What is the trajectory that minimizes the radiation received by a spacecraft performing a flyby around a star?

The idea for this exploration came from my passion for space exploration, as I was thinking about analysing interplanetary missions, both manned and unmanned. I was visualising the scenario in which a spacecraft has to perform a flyby of a star or another object, while receiving the least possible radiation from it. Radiation might be in the form of x-rays, which have been found emitted from cores of dead stars , or might be radiated heat in the form of electromagnetic waves, such as in the sun’s case (mostly in the infrared, visible and ultraviolet parts of the spectrum), or in any other active star. In the case of a manned mission, the radiation might be harmful for the humans, or in both cases, the exposure might cause damage to the spacecraft itself or to its instruments. For this reason, I wanted to investigate how the trajectory of a spacecraft around a star changes the amount of radiation received, and then finding the trajectory that minimises it, thus providing a solution to the problem of radiation absorption.

 

Because the scenario consists of a flyby, the problem will be limited to hyperbolic orbits, or orbits that follow a conic section with eccentricity greater than one. Eccentricity is a non-negative real number that characterises the shape of all conic sections, which in turn are all geometric curves that can be obtained from the intersection of two cones, with a common vertex and same axis, and a plane. These curves are circles, ellipses, parabolas and hyperbolas. These orbits take place when the concerned spacecraft travels with a velocity that is greater than the escape velocity (the velocity required to escape the gravitational attraction of an object) of the system. For this reason hyperbolic orbits, as well as parabolic orbits, are called open orbits, as the spacecraft leaves the system of the planet or star after having performed the flyby. The other type of orbit is a closed orbit, such as an elliptical or circular orbit, which occurs when the orbiting body travels with a velocity lower than the escape velocity of the system. These orbits will not be taken into consideration, as they would not fit in the hypothetical model. Having said this, the problem to be solved is -

 

What is the trajectory that minimises the radiation received by a spacecraft performing a flyby around a star?

 

The maths that will be used throughout this investigation will include topics from calculus, parametric representations of curves, and cartesian geometry.

A coordinate system has to be chosen and defined in order to describe the possible trajectories of the spacecraft. A two-dimensional cartesian plane has been chosen, as the problem will be limited to the two dimensions because all two-body interactions in space can be modelled as such. The only aspect that would change is the inclination of the plane of the orbit, but that doesn’t have an impact on the total radiation absorbed.

 

The cartesian plane will be organised as follows: the star will be put at the origin, while being one of the two foci of the hyperbola. Furthermore, two points will be chosen as the start and end points of the flyby. These will be assumed as the points in which the spacecraft enters into the star’s sphere of influence, or the area in which the star is the main gravitational force acting on an orbiting object. We can assume that the spacecraft begins experiencing significant radiation exposure from the star at that distance.

The starting point will be taken as Ps and the ending point will be Pf. These points have been put at a distance of 1 unit from the star (the origin) for ease of calculation, but can be scaled up to match a realistic quantity, and won’t interfere with the trajectory. Note that the star will be taken as point-like, and the distances will be calculated from the origin. In a real scenario, this would be accurate as the ratio between the star’s radius and the probe-star distance is very small, as a close flyby would most likely destroy the spacecraft.

A hyperbola is a conic section with eccentricity greater than one, and in orbital mechanics is classified as an open orbit with positive energy, meaning that the orbiting particle has enough velocity to escape the body it is orbiting. Additionally, a hyperbola can be defined as a curve in which the difference of distance from the two foci is constant . From a mathematical point of view, a hyperbola has two opposing and symmetrical branches. However, in the context of orbital mechanics, only one branch is relevant, as it represents the trajectory of a spacecraft that performs a flyby near an object that lays on one focus of the hyperbola. The probe’s trajectory will follow the branch that is closest to the focus occupied by the object, while the other branch and focus are left empty. This is because the focus nearer to the hyperbolic branch that represents the trajectory of the spacecraft exerts a gravitational attractive force on the spacecraft. If the object on the focus were to exert a repulsive force on the particle, then the opposing branch of the hyperbola would represent the trajectory of that particle. The foci are defined as two fixed points that are inside of the curve of the hyperbola, one for each branch. Their location can be found using a hyperbola’s characteristic parameters, which will be illustrated later on.

Conventionally, a hyperbola has some parameters that are represented in its mathematical equation. c is the distance between the foci and the intersection of the asymptotes, also called centre. a is the distance from the centre to the vertex, which is the closest point on the hyperbola to the focus, and b is the distance from the vertex to the length c on one of the asymptotes. M is the centre of the hyperbola, or the intersection of the asymptotes, and F₁ and F₂ are the two foci.

These parameters are used to calculate one of the defining characteristics of a hyperbola - the eccentricity. The formula for the eccentricity is -

 

\(e=\frac{c}{a}\)

 

Since c will always be greater than a, the eccentricity will always be greater than one for all hyperbolae. It can also be deduced that a parabola, the eccentricity of which is one, is a special case of a hyperbola, in which c and a have the same value.

 

Furthermore, these parameters can also be found in the general form for the equation of a hyperbola, which is -

 

\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)

The instantaneous intensity of radiation I an object is exposed to, regardless of type, is inversely proportional to the distance from the source of said radiation, r, squared5 . This proportionality is known as the inverse square law.

 

\(I∝\frac{1}{r^2}\)

 

The law holds true because as the source emits radiation in all directions, it creates a sphere. This sphere’s radius increases as the radiation gets further from the source, and so does the surface area of the sphere. However the area does not decrease linearly, because it is proportional to the radius squared -

 

\(A=4\pi\,r^2\)

 

\(A∝r^2\)

 

However, as the total energy emitted by the source remains the same, the intensity per unit area must decrease.

In the analysed problem, the intensity of the radiation at a specific moment of time can be expressed as the derivative of the total intensity I over time, which can be then equated to the inverse square law. A coefficient will also be added, that accounts for the star’s luminosity L_0, measured in watt (W), and S, the exposed surface area of the spacecraft, which will be taken as constant.

 

\(\frac{dI}{dt}=s\frac{l_o}{4\pi}\frac{1}{r^2}\)

 

To find I, the derivative of the instantaneous intensity has to be integrated for the time it takes the spacecraft to travel from the starting point to the finishing point, or the time it takes the spacecraft to perform the flyby. This value will be denoted by s. Throughout this period of time the craft is exposed to the star’s radiation. Because the distance from the star changes during the manoeuvre, r_t represents the radius at the specific time t. As the coefficient of radiation is constant, it can be taken out of the integral and ignored, as it has no influence on the final function. The resulting integral is -

 

\(\int_{0}^{s} \frac{dI}{dt} \, dt = \int_{0}^{s} \frac{1}{(r_t)^2} \, dt \)

 

The integral cannot be solved as is - the bounds of integration are not known. For this reason, further analysis has to be carried out, focussing on the trajectory that the spacecraft follows.

In the analysed problem, the intensity of the radiation at a specific moment of time can be expressed as the derivative of the total intensity I over time, which can be then equated to the inverse square law. A coefficient will also be added, that accounts for the star’s luminosity L_0, measured in watt (W), and S, the exposed surface area of the spacecraft, which will be taken as constant.

 

\(\frac{dI}{dt}=s\frac{l_o}{4\pi}\frac{1}{r^2}\)

 

To find I, the derivative of the instantaneous intensity has to be integrated for the time it takes the spacecraft to travel from the starting point to the finishing point, or the time it takes the spacecraft to perform the flyby. This value will be denoted by s. Throughout this period of time the craft is exposed to the star’s radiation. Because the distance from the star changes during the manoeuvre, r_t represents the radius at the specific time t. As the coefficient of radiation is constant, it can be taken out of the integral and ignored, as it has no influence on the final function. The resulting integral is -

 

\(\int_{0}^{s} \frac{dI}{dt} \, dt = \int_{0}^{s} \frac{1}{(r_t)^2} \, dt \)

 

The integral cannot be solved as is - the bounds of integration are not known. For this reason, further analysis has to be carried out, focussing on the trajectory that the spacecraft follows.

Hyperbolae, like most other geometric shapes, can be represented in parametric form. This type of representation will be used for the rest of the investigation, as it will be easier to work with, and will enable the integral to be solved. The parametric representation, as the name suggests, introduces a parameter, in this case t, and gives a function for the x and y coordinates of the hyperbola. This is so that for every value of t, the coordinates (x(t), y(t)) will lay on the curve. The representation used will be -

 

\(x=acosh(t)\)

 

\(y=bsinh(t)\)

 

This representation uses the hyperbolic sine and cosine, which, just as their trigonometric analogues, are defined from the unit hyperbola, which has equation x² − y² = 1. In these functions, a and b represent the same characteristics of a hyperbola that have previously been mentioned. The parameter t varies throughout the set [- ∞;+ ∞]. It is not time, but it can be taken as such, as the speed is assumed to be constant for the sake of simplicity. This is because, in a real world situation, the probe would accelerate whilst approaching the planet, reaching the highest speed at periapsis (the closest point to the star that lays on its trajectory). However, the speed throughout the manoeuvre doesn’t have a big impact on the radiation received from the spacecraft, as this simplification is applied to all the hyperbolae considered. Furthermore, as the trajectory is symmetric, the speed balances out and can be taken as an average value.

Now that the problem has been analysed and described, the integral to be solved defined, and the parametric representation used set, some further mathematics is needed in order to evaluate the integral that has been previously expressed. The first step is to calculate rt , the distance the spacecraft has from the star, or, equally, from the focus, at a specific time t. Firstly, the coordinates of the focus F₁ (the right focus and consequently the right branch of the hyperbola will be used, but the calculations apply also to the left focus) can be found as -

 

\(c=\sqrt{a^2+b^2}\)

 

With c being the centre to focus distance expressed through Pythagoras theorem and parameters a and b.

Consequently, as -

 

F(c,o)

 

Substituting for c yields -

 

\(F_1(\sqrt{a^2+b^2,o)}\)

 

For this reason, the square of distance from a generic point on the hyperbola P( x(t), y(t)) to F₁ , \(\frac{r^ 2}{t} \), can be calculated as -

 

\(r_t^2 = (x_t - x_{F_t})^2 + (y_t - y_{F_t})^2 \)

 

\(r_t^2 = \left(a \cdot \cosh(t) - \sqrt{a^2 + b^2}\right)^2 + \left(b \cdot \sinh(t) - o\right)^2 \)

 

\(r_t^2 = a^2 \cosh^2(t) + a^2 + b^2 - 2a \sqrt{a^2 + b^2} \cosh(t) + b^2 \sinh^2(t) \)

 

Resulting from Pythagoras theorem. Applying the fundamental identity for hyperbolic sine and cosine -

 

cosh2(t) - sinh2(t)=1

 

And multiplying both sides by b² , then -

 

\(b^2 \cosh^2(t) - b^2 \sinh^2(t) = b^2 \)

 

\(b^2 \cosh^2(t) = b^2 + b^2 \sinh^2(t)\)

 

Thus, the previous equation can be re-written as -

 

\(r_t^2 = a^2 \cosh^2(t) + a^2 - 2a\alpha \sqrt{a^2 + b^2} \cdot \cosh(t) + b^2 \cosh^2(t) \)

 

Collecting cosh² (t) -

 

\(r_t^2 = \left[ a^2 + b^2 \right] \cosh^2(t) + a^2 - 2a \sqrt{a^2 + b^2} \cdot \cosh(t) \)

 

Which can be re-written as a square of a binomial -

 

\(r_t^2 = \left( \sqrt{a^2 + b^2 \cdot \cosh(t)} - a \right)^2 \)

 

Further rearranging this equation -

 

\(r_t^2 = \left( a \sqrt{1 + \frac{b^2}{a^2} \cdot \cosh(t)} - a \right)^2\)

 

\(r_t^2 = a^2 \left( \sqrt{1 + \frac{b^2}{a^2} \cdot \cosh(t)} - 1 \right)^2\)

Now that the distance \(r\frac{t}{ 2}\) has been found, the previously expressed integral can be solved. However, some considerations on the parameter t and on distance have to be made. The spacecraft enters the sphere of influence of the star at a distance R from the focus, such that -

 

\(R = a \left( \sqrt{1 + \frac{b^2}{a^2} \cdot \cosh(t)} - 1 \right) \)

 

As the hyperbolic cosine is a symmetric function, the values of t at entry and exit from the sphere of influence are the same, but opposite in sign.

 

\(t_{\text{in}} = t\)

 

\(t_{\text{out}} = -t\)

 

For this reason, the integral that has to be solved is -

 

\(r_t^2 = a^2 \left( \sqrt{1 + \frac{b^2}{a^2} \cdot \cosh(t) - 1} \right)^2 \)

 

\(f(a, b) = \int_{-t}^{t} \frac{1}{a^2 \left( \sqrt{1 + \frac{b^2}{a^2} \cdot \cosh(t) - 1} \right)^2} \, dt\)

 

After ruling out an analytical evaluation of the integral, through substitution or integration by parts, it has been decided that the integral should be evaluated numerically. I tried with online evaluation tools, such as Symbolab or Photomath, but they couldn't handle the equation. I then moved on to Wolfram Mathematica, a much more powerful tool. However, not even that could solve the equation. This is because, when graphing the original function that had to be integrated on Desmos online graphing calculator, it yields a convergent function with an asymptote on the x-axis at zero. It also approaches a maxima at around 4200 on the y-axis, slowly approaching the axis itself. The graph was taken with two arbitrary values for a and b (0,5 and 2).

For this reason, it is very difficult for the software to evaluate the integral (i.e. find the area under the graph). This could be improved by using techniques such as the Hadamard regularisation, that would allow to remove the convergent parts of the function and evaluate only the finite ones. However, these techniques are far beyond the scope of this paper, and involve a knowledge of mathematics that is far out of my reach. Given the shape of this function its integral would not give any maxima or minima, but would converge itself to a value. This result will be commented on in the conclusion.

When I first started this investigation, I felt confident about being able to solve the problem I had framed and find a result. However, as I got into calculating it with maths, I quickly realised the task was harder than expected. The hard part proved working with the initial integral, that, despite being quite simple in its form, could not be evaluated as it was. For this reason, I had to try and look for different ways of representing curves, and stumbled upon the parametric representations, which I had never studied. This new knowledge I had found enabled me to move forward in solving the problem, while learning something about a topic of mathematics I previously didn’t know existed. Another setback occurred when I had to evaluate the found integral, because the only method I knew to do so was analytical evaluation, which, as it turned out, wasn’t viable. For this reason, I had to find another method to calculate it, which had to be numerical. I looked online, but the few programs I was accustomed to, such as Symbolab, weren't capable of evaluating the integral. After some more research online, I found Wolfram Mathematica, a software that looked like the right one for me. I had to understand how it works, its functioning is not immediately clear, and despite not being able to find a usable result with it, I still managed to learn how to use a very powerful tool that might be helpful in the future.

 

The final result I obtained didn’t surprise me that much, because, after realising I couldn’t frame the problem as I initially intended, as only one hyperbola passes through two points and one focus, I understood that solving it using the distance R from the star wouldn’t work. This is because the function takes into consideration hyperbolae that are extremely small in length and far from the star, as they are at the edge of the sphere of influence. These hyperbolae yield a very small amount of radiation, and for this reason the final function is always decreasing. As the values of a and b tend to infinity, the hyperbolae stretch more and more, becoming close to straight lines in shape. For this reason, the trajectories yield a minimal amount of radiation. To improve the model, the bounds of integration have to be improved. As the distance R from the star cannot be used, so is t. Using limits that tend to infinity might work, as the integral would take into consideration the whole hyperbola, adding up the total radiation absorbed from the spacecraft. However, this model wouldn’t account for the gravitational field of the star, which significantly decreases as the distance from it increases.

In conclusion, through this investigation, I realised that orbital mechanics is a very complicated field, and especially that very complex maths is required in order to precisely evaluate scenarios related to interactions between man made spacecraft and celestial objects, as well as interactions between celestial objects. Furthermore, a very large computational capacity is required to evaluate these complex equations. If Mathematica, which is quite a powerful tool, struggled in evaluating the relatively simple equation yielded from my investigation, I can’t imagine the complexity of the software and hardware required to compute integrals that give the level of precision required by space agencies. Nonetheless, while writing this paper, I had the opportunity to solve problems and discover new applications and areas of maths, through which I managed to arrive at a solution, and, all in all, I am satisfied with my work.

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All the diagrams have been created by me in Adobe Illustrator.