2

ORBITS AS CONIC SECTIONS

In Chapter 1, the Two Body Equation of Motion was developed and we discussed how the elliptical orbit was one possible solution. In general, however, the solution can be any of the four conic sections: circles, ellipses, parabolas and hyperbolas. We will start this chapter with a discussion of how a satellite gets into orbit and relate it to the conic sections. We will then review the elliptical orbit and its parameters and then extend these results to consider the other conic sections.

First, how does a satellite get into orbit? Imagine you are at the top of a mountain and start firing a cannon from it. We can think of the cannonball as a projectile, an object that, when it is set in motion, continues in its path by its own inertia and is influenced only by the downward force of gravity. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at a constant velocity. The vertical motion of the projective, or cannonball, is influenced by the gravitational acceleration, g, pulling downward. The horizontal and vertical motions are independent, hence, the faster you fire the cannonball, the farther it will go. As an example, try the Newton’s Cannon simulator below. In this simulation, the cannon was fired at 3000 m/s, 4000 m/s, 5000 m/s and 6000 m/s.  Each time the cannonball goes a little farther before hitting the earth. Try the simulator yourself and see what happens! (Schroeder, 2020).

In this image, a simulation of Newton's thought experiment, commonly known as "Newton's Cannon," is depicted. The illustration illustrates the concept of projectile motion and gravity. Newton imagined firing a cannon horizontally from a hypothetical mountain. The cannonball's initial horizontal velocity, combined with the gravitational pull of the Earth, results in a curved trajectory. This thought experiment helped Newton conceptualize the principles of gravity and motion, eventually leading to the development of his laws of motion and the law of universal gravitation.
Figure 1: Newton’s Cannon Simulator (Source: Fowler and Dolgert, 2023).

 

Since the earth is approximately spherically shaped (we will find out later that this it not entirely true), its surface drops about 5 meters for every 8 km the cannonball travels horizontally.  So, if we fire the cannonball faster and faster (assuming no air resistance), its path would eventually match the rate of curvature of the earth. Here are the results of the simulator firing the cannon at 7300 m/s, 7600 m/s, and 8000 m/s.  You will notice that at the lower velocity, the orbit is nearly circular; however, as the cannonball is fired faster and faster, the orbital shape becomes more elliptical. In fact, as more and more horizontal velocity is added, the orbit will eventually become parabolic, then hyperbolic, eventually breaking away from the earth’s sphere of influence entirely. We will consider each of these types of orbits in this chapter.

 

In the provided image, the results of Newton's cannon thought experiment are simulated at three different velocities: 7300 m/s, 7600 m/s, and 8000 m/s. The simulation illustrates the impact of varying horizontal velocities on the cannonball's orbital path. At a lower velocity, the orbit appears nearly circular, while at higher velocities, the orbital shape becomes more elliptical. With increasing horizontal velocity, the orbit transitions through circular, elliptical, parabolic, and eventually hyperbolic, representing the cannonball breaking away from the Earth's sphere of influence. This demonstration highlights the influence of velocity on the type of orbit a projectile can achieve in the context of Newtonian physics.
Figure 2: Types of Orbits Seen in Newton’s Canon Simulation.

A rocket launches vertically to get as quickly as possible out of the earth’s thick atmosphere and minimize drag. However, you will notice that it begins to tilt horizontally and gradually increases this tilt until it achieves orbit around the earth. This technique of optimizing a trajectory of a rocket so that it attains the desired path is called a gravity turn. This technique allows the rocket to use Earth’s gravity rather than its own fuel in order to change its direction. The fuel that the rocket saves can then be used to accelerate it horizontally in order to attain high speed and more easily enter into orbit.

The image depicts a rocket launch strategy known as a gravity turn. Initially launching vertically to swiftly exit the Earth's dense atmosphere and minimize drag, the rocket gradually tilts horizontally during its ascent. This technique optimizes the rocket's trajectory, utilizing Earth's gravity to redirect its path, conserving fuel. The saved fuel can later be employed to accelerate the rocket horizontally, facilitating the attainment of high speed and the transition into orbit around the Earth. The gravity turn is an efficient approach to leverage gravitational forces for a more fuel-efficient and effective journey into space.
Figure 3: A Rocket’s Horizontal Gravity Turn (Source: Arjit Raj, 2017).

ELLIPTICAL ORBITS

Kepler’s First Law said the orbits of the planets are ellipses. These are the most common orbits because one object is ‘captured’ and orbits another larger object. Not only do the planets and minor planets have elliptical orbits, most comets and binary stars also do. The velocity in elliptical orbits is always less than that needed to escape from the central object’s influence. Let us take a closer look at the geometry of an orbital ellipse, which we considered in Chapter 1, and describe each of its parameters more precisely.

The provided image illustrates the geometry associated with the analytical solution for the two-body equation of motion in terms of polar coordinates. The equation expresses the position vector (R) as a function of the semimajor axis (a), eccentricity (e), true anomaly (ν), and the cosine of the true anomaly. The formula, R = a(1 - e^2) / (1 + e * cos ν), is applicable when the eccentricity (e) falls within the range of 0 to 1. The depiction highlights the geometrical relationships involved in describing the orbit of a satellite around a central body using these polar coordinates.
Figure 4: Associated Geometry for the Two-Body Equation of Motion With Emphasis on Angle Beta (Source: Siegenthaler and Saylor, 2017).

[latex]\vec{R}[/latex] = satellite position vector, measured from the center of the Earth

[latex]R[/latex] is the radius from the focus of the ellipse (center of the earth) to the satellite, or [latex]R=\left | \vec{R} \right |[/latex], the magnitude of the [latex]\vec{R}[/latex] vector

[latex]\vec{V}[/latex] = satellite velocity vector, tangent to the orbital path

[latex]V[/latex] is the magnitude of the [latex]\vec{V}[/latex] vector

[latex]F[/latex] and [latex]F'[/latex] = primary (occupied) and vacant (unoccupied) foci of ellipse

[latex]R_{p}[/latex]= Radius of periapsis (closest point) = Radius of perigee when the satellite is around the earth = a(1-e)

[latex]R_{a}[/latex]= Radius of apoapsis (farthest point) = Radius of apogee when the satellite is around the earth = a(1+e)

[latex]=\frac{R_{a}+R_{p}}{2}[/latex]

[latex]a[/latex] = semi-major axis

[latex]b[/latex] = semi-minor axis

[latex]2c[/latex] is the distance between the foci [latex]R_{a}-R_{p}[/latex]

[latex]e[/latex] = eccentricity, the ratio of the distance between the foci (2c) to the length of the ellipse (2a), or [latex]e=\frac{c}{a}[/latex].  Thus

[latex]e=\frac{R_{a}-R_{p}}{R_{a}+R_{p}}[/latex]

Eccentricity defines the shape of the conic section. For ellipses,

0 < [latex]e[/latex] < 1

[latex]V[/latex] = true anomaly = the polar angle measured from perigee to the satellite position vector, [latex]\vec{R}[/latex],  in the direction of satellite motion

[latex]\phi[/latex] = flight path angle, measured from the local horizontal to the velocity vector, [latex]\vec{V}[/latex].

When the satellite is traveling from perigee to apogee, the velocity vector will always be above the local horizon (gaining altitude) so [latex]\phi[/latex] > 0 for [latex]0° < V < 180°[/latex].

When travelling from apogee to perigee, the velocity vector will always be below the local horizon (losing altitude) so [latex]\phi[/latex] < 0 for [latex]180° < V < 360°[/latex].

At exactly apogee and perigee on an ellipse, the position and velocity vectors will be perpendicular so the velocity vector is parallel to the local horizon, hence [latex]\phi[/latex] = 0.

p = semi-latus rectum = the magnitude of the position vectors at [latex]V = 90[/latex] degrees and 270 degrees

Since ellipses are closed curves, an object in an ellipse repeats its path over and over.  The time for a satellite to go once around its orbit is called the period.  For a detailed derivation of the period equation, see Bate.

[latex]Period=2\pi \sqrt{\frac{a^3}{\mu}}[/latex] seconds

Let us review what we know about orbits so far before taking a look at special types of orbits.

  1.  Conic sections are the only possible paths for an orbiting object governed by the ideal two body equation of motion.
  2. The focus of the conic section is the center of the central body.
  3. The specific mechanical energy, [latex]\varepsilon[/latex] is constant, so potential energy and kinetic energy are traded off according to the relationship.

[latex]\varepsilon = \frac{V^2}{2}-\frac{\mu}{R}=-\frac{\mu}{2}[/latex]

 

This relationship also tells us that the orbit size is fixed.

The velocity can then be found from by rearranging this equation to yield:

[latex]V = \sqrt{2(\frac{\mu}{R} + \varepsilon)}[/latex]

 

4.  The specific angular momentum,

[latex]\vec{h} = \vec{R} \times \vec{V}[/latex]

 

is constant, so the orbital plane is fixed in orientation.  Also

[latex]h = |\vec{h}| = RVsin\beta[/latex]

 

where [latex]\beta[/latex] is the angle between [latex]\vec{R}[/latex] and [latex]\vec{V}[/latex] so

[latex]h=RVcos\phi[/latex]

 

Recall at at apogee and perigee, [latex]\vec{R}[/latex] and [latex]\vec{V}[/latex] are perpendicular so since [latex]\phi[/latex] = 0 degrees,

[latex]h = R_{p}V_{p}[/latex]

and

[latex]h = R_{a}V_{a}[/latex]

       Finally, from Chapter 1 we found another express for specific angular momentum,

[latex]h=\sqrt{\mu a(1-e^2)}[/latex]

 

An example of an elliptical orbit is the earth’s orbit around the sun. It has a semi-major axis, a, of approximately 149,598,260 km and an eccentricity of 0.0174. This is a very low eccentricity, which results in only a 3% difference in the distance to the sun at perihelion and aphelion.

The image illustrates Earth's orbit around the Sun, highlighting the elliptical nature of the orbit. Earth completes one orbit around the Sun in a year. The image notes that July 4th is Aphelion Day when Earth is at its farthest distance from the Sun in its orbit. It explains that Earth's orbit is not a perfect circle but rather an ellipse. During summer, Earth is at Aphelion, the farthest point, and during winter, it reaches Perihelion, the closest point to the Sun. The text clarifies that despite being counterintuitive, Earth's seasons are not solely determined by its proximity to the Sun but are influenced by its axial tilt. During Northern Hemisphere summer, the tilt is towards the Sun, while in winter, it tilts away from the Sun.
Figure 5: Earth’s Orbit Around the Sun (Source: Encyclopædia Britannica Inc., 2015).

An example of a highly elliptical earth orbiting satellite is one travelling in a Molniya orbit (see Figure below). It is a 12-hour orbit with semi-major axis, a, of 26,571 km, an eccentricity, e, of about 0.7. Its perigee is located in the Southern hemisphere and the satellite spends 11 hours, or about 92% of its time, in the northern hemisphere. We will study more about Molniya orbits when we discuss perturbations later in this book.

The image depicts a Molniya orbit, showcasing a highly elliptical Earth orbiting satellite. This specific orbit is a 12-hour orbit with a semi-major axis (a) of 26,571 km and an eccentricity (e) of approximately 0.7. The perigee (closest point to Earth) is situated in the Southern hemisphere, and interestingly, the satellite spends approximately 92% of its 12-hour orbit, or 11 hours, in the northern hemisphere. Molniya orbits are known for their unique characteristics and are particularly useful for providing extended coverage to high latitudes. The intricacies of Molniya orbits will be explored further when discussing perturbations later in this book.
Figure 6: Molniya Satellite Orbit Around the Earth (Source: GIFER, 2021).

In the next section, we will consider a special case of elliptical orbits, known as circular orbits. But first, work through this example problem to see how well you understand the concepts from this section. The answers and solutions are at the end of this chapter.

EXAMPLE 1

The altitude of a satellite at perigee is 500 km and its orbital eccentricity is 0.1.  Find:

a)  The satellite’s altitude at apogee

b) The orbit’s specific mechanical energy, [latex]\varepsilon[/latex]

c)  The magnitude of the orbit’s specific angular momentum, [latex]h[/latex]

d)  The satellite’s speed at apogee

 

CIRCULAR ORBITS

The circular orbit is just a special case of an elliptical orbit, so the relationships for an elliptical orbit described above are valid.

The image illustrates a circular orbit around Earth. In a circular orbit, the satellite follows a path that forms a perfect circle. The orbit is characterized by a constant distance from the center of the Earth, maintaining the same altitude throughout its trajectory. Circular orbits are one of the fundamental orbital configurations and are commonly employed for various purposes, including satellite communication and Earth observation. The simplicity and stability of circular orbits make them a practical choice for certain missions.
Figure 7: A Schematic of an Orbit (Source: NASA, 2000).

The semi-major axis, a, is just it’s radius, or [latex]a = R[/latex] and the eccentricity of a circular orbit, [latex]e = 0[/latex].  This makes the period equation, in seconds,

[latex]P = 2\pi \sqrt{\frac{R^3}{\mu}}[/latex]

 

Also notice that the Velocity vector, [latex]\vec{V}[/latex], is always perpendicular to the Position vector, [latex]\vec{R}[/latex], so the flight path angle, [latex]\phi[/latex] is always equal to zero.  We can calculate the speed required for a circular orbit of radius [latex]R[/latex] from the velocity equation:

[latex]V = \sqrt{2(\frac{\mu}{R} + \varepsilon)}[/latex]

 

Now we know

[latex]\varepsilon = -\frac{\mu}{2a} = -\frac{\mu}{2R}[/latex]

 

So now

[latex]V = \sqrt{2(\frac{\mu}{R} - \frac{\mu}{2R})} = \sqrt{2(\frac{\mu}{2R})}[/latex]

 

So the velocity of a satellite in a circular orbit simplifies to a constant:

[latex]V=\sqrt{\frac{\mu}{R}}[/latex]

 

Notice that the greater the radius of the circular orbit, the less speed is required to keep the satellite in this orbit. For a low altitude earth orbit, the circular speed is about 7.8 km/s, while the speed required to keep the moon in its orbit around the earth is only about 0.9 km/s.

Although satellite orbits never have eccentricities exactly equal to zero, there are some satellites that come close. Global Positioning System, GPS, satellites provide navigation services worldwide.  The satellite constellation’s main functions are to transmit radio-navigation signals and to store and retransmit navigation messages. These transmissions are controlled by highly stable atomic clocks that are on board the satellites. The constellation is designed so users will have at least four simultaneous satellites in view from any point on the earth’s surface at any time. The United Sates is committed to maintaining the availability of at least 24 operational GPS satellites 95% of the time.  This means at any one time there are about 30 operational GPS satellites (ESA, 2020).

The image depicts the constellation of Global Positioning System (GPS) satellites in Earth's orbit. The GPS satellites are crucial for providing global navigation services, transmitting radio-navigation signals, and storing and retransmitting navigation messages. These satellites are equipped with highly stable atomic clocks to ensure accurate timing. The GPS constellation is strategically designed to ensure that users on Earth have a minimum of four satellites in view simultaneously from any location, facilitating precise navigation and positioning. The commitment to maintaining at least 24 operational GPS satellites 95% of the time ensures continuous and reliable global coverage for navigation purposes.
Figure 8: GPS Satellite Constellation (Source: ESA, 2011).

GPS satellites fly in medium space orbit (MEO) at an altitude of approximately 20,200 km, which results in a period of 12 hours and an orbital speed of 3.9 km/s. They are in 6 planes spaced equally (60 degrees apart) with four satellites in each. These orbits are nearly circular, with eccentricity less than 0.02.

Another example of a nearly circular orbit is the one used for the International Space Station, ISS.  The ISS is a modular space station, or habitable artificial satellite, and is a multinational collaborative project involving five participating space agencies:  NASA (United States),  Roscosmos (Russia), JAXA (Japan), ESA (Europe), an CSA (Canada). It serves as a microgravity and space environment research laboratory and has been continuously occupied by humans since November 2000. The ISS flies in Low Earth Orbit (LEO) at an altitude of approximately 420 km, which results in a period of just over 90 minutes. This means the ISS circles the earth 15 to 16 times a day! Its orbit is also nearly circular, with an eccentricity of 0.0003 and an orbital speed of 7.7 km/s. It is at a relatively high orbital tilt of 51.6 degrees relative to earth’s equator, resulting in it passing over a large portion of the earth’s surface each day.  

The image provides a visualization of the International Space Station's (ISS) orbit over approximately 4.5 days. The ISS coordinates were recorded using the open-notify.org API and visualized using "Processing" on both a 3D globe and a flat earth map in azimuthal equidistant projection. An interesting observation is the apparent change in speed due to the distortion of space at the edges of the map projection, causing an increase in speed. Despite slight deviations caused by the recording method, the ISS typically orbits in Low Earth Orbit (LEO) at an altitude of around 420 km with a period of just over 90 minutes, completing 15 to 16 orbits daily. Its nearly circular orbit has a small eccentricity of 0.0003, and it travels at an orbital speed of 7.7 km/s. The ISS's high orbital tilt of 51.6 degrees relative to Earth's equator allows it to pass over a large portion of the Earth's surface each day.
Figure 9: ISS Orbit Visualized on the Flat Earth Map (Source: Janosh Gaia, 2019).

In the next section, we will consider parabolic orbits.  But first, work through this example problem to see how well you understand the concepts from this section. The answers and solutions are at the end of this chapter.

EXAMPLE 2

A geostationary orbit, GEO, is one in which a satellite always remains above the same point on the earth’s equator. For a GEO satellite, the radial from the center of the earth to the satellite must have the same angular velocity as the earth itself, or a period of 24 hours.

1.  Calculate the altitude of a GEO orbit.

2. Calculate the specific mechanical energy, [latex]\varepsilon[/latex], of a GEO satellite’s orbit.

3. Calculate the speed of a GEO satellite.

PARABOLIC ORBITS

Parabolic orbits are rarely found in use, but they are important and interesting because they are the borderline case between the elliptical (closed) orbit and the hyperbolic (open) orbit.

The provided image is an animation illustrating a parabolic orbit. In orbital mechanics, a parabolic orbit is one of the three conic sections (along with elliptical and hyperbolic orbits) that a celestial body can follow under the influence of gravity. Unlike elliptical and hyperbolic orbits, which are closed and open, respectively, a parabolic orbit has exactly enough energy to escape from a massive body (like a planet or star), but not enough to reach an infinite distance. The trajectory of a parabolic orbit is a curved path that approaches the massive body but never closes into a complete orbit. This type of orbit is characteristic of objects with a specific energy level, such as those on escape trajectories.
Figure 10: Parabolic Orbit Visualization (Source: Brandir, 2006).

The semi-major axis of a parabolic orbit is infinitely large, so an object traveling in a parabolic orbit is on a one-way trip to infinity and will never retrace the same path again. A unique property of a parabolic orbit is that the two arms of the parabola become more and more nearly parallel as they are extended further and further to the right of the focus.

Also, since the eccentricity of a parabolic orbit is 1, the radius to perigee equation used for elliptical orbits, [latex]R_{p} = a(1-e)[/latex], cannot be used so the more generic form will be used:

[latex]R_{p} = \frac{p}{1 + escov}[/latex]

 

Thus, for [latex]e = 1[/latex] and [latex]V = 0[/latex] degrees, this becomes:

 

[latex]R_{p} = \frac{p}{2}[/latex]

 

Of course, there is no apogee for a parabola and it may be thought of as an infinitely long ellipse.

Even though a planet’s gravitational field strength theoretically extends to infinity, its strength decreases rapidly with distance from the center body, such as the earth. Thus, there is a finite amount of kinetic energy that is needed to overcome the effects of gravity and allow an object to coast to an “infinite” distance without being pulled back toward the earth. The speed at which this occurs is called the escape speed, [latex]V_{esc}[/latex].  Theoretically, as the object’s distance from the earth approaches infinity, it’s speed approaches zero. This results in a total specific mechanical energy, [latex]\varepsilon = 0[/latex] for a parabolic orbit.

We can then calculate the speed necessary to escape by writing the energy equation for this point:

[latex]\varepsilon = \frac{V^2_{esc}}{2} - \frac{\mu}{R} =0[/latex]

 

Thus

[latex]V_{esc}= \sqrt{\frac{2\mu}{R}}[/latex]

 

As you would expect, the farther away you are from the earth, the less speed it takes to escape from the remainder of the gravitational field. The escape speed from the surface of the earth is about 11.2 km/s, while from a point 6300 km above the surface, it will only take about 7.9 km/s.

Although parabolic orbits are not used in practice, some comets approach parabolic orbits. For example, Comet McNaught, also known as the Great Comet of 2007 and given the designation C/2006 P1, is a non-periodic comet. It was discovered on 7 August 2006 by British-Australian astronomer Robert McNaught. It was the brightest comet in over 40 years and it was visible to the naked eye for observers in the Southern Hemisphere in January and February 2007.  Around perihelion on 12 January, it was visible in broad daylight. The comet, as seen from Swifts Creek, Victoria, Australia, may be seen below.

The image depicts Comet C/2006 P1 McNaught, a bright comet that became visible to the naked eye in the southern hemisphere during its 2007 perihelion passage. This photograph captures the comet's impressive tail and bright coma against the dark sky. Comet McNaught was one of the brightest comets visible from Earth in recent decades and gained widespread attention and observation due to its brightness and striking appearance.
Figure 11: Comet C/2006 P1 McNaught (Source: fir0002, 2007).

Its perihelion is 0.17 Astronomical Units (AU), where one AU is the distance of the earth from the sun, or about 149,597,871 km. This means it came as close as 25,544,000 km to earth. However, with an eccentricity of 1.000019, its aphelion is about 4100 AU and it has a period of about 92,600 years! Although we know its last perihelion was in 2007, it is unknown when it will return, or even where humans may be settled in our galaxy that far in the future.

In the next section, we will consider hyperbolic orbits. But first, work through this example problem to see how well you understand the concepts from this section. The answers and solutions are at the end of this chapter.

EXAMPLE 3

A Jupiter probe is in a circular orbit around Earth with a radius of 25,000 km. The next step on the way to Jupiter is to thrust so the probe can enter into an escape orbit.

  1. Determine the probe’s velocity in this circular orbit.
  2. Determine the minimum velocity required to enter a parabolic trajectory at that radius.
  3. Determine the difference in the specific kinetic, potential, and mechanical energies between the two orbits.

HYPERBOLIC ORBITS

Hyperbolic orbits are very useful, especially for interplanetary travel. An interplanetary probe must have some speed left over after it escapes the earth’s gravitational field.

The image illustrates a hyperbolic orbit. In celestial mechanics, a hyperbolic trajectory is the path of an object that has more than enough velocity to escape the gravitational influence of another massive body. Unlike an elliptical orbit, which is closed and bound, a hyperbolic trajectory is open and unbound. Objects following a hyperbolic orbit are typically considered to be on a trajectory that will not bring them back to the body they are currently orbiting. The gif visually represents how an object on a hyperbolic trajectory would travel through space.
Figure 12: Hyperbolic Orbit Visualization (Source: Brandir, 2006).

The semi-major axis of a hyperbolic orbit is negative. This seems like a strange phenomenon, but it makes sense if you consider the specific mechanical energy of a hyperbolic orbit:

[latex]\varepsilon = -\frac{\mu}{2a}[/latex]

If the semi-major axis, a, is negative, this will make the specific mechanical energy positive.  Thus in the energy equation,

[latex]\varepsilon = \frac{V^2}{2}-\frac{\mu}{R}[/latex]

as R gets very large, the potential energy term gets smaller and smaller, leaving only kinetic energy!

In addition, the eccentricity of a hyperbolic orbit is greater than one. The eccentricity gets larger the more spread out the arms are. For some examples of how eccentricity is affected based on the shape of the hyperbola, see the figure below. For more details on hyperbolic orbit geometry, see Bate.

The image illustrates a hyperbola, a geometric curve defined by the difference in distances from two fixed points, called foci, which are located inside the curve. In the context of orbital mechanics, hyperbolic trajectories often describe the paths of celestial bodies, spacecraft, or other objects with sufficient velocity to escape the gravitational pull of a massive body. The animation shows how a hyperbola can be constructed based on its definition using the foci.
Figure 13: Hyperbolic Orbit Eccentricities (Source: American Math Society, 2005).

Also, a similar form to find radius of perigee for a parabolic orbit is used, while radius to apogee is infinitely large.

[latex]R_{p}=\frac{p}{2}[/latex]

 

So now let’s consider how hyperbolic trajectories are used for interplanetary travel and some terminology associated with them. If you give a spacecraft the exact escape speed, it will just barely escape the gravitational field. This means that its speed will approach zero as its distance from the earth approaches infinity. However, if we give our spacecraft more than the escape speed at a point near the earth, the speed at a great distance from the earth will have some positive value. This residual speed that the spacecraft would have left over is called the “hyperbolic excess speed,” or [latex]V_{\infty}[/latex].

The image illustrates the concept of hyperbolic excess speed in the context of interplanetary travel. When a spacecraft is given more than the escape speed at a point near Earth, it will have a residual speed at a great distance from Earth. This residual speed, represented by the hyperbolic excess speed, is the extra velocity that the spacecraft carries beyond the minimum needed to escape Earth's gravitational field. The hyperbolic trajectory is a result of this excess speed, allowing the spacecraft to follow a path that extends to infinity while still having a positive velocity.
Figure 14: Hyperbolic Departure Visualized.

If a satellite starts in a parking orbit near earth, we can find the speed needed to depart from the parking orbit to achieve  [latex]V_{\infty}[/latex] at an infinite distance from the earth, which we will consider the earth’s Sphere of Influence, or SOI.  Although it is difficult to find an agreed-upon definition, for our purposes we can consider it approximately 1.5 million km from the earth.

We can then calculate the boost needed to get on the hyperbolic departure trajectory to the edge of the earth’s SOI by writing the energy equation at two points on this orbit.  The first point is at the parking orbit, called Burnout Velocity, [latex]V_{BO}[/latex].  The second point will be at the edge of the earth’s SOI, or [latex]V_{\infty}[/latex].  Since:

[latex]\varepsilon = \frac{V^2}{2}-\frac{\mu}{R}[/latex]

 

Since the energy needed at the edge of the earth’s SOI, or

[latex]\varepsilon_{\infty}=\frac{V^2_{\infty}}{2}-\frac{\mu}{R_{\infty}}=\frac{V^2_{\infty}}{2}[/latex]

 

This energy must be equal to the energy needed to get on the hyperbolic escape trajectory at the parking orbit, or

[latex]\varepsilon_{h} = \frac{V^2_{BO}}{2}-\frac{\mu}{R_{BO}}[/latex]

 

So

[latex]\frac{V^2_{BO}}{2}-\frac{\mu}{R_{BO}} = \frac{V^2_{\infty}}{2}[/latex]

 

Rearranging this equation yields:

[latex]V^2_{BO}=V^2_{\infty} - \frac{2\mu}{R_{BO}}[/latex]

 

Notice the last term:

[latex]\frac{2\mu}{R_{BO}} = V^2_{esc}[/latex]

 

So our final equation relating these velocities become:

[latex]V^2_{BO}=V^2_{\infty} + V^2_{esc}[/latex]

 

Now let us recap the meaning of these velocity terms.

[latex]V_{BO} =[/latex] the velocity needed at the parking orbit to achieve the velocity needed at the edge of the Earth’s SOI

[latex]V_{\infty} =[/latex] the velocity needed at the edge of the Earth’s SOI to break away from the earth’s influence

[latex]V_{esc} =[/latex]  the velocity need to barely get to the Earth’s SOI with no excess velocity

Finally, to get from an initial parking orbit (assuming circular) to a hyperbolic escape orbit, we must first know the spacecraft’s velocity in the parking orbit:

[latex]V_{Park} = \sqrt{\frac{\mu}{R_{Park}}}[/latex]

 

Then

[latex]V_{BO} = \sqrt{V^2_{\infty}+V^2_{esc}}[/latex]

 

So the change in velocity, or boost needed to escape the earth’s sphere of influence, is given by:

[latex]\Delta V_{needed}=|V_{BO}-V_{Park}|[/latex]

 

Where the vertical lines represent the absolute value of the result.

Although we do not encounter hyperbolic trajectories much on the earth or for the earth-orbiting satellites, they are very common in interplanetary travel and among bodies in our solar system.  Meteors that strike the earth travel on hyperbolic paths relative to the earth. The Perseids Meteor Showers are one example. These meteors, which peak during mid-August, originate from comet 109P/Swift-Tuttle. Swift-Tuttle is in an elliptical orbit around the sun and has a period of 133 years.  Comet Swift-Tuttle last visited the inner solar system in 1992. When it came around the sun, it left a dusty trail behind it. Every year the earth passes through these debris trails, bits of leftover comet particles collide with our atmosphere and disintegrate, creating fiery and colorful streaks in the sky. At its peak, it can result in up to 100 meteors per hour at speeds of approximately 59 km/s (NASA 2019).

The image describes the prevalence of hyperbolic trajectories in interplanetary travel and within our solar system. It highlights that meteors striking Earth travel on hyperbolic paths relative to the planet. The specific example of 109P/Swift-Tuttle, a comet with an elliptical orbit around the sun and a 133-year period, is mentioned. The comet's debris trail, left behind during its last visit in 1992, results in an annual occurrence where Earth passes through the debris, causing meteor showers. The image emphasizes the connection between hyperbolic trajectories, celestial bodies, and the visual spectacle of meteor showers on Earth.
Figure 15: Comet Swift-Tuttle Snapshot (Source: Takoda Edlund, 2007).
The provided image shows a visually striking depiction of a meteor shower, likely the Perseids, one of the most famous and reliable meteor showers that occurs annually. The image captures the streaks of light created by the entry of meteoroids into Earth's atmosphere. Meteor showers occur when the Earth passes through the debris left behind by a comet, such as Swift-Tuttle in the case of the Perseids. The bright trails and numerous meteors in the image illustrate the beauty and celestial spectacle associated with meteor showers.
Figure 16: Perseid Meteor Shower Snapshot (Source: University of Nottingham, 2016).

Interplanetary missions must use hyperbolic trajectories that are relative to the earth if we want the interplanetary probe to break away from the earth’s gravitational field. An interesting interplanetary spacecraft is New Horizons, which was sent to explore far into our solar system. By the time it reached Pluto, the spacecraft had travelled farther away and for a longer time period, more than nine years, than any previous deep space spacecraft ever launched. It observed Pluto close up, flying by the dwarf planet and its moons in July 2015 (NASA 2020).

The image depicts the New Horizons spacecraft passing by the dwarf planet Pluto and its largest moon Charon. This historic moment occurred during the New Horizons mission, which was the first spacecraft to explore Pluto up close. The image captures the distant and icy world of Pluto, revealing details of its surface features. The New Horizons mission provided valuable data and insights into the outer reaches of our solar system, expanding our understanding of these distant celestial bodies.
Figure 17: New Horizons Spacecraft Begins First Stages of Pluto Encounter (Source: NASA, 2015).

In early 2019, New Horizons flew past its second major science target, 2014 MU69, the most distance object ever explored up close. New Horizons discovered the object is a contact binary, two different spherical objects. The lobes were formed separately and then eased together over four billion years ago at speeds between one to two miles per hour. These two rocks were then pulled together by the extremely weak gravitational forces of each other, nudging together at such slow speeds that initial observations from New Horizons do not reveal any stress fractures or damage from the ‘collision.’ More discoveries by New Horizons are yet to come (Gebhart, 2019).

The image provides a diagram describing the size and shape of Ultima Thule, the nickname for the Kuiper Belt object 2014 MU69, as discovered by NASA's New Horizons spacecraft. This distant object, located about 1 billion miles beyond Pluto, is revealed to be a contact binary, consisting of two separate spherical objects that merged together over four billion years ago. The lobes were formed independently and then gently brought together at extremely slow speeds, approximately one to two miles per hour, by the weak gravitational forces between them. The image is part of the groundbreaking discoveries made by the New Horizons mission during its exploration of the outer reaches of our solar system. Studying binary systems helps scientists understand the complexities of stellar processes and the broader dynamics of celestial bodies in the cosmos.
Figure 18: New Horizons Spacecraft Encounter With 2014 MU69, aka Ultima Thule (Source: NASA/JHUAPL/SwRI).

In order to get on this interplanetary trajectory, New Horizons was launched in January 2006 from Cape Canaveral Air Force Station. (You will learn later in this book why the Cape is an ideal location from which to launch an interplanetary probe). It was launched into a hyperbolic departure trajectory, which took it on a path by Jupiter. The giant planet’s gravitational pull helped it “slingshot” the spacecraft toward the outer solar system.

There are two reasons why the New Horizons science team wanted to reach Pluto as soon as possible. The first had to do with Pluto’s atmosphere. Since 1989, Pluto has been moving farther from the Sun, getting less heat every year. As Pluto gets colder, scientists expect its atmosphere will “freeze out,” so the team wanted to arrive while there was a chance to study a thicker atmosphere.

The second reason was to map as much of Pluto and its largest moon, Charon, as possible. On the earth, the North Pole and other areas above the Arctic Circle have half a year of night and half a year of daylight. In the same way, parts of Pluto and Charon never see the Sun for decades at a time. The longer the wait, the more of Pluto and Charon would have been shadowed in a long “arctic night,” impeding the spacecraft’s ability to take pictures of the surface in reflected sunlight.

In February 2007, New Horizons passed through the Jupiter system at more than 50,000 mph, ending up on a path that got it to Pluto on July 14, 2015, about nine and a half years later, compared to the 30 years or so it would take to get there directly. The figure below shows the path New Horizons took to get to Jupiter and the Kuiper Belt Objects (KBOs).  You will learn about gravitational assist trajectories later. Without the use of hyperbolic trajectories, however, New Horizons would have never left the earth’s gravitational field at all!

The image illustrates the trajectory of NASA's New Horizons spacecraft as it utilized a gravitational assist from Jupiter to reach the Kuiper Belt Objects (KBOs). Launched at a speed exceeding 50,000 mph, New Horizons passed through the Jupiter system, capitalizing on the planet's gravity to gain the necessary velocity for its extended journey. This gravitational assist significantly reduced the travel time to Pluto, allowing New Horizons to reach its destination on July 14, 2015, in approximately nine and a half years, compared to the approximately 30 years it would have taken with a direct route. The mission's use of gravitational assists and hyperbolic trajectories highlights the sophisticated orbital mechanics involved in space exploration.
Figure 19: New Horizons Spacecraft Path Through the Cosmos (Source: NASA/New Horizon, 2015).

Now, work through this example problem to see how well you understand the concepts from this section. The answers and solutions are at the end of this chapter.

EXAMPLE 4

A spacecraft is launched from a parking orbit around earth to Mars.

Given:

[latex]V_{\infty} = 2.94 km/s[/latex]

[latex]R_{Park\_at\_Earth} = 6697[/latex] (circular)

Find:

a)  [latex]\varepsilon_{\infty}[/latex] energy needed at the edge of the Earth’s sphere of influence

b) [latex]V_{BO} =[/latex] the velocity needed at the parking orbit to achieve the velocity needed at the edge of the Earth’s SOI

c) [latex]\Delta V_{needed} =[/latex] The boost needed to get the spacecraft from its parking orbit onto the hyperbolic departure trajectory

SUMMARY

In this chapter, we first discussed how a satellite gets into orbit and related it to the conic sections.  We then reviewed elliptical orbits and its parameters, then extended these results to consider the other conic sections. Below is a summary of the orbits types and the value, or range of values, for some of their orbital parameters.

Orbit Type Semi-major axis, [latex]a[/latex] eccentricity, [latex]e[/latex] Specific mechanical energy, [latex]\varepsilon[/latex]
Ellipse [latex]a > 6578 km[/latex]* [latex]0 < e < 1[/latex] [latex]\varepsilon < 0[/latex]
Circle [latex]a = R[/latex] (constant) [latex]e = 0[/latex] [latex]\varepsilon < 0[/latex]
Parabola [latex]a = \infty[/latex] [latex]e = 1[/latex] [latex]\varepsilon = 0[/latex]
Hyperbola [latex]a < 0[/latex] [latex]e > 1[/latex] [latex]\varepsilon > 0[/latex]

* Earth’s radius + 200 km

Check your understanding with this quick quiz!

SOLUTIONS TO EXAMPLES:

***EXAMPLE 1 SOLUTION***

The altitude of a satellite at perigee is 500 km and its orbital eccentricity is 0.1.  Find:

a)  The satellite’s altitude at apogee

Since

[latex]R_{p}=a(1-e)[/latex]

[latex]6378[/latex] km [latex]+[/latex] [latex]500[/latex] km [latex]= a(1-0.1)[/latex]

[latex]a = 6878/0.9[/latex] km

[latex]a = 7642[/latex] km

and

[latex]R_{a} = a(1+e) = 7642(1+0.1)[/latex]

[latex]Ra = 8406 km[/latex]

So the satellite’s altitude at apogee [latex]= 8406 - 6378[/latex] km

Altitude at apogee = 2028 km

b)  The orbit’s specific mechanical energy, [latex]\varepsilon[/latex]

Since

[latex]\varepsilon = -\frac{\mu}{2a}[/latex]

[latex]\varepsilon = -\frac{398600.5\,\frac{km^3}{s^2}}{2(42241\,km)} = -4.718\,\frac{km^2}{s^2}[/latex]

so

[latex]\varepsilon =[/latex] -26.08 km2/s2

c) The magnitude of the orbit’s specific angular momentum, h

Since

[latex]h=\sqrt{\mu a(1-e^2)}[/latex]

[latex]h=\sqrt{398600.5*7642*(1-0.1^2)}[/latex]

so

  h = 54915 km2/s

d)  The satellite’s speed at apogee

Since [latex]h = R_{a}V_{a}[/latex]

[latex]V_{a} = \frac{h}{R_{a}}[/latex]

so

Va = 6.53 km/s

 

***EXAMPLE 2 SOLUTION***

A geostationary orbit, GEO, is one in which a satellite always remains above the same point on the earth’s equator. For a GEO satellite, the radial from the center of the earth to the satellite must have the same angular velocity as the earth itself, or a period of 24 hours.

a)  Calculate the altitude of a GEO orbit.

Since

[latex]Period = 2 \pi \sqrt{\frac{a^3}{\mu}}=24\:hours[/latex]

So

[latex]a=\sqrt[3]{ \left( \frac{24 \, \times \, 3600 \frac{sec}{hr}}{2\pi} \right) ^2 \, (398600.5 \, \frac{km^3}{s^2})}[/latex]

[latex]a = 42241[/latex] km

        altitude of a GEO satellite = 35,863 km 

b) Calculate the specific mechanical energy, [latex]\varepsilon[/latex], of a GEO satellite’s orbit.

[latex]\varepsilon = -\frac{\mu}{2a}[/latex]

[latex]\varepsilon = - \frac{398600.5\:\frac{km^3}{s^2}}{2(7642\:km)}[/latex]

[latex]\varepsilon[/latex] = -4.718 km3/s2

c) Calculate the speed of a GEO satellite.

[latex]V = \sqrt{\frac{\mu}{R}}[/latex]

[latex]V = \sqrt{\frac{398600.5\:\frac{km^3}{s^2}}{42241\:km}}[/latex]

so

V = 3.07 km/s

***EXAMPLE 3 SOLUTION***

A Jupiter probe is in a circular orbit around the earth with a radius of 25,000 km. The next step on the way to Jupiter is to thrust so the probe can enter into an escape orbit.

  1. Determine the probe’s velocity in this circular orbit.

Since [latex]R = 25,000 \, km[/latex]

[latex]V = \frac{\mu}{R}=\sqrt{\frac{398600.5\:\frac{km^3}{s^2}}{25000\:km}}[/latex]

So

 V = 3.99 km/s

2. Determine the minimum velocity required to enter a parabolic trajectory at that radius.

[latex]V_{esc} = \sqrt{\frac{2\mu}{R}}=\sqrt{\frac{2(398600.5\:\frac{km^3}{s^2})}{25000\:km}}[/latex]

So

V = 5.65 km/s

3.  Determine the difference in the specific kinetic, potential, and mechanical energies between the two orbits.

Orbit 1 (circular parking orbit)

[latex]KE = \frac{V^2}{2}=\frac{3.99^2}{2}=7.96\:\frac{km^2}{s^2}[/latex]

[latex]PE = -\frac{\mu}{R} = - \frac{398600.5\frac{km^3}{s^2}}{25000\:km} = -15.94\:\frac{km^2}{s^2}[/latex]

[latex]\varepsilon[/latex] = -7.98 km2/s2

Orbit 2 (Escape Orbit)

[latex]KE = \frac{V^2}{2}=\frac{5.65}{2}=15.96\:\frac{km^2}{s^2}[/latex]

[latex]PE = -15.96\:\frac{km^2}{s^2}[/latex]

[latex]\varepsilon = 0[/latex]

So PE doesn’t change (any small difference is due to round-off error) while [latex]KE[/latex] changes by [latex]8\;km^2/s^2[/latex].  Since the Specific Mechanical Energy changes by approximately [latex]8\;km^2/s^2[/latex], all of the energy added is strictly due to the change in [latex]KE[/latex].

***EXAMPLE 4 SOLUTION***

A spacecraft is launched from a parking orbit around earth to Mars.

Given:

[latex]R_{Park\_at\_Earth} = 6697[/latex] (circular)

Find:

a)  [latex]\varepsilon_{\infty}[/latex] energy needed at the edge of the Earth’s sphere of influence

[latex]\varepsilon_{\infty} = \frac{V^2_{\infty}}{2} = 0.5 * (2.94\:km/s)^2[/latex]

[latex]\varepsilon_{\infty}[/latex] = 4.32 km2/s2

b) [latex]V_{BO}=[/latex]the velocity needed at the parking orbit to achieve the velocity needed at the edge of the Earth’s SOI

[latex]\bf{V^2_{BO} = V^2_{\infty}+V^2_{esc}}[/latex]

So

[latex]V_{BO} = \sqrt{(2.94)^2+\frac{2\,*\,398600.5\:\frac{km^3}{s^2}}{6697\:km}}[/latex]

[latex]V_{BO} = 11.3\:km/s[/latex]

 

c) [latex]\Delta V_{needed} = |V_{BO}-V_{park}| =[/latex] The boost needed to get the spacecraft from its parking orbit onto the hyperbolic departure trajectory

The satellite’s velocity in its original parking orbit is

[latex]V = \sqrt{\frac{\mu}{R}}[/latex]

so

[latex]V_{Park} = \sqrt{\frac{398600.5\:\frac{km^3}{s^2}}{6697\:km}}[/latex]

[latex]V_{Park}=7.71\:km/s[/latex]

and

[latex]\Delta V_{needed} = |11.3-7.715|\:km/s[/latex]

So

[latex]\Delta V_{needed}[/latex] = 3.585 km/s

 

REFERENCES

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European Space Agency navipedia. (2020, 20 November). GPS Space Segment.  https://gssc.esa.int/navipedia/index.php/GPS_Space_Segment

Gebhardt, W. (2019, January 02). 2014 MU69 Revealed as a Contact Binary in First New Horizons Data Returns. https://www.nasaspaceflight.com/2019/01/2014-mu69-contact-binary-first-new-horizons-returns/

NASA Solar System Exploration.  (2019, 19 December).  Perseids:  In Depth.  https://solarsystem.nasa.gov/asteroids-comets-and-meteors/meteors-and-meteorites/perseids/in-depth/

NASA Solar System Exploration. (2020, 7 October). New Horizons:  In Depth.  https://solarsystem.nasa.gov/missions/new-horizons/in-depth7

The Physics Classroom.  What is a projectile? (n.d.) https://www.physicsclassroom.com/Class/vectors/

Schroeder, Daniel V.   Newton’s cannon (2020, 20 January).  https://physics.weber.edu/schroeder/software/NewtonsCannon.html

Sellers, Jerry Jon.  (2005).  Understanding Space:  An Introduction to Astronautics, 3rd edition.  McGraw-Hill.

 

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