpha_{\text{in}}) \hat{i} + |\vec{V}_{\infty \text{in}}| \sin(\alpha_{\text{in}}) \hat{j} = |\vec{V}_{\text{sun in}}| \cos(\alpha_{\text{in}}) \hat{i} - |\vec{V}_{p}| \hat{i} + |\vec{V}_{\text{sun in}}| \sin(\alpha_{\text{in}}) \hat{j} \)
Or, in a simpler way -
\(|\vec{V}_{\infty \text{in}}| \hat{i} + |\vec{V}_{\infty \text{in}}| \hat{j} = |\vec{V}_{\text{sun in}}| \hat{i} + |\vec{V}_{\text{sun in}}| \hat{j} - |\vec{V}_{p}|_{ \hat{i}}\)
With the equation only in horizontal components,\( [\vec{V}_{ ∞ \,in} ]_{i}\) can be expressed as -
\([\vec{v}_{\infty in}]_i = [\vec{v}_{sun in}]_i - [\vec{v}_p]_{i}\)
\([\vec{v}_{\infty in}]_i = |\vec{v}_{sun in}|\cos(\alpha_{in})\hat{i} - |\vec{v}_p|\hat{i}\)
Writing the equation only in vertical components, \([\vec{V}_{ ∞\, in}] j\) can be expressed as -
\( [\vec{v}_{\infty in}]_j = [\vec{v}_{sun in}]_j \)
\( [\vec{v}_{\infty in}]_j = |\vec{v}_{sun in}|\sin(\alpha_{in})\hat{j} \)
Thus, the inbound hyperbolic excess velocity \(\vec{V}_{ ∞\, in}\) can be expressed in base vectors as -
\(\vec{v}_{\infty in} = (|\vec{v}_{sun in}|\cos(\alpha_{in}) - |\vec{v}_p|)\hat{i} + |\vec{v}_{sun in}|\sin(\alpha_{in})\hat{j}\)
From this, the magnitude of the hyperbolic excess velocity can be found using Pythagoras’ theorem.
\(|\vec{V}_{ ∞ \,in}| = \sqrt{(|\vec{V}_{ sun \,in}|cos(α_{in}) − |\vec{V}_{ p}|)^ 2 + (|\vec{V}_{ sun \,in}|sin(α_{in}))^2}\)
Expanding the square -
\(|\vec{V}_{ ∞\, in}| = \sqrt{(|\vec{V} _{sun \,in}|^ 2 ⋅ cos^ 2(α_{in}) − 2|\vec{V}_ {sun \,in}|cos(α_{in}) |\vec{V}_ {p}| + |\vec{V}_ {p}|^ 2 + |\vec{V}_{ sun\, in}|^ 2 sin^ 2(α_{in})}\)
Collecting \(|V_{sun \,in}|^ 2\) -
\(|\vec{V}_{ ∞\, in}| = \sqrt{|\vec{V}_{ sun\, in}|^ 2 [cos^ 2(α_{in}) + sin^ 2(α_{in})] − 2|\vec{V}_ {sun\, in}|cos(α_{in}) |\vec{V}_ {p}| + |\vec{V}_{ p}|^2}\)
Applying the Pythagorean identity -
\(|\vec{V}_{ ∞\, in}| = \sqrt{|\vec{V}_ {sun \,in}|^ 2 − 2 (|\vec{V}_ {sun\, in}| cos(α_{in})|\vec{V}_{ p}| + |\vec{V}_{ p}|^ 2}\)
It can be seen that increasing the magnitude of the spacecraft’s inbound heliocentric velocity \(\vec{V} _{sun\, in}\) or of the planet’s heliocentric velocity \(\vec{V}_{ p}\) increases the magnitude of the inbound hyperbolic excess \(\vec{V}_{ ∞ \,in}\) Furthermore, as the angle αin increases, so does the magnitude of \(\vec{V} _{∞\, in} \)As the magnitude of the inbound hyperbolic excess velocity \(\vec{V}_{ ∞\, in}\) has now been expressed, its angle θin in relation to the horizontal direction will be found to complete its representation.
Finding the angles θin and γ
Knowing the magnitudes of the planet’s velocity\(\vec{ V}_{ p}\) and of the heliocentric velocity\( \vec{V}_{ sun\, in}\) expressed in both the vertical and horizontal components (note that [ V∞ in]i is taken as absolute value because it would be negative), the angle θin can be found as -
\(\tan(\pi - \theta_{\text{in}}) = \frac{\left[ \vec{V}_{\text{sun in}} \right]_j}{\left[ \vec{V}_{\infty \text{ in}} \right]_i}\)
\(\begin{equation} \tan(\pi - \theta_{in}) = \frac{|\vec{V}_{\text{sun in}}|\sin(\alpha_{in})}{\biggl ||\vec{V}_{\text{sun in}}|\cos(\alpha_{in}) - |\vec{V}_{p}|\biggl |} \end{equation} \)
\(\theta_{in} =\pi- \tan^{-1}\left(\frac{|\vec{V}_{sun\ in}|\sin(\alpha_{in})}{|\vec{V}_{sun\ in}|\cos(\alpha_{in}) - |\vec{V}_{p}|}\right) \)
Note that it is assumed that\( θ_{in} >\frac{ π}{ 2} \). A different formula, displayed hereafter, has to be used if that is not the case \((i.e., θin < \frac{π}{ 2} )\)
\(\theta_{in} = \tan^{-1}\left(\frac{|\vec{V}_{sun\ in}|\sin(\alpha_{in})}{|\vec{V}_{sun\ in}|\cos(\alpha_{in}) - |\vec{V}_{p}|}\right) \)
Visualisation Of The Angle Θin And Its Components.
The angle through which the velocity vector changes will be found. This angle will be labelled as γ. The change in the spacecraft’s heliocentric velocity ΔVsun is also shown, resulting from the difference in hyperbolic excess velocities.
This angle can be found in figures 15 and 10, as it is the angle, on the planet’s side, that the asymptotes create. This is because, as previously mentioned, the inbound and outbound hyperbolic velocities lay on the hyperbola’s two asymptotes. The angle φ is also shown, which is half of the supplementary angle of γ and will be used in order to calculate θin. The angle γ will be taken as negative, and would be positive in a leading side flyby (figure
From this visualisation it is possible to see that
φ = θout
And that -
φ = θin + γ
Thus, equating φ gives -
θout = θin + γ
In order to have an expression for θout, γ has to be substituted. The visualisation in figure 15 allows for a simple equation for γ to be written -
γ = π − 2φ
Thus, to find a more valuable expression for γ, φ must be substituted. To derive φ, the eccentricity e of the hyperbola has to be introduced, calculated using the formula \(e =\frac{ c}{ a}\) , where a is the semi-transverse axis and c is the centre to focus distance.
As the centre to focus distance (c) is always bigger than the semi-transverse axis (a) for all hyperbolae, the eccentricity will always result greater than 1, in accordance with the previously given definition of a hyperbola. Since -
\(e= \frac{c}{a}\)
Then, from figure 17 -
\(cos (φ)=\frac{a}{c}=\frac{1}{e}\)
\(φ = cos^ {−1} (\frac{ 1} {e} )\)
However, the parameters c and a cannot be easily calculated or defined, as they are applied to the mathematical concept of a hyperbola, whereas in this paper, the trajectory, being made up of relevant angles and distances, is the unknown that has to be defined. This equation will be useful later on in the analysis. The more relevant equation for e in an orbital mechanics context is9 -
\(e = 1 + \frac{r_p \left| \vec{V}_\infty \right|^2}{\mu} \)
the change in velocity. The gravitational parameter10 μ is commonly expressed as the product of G, the gravitational constant, and M, the sum of the masses of the spacecraft and the planet, however, in this analysis, only the mass of the planet will be considered, as the spacecraft is considered to be point-like (assumption 6) -
μ = Gmp
Having calculated the value of eccentricity given the parameters of the specific manoeuvre, the angle of change γ of the hyperbolic excess velocity \(\vec{V}_{ ∞}\) can be found.
\(\gamma = \pi - 2 \cos^{-1} \left( 1 + \frac{\mu}{r_p |\vec{V}_{\infty}|^2} \right) \)
Finding magnitude and direction of \(\vec{V}_{ sun\, out}\)
Having found the angle γ through which the spacecraft is rotated while performing the manoeuvre, and already having an expression for the angle of elevation from horizontal θin of the inbound hyperbolic excess velocity \(\vec{V}_{ ∞ \,in}\), the angle θout the outbound hyperbolic excess velocity \(\vec{V}_{ ∞}\) out has from horizontal can also be found, as it is simply the sum of the two angles. This operation can be done because the inbound hyperbolic excess velocity \(\vec{V}_{ ∞\, in}\) lays on the entry asymptote of the hyperbola, and γ represents the angle in-between asymptotes. Hence -
θout = θin + γ
As angle γ is negative, the equation holds for the model analysed in this investigation, which is a trailing side flyby.
visualisation of angle θout as a subtraction of two angles in a trailing side flyby.
Having found the angle θout, also the outbound hyperbolic excess velocity is\( \vec{V⃗}_{∞\, out} \)has been fully defined, making it possible to evaluate the outbound heliocentric velocity\( \vec{V}_{ sun\, out}\) The components can be expressed as follows -
\( \left[ \vec{V}_{\text{sun out}} \right]_t = \left| \vec{V}_{\infty \text{ out}} \right| \cos (\theta_{\text{out}}) + \left| \vec{V}_p \right| \)
\(\left[ \vec{V}_{\text{sun out}} \right]_t = \left| \vec{V}_{\infty \text{ out}} \right| \sin (\theta_{\text{out}}) \)
Finally, the magnitude of the outbound heliocentric velocity\( \vec{V}_{sun\, out }\)can be calculated through the use of Pythagoras’ theorem -
\( |\vec{V}_{\text{sun out}}| = \sqrt{\left( |\vec{V}_{\infty \text{ out}}| \cos(\theta_{\text{out}}) + |\vec{V}_p| \right)^2 + \left( |\vec{V}_{\infty \text{ out}}| \sin(\theta_{\text{out}}) \right)^2} \)
Expanding the square -
\(|\vec{V}_{\text{sun out}}| = \sqrt{ |\vec{V}_{\infty \text{ out}}|^2 \cos^2(\theta_{\text{out}}) + 2|\vec{V}_{\infty \text{ out}}| \cos(\theta_{\text{out}}) |\vec{V}_p| + |\vec{V}_p|^2 + |\vec{V}_{\infty \text{ out}}|^2 \sin^2(\theta_{\text{out}}) }\)
Collecting \(|\vec{V}_{∞\, out}| ^2 \) -
\(|\vec{V}_{\text{sun out}}| = \sqrt{ |\vec{V}_{\infty \text{ out}}|^2 [ \cos^2(\theta_{\text{out}}) + \sin^2(\theta_{\text{out}}) ] + 2|\vec{V}_{\infty \text{ out}}| \cos(\theta_{\text{out}}) |\vec{V}_p| + |\vec{V}_p|^2 }\)
Applying the Pythagorean identity -
\(|\vec{V}_{\text{sun out}}| = \sqrt{|\vec{V}_{\infty \text{out}}|^2 + 2(|\vec{V}_{\infty \text{out}}| \cos(\theta_{\text{out}})) |\vec{V}_{p}| + |\vec{V}_{p}|^2} \)
The angle αout is now needed to complete the analysis of the outbound heliocentric velocity \(\vec{V}_{ sun \,out}\) It can be calculated through the use of trigonometry.
\(\sin(\alpha_{\text{out}}) = \frac{|\vec{V}_{\infty \text{out}}| \sin(\theta_{\text{out}})}{\sqrt{|\vec{V}_{\infty \text{out}}|^2 + 2(|\vec{V}_{\infty \text{out}}| \cos(\theta_{\text{out}})) |\vec{V}_{p}| + |\vec{V}_{p}|^2}}\)
Having found the magnitude and the direction of the outbound heliocentric velocity \(\vec{V}_{ sun\, out}\), the investigation is now complete. The trajectory of the spacecraft is also known, thanks to the relevant parameters of the hyperbola.
Now that all the relevant formulas have been derived, an example will be shown, in which the data will be taken from real-world data, taken from the Voyager 2 mission. Specifically, the data used will be of Voyager’s first flyby, the one on Jupiter, which happened in the first days of July 1979. The results obtained will be then checked with the real-world data from Voyager 2. The speeds are retrieved from the Horizon System , an online database developed by NASA’s JPL, and the values for Jupiter from the Jupiter Fact Sheet provided by NASA The angle αin will be taken from graphs of the spacecraft’s trajectory. Firsty, the initial parameters required for the calculations to be brought forward have to be defined. From the previous analysis, they are -
M - the planet’s mass
rp - the periapsis radius
\(|\vec{V}_{ sun \,in}|\) - the spacecraft’s heliocentric velocity when entering the planet’s sphere of influence. The magnitude and direction (αin) of this vector are both needed. The magnitude of the velocity is obtained from the Horizons System data for Voyager 2 at 08:00 July 9, 1979, 36 hours before the spacecraft’s closest encounter with Jupiter.
\(|\vec{V}_{ p}| \)- the planet’s heliocentric velocity. In this case, the mean velocity will be used (see assumption 1).
Now that these parameters have been outlined, the relevant data will be specified.
M - 1.898 x 1027 kg
rp - 71,492km (Jupiter’s radius) + 645,000km (closest encounter) = 716492 km = 716500 km
\(|\vec{V}_{\text{sun in}}| \approx 10.52 \, \text{km/s}\)
\(\alpha_{\text{in}} = \frac{\pi}{3} \)
\(|\vec{V}_{p}| \approx 13.06 \, \text{km/s}\)
The operations outlined in the analysis will now be carried out, starting with defining the characteristics of the hyperbola, being the eccentricity and relevant angles, that are dependent from the inbound hyperbolic excess velocity \(\vec{V}_{ ∞\, in}\), the first value that will be expressed. All the results will be given to four significant figures.
\(|\vec{V}_{\infty \text{ out}}| = \sqrt{|\vec{V}_{\text{sun in}}|^2 - 2(|\vec{V}_{\text{sun in}}|\cos(\alpha_{\text{in}}))|\vec{V}_{p}| + |\vec{V}_{p}|^2} \)
\(|\vec{V}_{\infty \text{ out}}| = \sqrt{10.52\ldots^2 - 2\left(10.52\cos\left(\frac{\pi}{3}\right)\right)13.06\ldots + 13.06\ldots^2}\)
\(|\vec{V}_{\infty \text{ out}}| = \sqrt{143.8\ldots}\)
\(|\vec{V}_{\infty \text{ out}}| \approx 11.99\)
Before calculating the eccentricity, the gravitational parameter μ has to be calculated. This can be done by simply multiplying the planet’s mass M by the gravitational constant G.
\(\mu = GM\)
\(\mu = 6.674 \cdot 10^{-11} \cdot 1.898 \cdot 10^{27}\)
\(\mu = 1.267 \cdot 10^{17} \, \text{m}^3 \text{s}^{-2}\)
This value matches the value given in JPL’s “Astrodynamic Parameters”, confirming the calculations made. However, as all the speeds are expressed in km/s, μ has to be expressed in km too.
μ = 1.267 ⋅ 108km3s−2
The next step is to calculate the eccentricity that will enable the calculation of γ and φ.
\(e = 1 + \frac{r_p \left| \vec{V}_{\infty} \right|^2}{\mu}\)
\(e = 1 + \frac{716500 \cdot 11.99^2}{1.267 \cdot 10^8} \)
e = 1 + 0.8134. ..
e ≈ 1.813
Having calculated the eccentricity e, the value can be plugged in to find the relevant angles of the hyperbola.
\(\phi = \cos^{-1}\left(\frac{1}{e}\right)\)
\(\phi = \cos^{-1} \left( \frac{1}{1.813\ldots} \right)\)
φ ≈ 0.9867 rad
And, consequently -
γ = π − 2φ
γ = π − 2 ⋅ 0.9867. ..
γ ≈ 1.168 rad
Now, the entry angle θin of the inbound hyperbolic excess velocity has to be found, in order to calculate the exit angle θout of the outbound hyperbolic excess velocity.
\(\theta_{in} = \pi - \tan^{-1} \left( \frac{|\vec{V}_{sun\ in}| \sin(\alpha_{in})}{|\vec{V}_{sun\ in}| \cos(\alpha_{in}) - |\vec{V}_p|} \right) \)
\(\theta_{in} = \pi - \tan^{-1} \left( \frac{10.52 \ldots \sin \left( \frac{\pi}{3} \right)}{10.52 \ldots \cos \left( \frac{\pi}{3} \right) - 13.06 \ldots} \right) \)
θin = π − tan −1 (1.168. . . )
θin ≈ 2.279 rad
Hence -
θout = θin − γ
θout = 2.279. . . −1.168. ..
θout ≈ 1.111 rad
Finally, the magnitude of the outbound heliocentric velocity \(\vec{V}_{ sun\, out}\) can be calculated -
\(|V_{\text{sun out}}| = \sqrt{|\vec{V}_{\infty \text{ out}}|^2 + 2 (|\vec{V}_{\infty \text{ out}}| \cos(\theta_{\text{out}})) |\vec{V}_p| + |\vec{V}_p|^2}\)
\(|V_{\text{sun out}}| = \sqrt{11.99^2 + 2 \cdot 11.99 \cos(1.111) \cdot 13.06 + 13.06^2}\)
\(|V_{\text{sun out}}| = \sqrt{453.3\ldots} \)
\(|V_{\text{sun out}}| \approx 21.29 \text{ km/s}\)
The spacecraft has almost doubled its speed, confirming that the hyperbolic flyby manoeuvre can significantly accelerate a spacecraft without using any fuel. Comparing the obtained result with the data from the horizons system, retrieved 36 hours after the closest encounter with Jupiter, the recorded speed is 24.47 km/s. The difference between the obtained result and the recorded velocity is about 3 km/s. The percentage error in the final velocity is11.5%. This error is quite significant, however, there are many reasons for it, which lay in the assumptions. The main reason is the 2 dimensionality of this analysis, and the fact that gravitational attraction from other nearby celestial bodies, such as Jupiter’s moons, is neglected. More considerations will be made in the conclusion.
In conclusion, I have reached my initial goal of defining the final velocity of a spacecraft after performing a hyperbolic flyby manoeuvre, given specific initial conditions. I also succeeded in finding out what these conditions are. I managed to answer my research question “What are the mathematical methods involved in defining the trajectory and speed of a spacecraft in a hyperbolic flyby manoeuvre?”, defining and applying the mathematics required to solve the problem.
Originally, the idea for my investigation was to analyse a far more complicated topic: gravity assists, however, after some research, I quickly realised that the problem was not feasible for my level of understanding in maths. For this reason, I tried looking for a simpler topic that would still in some way resemble my original idea. When I thought I found it, I started researching it, but at first it was quite difficult because I did not know where to start and where to aim. However, once I got started with planning and applying the mathematics to the scenario I had chosen, progress was easier. I had to base my research on secondary sources which had done something similar, and I went through a great amount of books and online sources in order to manage to complete this investigation. Nonetheless, I sometimes found myself at a dead end, where I could not find any way to go on with the investigation, for example when trying to find a way to define the relevant angles of the hyperbola. During these moments I had to take a wider look at the problem and find other solutions, as it was often the case that I found one way of solving that particular part and tried to stick with it, even though it wasn't working.
The methods used, mainly trigonometry and vector theory, turned out to be quite simple concepts, however, the difficulty laid in applying them correctly. This is because it was sometimes confusing to try and fit concepts which I had studied theoretically in class and on my own to a complex real world situation as was the one I chose to analyse. Thanks to these difficulties, I learned how mathematical topics that I once saw theoretical and distant from the real world can be applied and used to describe complicated scenarios in our universe. In my opinion, this is a skill that is crucial, because it enables me to make links between what I study and my everyday life, solidifying the concepts that I learn at school. Furthermore, it was very interesting and fascinating to discover how mathematics that I could understand is capable of describing and analysing phenomena that seem very complex and out of my reach.
My calculations from real-world data yielded an error of 3 km/s out of a calculated value of 21.29 km/s. It is a significant error, 11.5%, however there are many reasons behind it. The main one is the many assumptions that have been made prior to the analysis, especially assumption 5, as Voyager 2’s trajectory will have been affected, even though slightly, by Jupiter's moons or by Saturn, which was nearby at the time of the flyby, causing more acceleration compared to the Voyager 2 - Jupiter interaction alone. However, calculating the attraction from moons and other planets would have definitely been out of my reach. Furthermore, the model describes the manoeuvre in two dimensions, whilst the real flyby happened in three. Despite the fact that all orbits, if not perturbed, lay on a two-dimensional plane, because of the influence of other planets and different orbital inclinations, velocity vectors often have a third component in the z direction, which wasn’t accounted for in this analysis. Another fault regarding the calculation is the angle αin of the inbound heliocentric velocity. As this angle could not be found specified in any document, it was taken from graphs of Voyager 2’s trajectory, thus is approximately correct, however lacks the precision required to obtain an accurate result.
If I were to further investigate this topic in the future, I would try to analyse it including the third dimension, making the results more accurate and representing a scenario which is nearer to reality. However, I think that trying to solve the problem without the other assumptions, for example keeping into consideration other gravitational attractions, would prove to be too difficult, even after further study of the topic and of mathematics. I think that analysing some different aspects of orbital mechanics, for example gravity assists themselves, would be interesting and feasible, working upon the skills and knowledge I have developed throughout this investigation.
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All the graphs have been created by me in Adobe Illustrator.