4
SATELLITE CONTROL NETWORKS
Now we know how to describe a satellite in orbit in terms of its position and velocity vectors, as well as its orbital elements. We will next consider how a satellite’s ground tracking station tracks a satellite. In other words, through observations from a tracking station on the ground, we can tell exactly where a satellite is. Later, we may want to predict where it will be at some point in the future so that we can acquire data or transmit commands to it. We will refer to these ground stations as part of the spacecraft tracking and communication network.
NASA has its own Space Network (SN) to track and communicate with satellites in orbit. The SN empowers missions with reliable and secure relay communications and tracking services. The SN’s global coverage enables near-continuous, bi-directional communications. The SN’s architecture is comprised of two segments. The space segment consists of a constellation of Tracking and Data Relay Satellites (TDRS) in geosynchronous orbit, which were introduced in this textbook’s introductory chapter (NASA). A series of ground-based antennas make up the network’s ground segment. A 70-meter-wide (230-foot-wide) radio antenna Deep Space Station in Canberra, Australia is shown in Figure 1 below.
![The provided image depicts crews conducting critical upgrades and repairs to the 70-meter-wide (230-foot-wide) radio antenna Deep Space Station 43 in Canberra, Australia. This antenna is part of NASA's Space Network (SN), a crucial component for tracking and communicating with satellites in orbit. The SN's architecture involves two segments: the space segment, featuring a constellation of Tracking and Data Relay Satellites (TDRS) in geosynchronous orbit, and the ground segment, which comprises a series of ground-based antennas. In this clip, one of the antenna’s white feed cones (which house portions of the antenna receivers) is being moved by a crane. The 70-meter-wide antenna in Canberra is a vital element of the network, enabling reliable and secure relay communications and tracking services for various space missions.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/42_voyager2-1280-300x169-1.jpg)
The Air Force also has a Satellite Control Network, which consists of satellite control centers, tracking stations, and test facilities located around the world. Satellite Operations Centers, SOCs, are located at Schriever Air Force Base near Colorado Springs, Colorado and various other locations around the continental United States. The SOCs are manned around the clock and are responsible for the command and control of their assigned satellite systems and are also linked to remote tracking stations, RTSs, around the world. The RTSs provide the link between the satellites and the SOCs. The worldwide locations of the AFSCN are shown below (Space News).
![The image displays the worldwide locations of the Air Force Satellite Control Network (AFSCN), a crucial component of the United States Air Force's satellite operations. The AFSCN consists of Satellite Operations Centers (SOCs) located at Schriever Air Force Base near Colorado Springs, Colorado, and various other locations across the continental United States. These SOCs operate around the clock, managing the command and control of their assigned satellite systems. The network is connected to remote tracking stations (RTSs) positioned globally, serving as the crucial link between the satellites and the SOCs. This distributed network ensures effective satellite communication, tracking, and control capabilities for the Air Force.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/R5149393bda4c732a45b2eb9953a9b276-300x222-1.jpg)
In this chapter, we will concentrate on the observation data we might obtain from a typical ground station and learn how to convert those measurements into position and velocity vectors for the satellite. This initial observation data, or a satellite’s tracking data, is called an ephemeris. Later in this book we will discover how to predict where a satellite will be in the future using an initial ephemeris and Kepler’s Predication problem. Assuming we have a set of future orbital elements, we will then determine how to find observation data for a ground station in order to communicate with the satellite at a future time. This concept is pictured below.
![Diagram illustrating the concept of obtaining observation data from a ground station and converting it into position and velocity vectors for a satellite. The initial observation data, known as an ephemeris, is crucial for tracking and understanding the satellite's current state. The image also hints at the future application of Kepler's Prediction problem, where future orbital elements are used to predict the satellite's position, allowing for the planning of future communications with ground stations.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/BIG-PICTURE-1-300x167-1.png)
We will first use a series of location parameters from the tracking station and their rates as measured from a coordinate frame centered at the tracking station. These parameters are given as:
[latex]\rho[/latex]: Range = distance from tracking station to satellite, usually in km.
Az: Azimuth = the angle measured from North, clockwise (as viewed from above) to the location beneath the object of interest. It can take values 0 Az
360 degrees
El: Elevation = the angle measured from the local horizon (tangent to the earth’s surface) to the radar line of site. Although it can take angles from -90 degrees El
90 degrees, only those satellites wit elevations greater than about 20 degrees can be “seen” from a ground station.
[latex]\dot{\rho}[/latex]: Range Rate = the rate at which the satellite is moving relative to to the ground station. It can have positive or negative values and is usually measured in km/s.
[latex]\dot{Az}[/latex]: Rate of Change of Azimuth = the rate at which the satellite is changing azimuth. It can have positive or negative values and is usually measured in degrees/second or radians/second.
[latex]\dot{El}[/latex]: Rate of Change of Elevation = the rate at which the satellite is changing elevation. It can have positive or negative values as is usually measured in degrees/ second or radians/second.
But notice where we are measuring them from. We are measuring them from the tracking site, therefore we must introduce a new coordinate frame, SEZ.
THE SEZ, SOUTH-EAST-ZENITH, COORDINATE FRAME
The SEZ coordinate frame, sometimes called the Topocentric Horizon Coordinate System, will be used to define the satellite’s initial location. It is shown in the figure below.
![The image depicts the SEZ coordinate frame, sometimes called the Topocentric Horizon Coordinate System. In this coordinate system, celestial objects' positions are measured relative to an observer's local horizon. The system uses zenith (the point in the sky directly above the observer), nadir (the point on the celestial sphere directly below the observer), and the local horizon as reference points. This system is commonly employed in astronomy and observational sciences to describe the apparent positions of stars, planets, and other celestial bodies as observed from a specific location on Earth.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/Picture3-300x215-1.jpg)
Now, we will consider a side view of the satellite in orbit. (Note: not to scale):
![The image illustrates a side view of a satellite in orbit, and it presents coordinate frames with attached vector (Rijk) components. These components are likely representing the position vector (R) in a specific coordinate system. The coordinate frames could be associated with various reference frames used in astrodynamics, such as inertial or orbital frames, and the vector components likely describe the satellite's position in terms of its radial (i), along-track (j), and cross-track (k) components. This representation is commonly used to analyze and describe the motion of satellites in space, considering the different axes and reference frames relevant to the satellite's orbit.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/RIJK-275x300-1.png)
So, we see that:
[latex]\vec{R} = \vec{R}_{\text{site}} + \vec{\rho}[/latex]
But is it really that easy? Not exactly, but it’s not that hard if you keep in mind Figure 1 and the coordinate frames attached to each vector. First, let us consider those coordinate frames, indicated by the subscripts in the following equation:
[latex]\vec{R}_{IJK} = \vec{R}_{\text{site}IJK} + \vec{\rho}_{SEZ}[/latex]
Can vectors in different coordinate frames be added? No more than units of feet and meters can directly be added together. There must be a conversion factor between the two references. The same holds true for vectors – there must be a conversion factor between the two reference frames. In vector analysis, this is called a coordinate transformation.
TRANSFORMATION BETWEEN SEZ AND IJK COORDINATE FRAMES
Why is a coordinate transformation needed in this instance? Well, the radar site observations ([latex]\rho[/latex], Ax, and El) are measured from the ground site. It is important to know that ground site moves with the earth. It is not an inertial coordinate frame, while the IJK frame is inertial. The relationship between the IJK and SEZ frames are shown in the figure below.
![The image depicts the relationships between Earth-Centered, Earth-Fixed (IJK) coordinates and the South-East-Zenith (SEZ) local coordinate system in terms of longitude and latitude. IJK is a global, fixed-coordinate system tied to Earth's center, while SEZ is a local coordinate system that varies based on a specific point on Earth's surface. The illustration provides a visual understanding of how changes in longitude and latitude correspond to the South, East, and Zenith (up) directions in the local SEZ frame.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/SEZ-300x280-1.jpg)
Where:
Rearth = Radius of the earth, the distance from the center of the earth to the ground station site (at 0 altitude), km
LST = Local Sidereal Time (degrees), time measured from the [latex]\hat{I}[/latex] vector, or principle axis, to current location on the earth. More about sidereal time will be discussed in Chapter 6, but we will use it for now because it is an inertial reference frame.
L = latitude (degrees)
In order to proceed further, we will need to find expressions for the quantities x and z in the figure, as can be seen from the orange highlighted right triangle.
![The image depicts the relationships between Earth-Centered, Earth-Fixed (IJK) coordinates and the South-East-Zenith (SEZ) local coordinate system in terms of longitude and latitude. IJK is a global, fixed-coordinate system tied to Earth's center, while SEZ is a local coordinate system that varies based on a specific point on Earth's surface. The unknown quantities (x) and (z) are defined later in the chapter. The illustration provides a visual understanding of how changes in longitude and latitude correspond to the South, East, and Zenith (up) directions in the local SEZ frame.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/triangle-1-300x273-1.jpg)
The equations for x and z hence become:
[latex]x = R_{\text{earth}} \cos L[/latex]
[latex]z = R_{\text{earth}} \sin L[/latex]
Now also considering the projection on the IJ plane:
[latex]\vec{R}_{\text{site}} = x \cos(\text{LST}) \hat{i} + x \sin(\text{LST}) \hat{j} + z \hat{k}[/latex]
Combining the last two equations yields:
[latex]\vec{R}_{\text{site}} = R_{\text{earth}}\cos L \cos(\text{LST}) \hat{i} + R_{\text{earth}}\cos L \sin(\text{LST}) \hat{j} + R_{\text{earth}}\sin L \hat{k}[/latex]
This equation would be a pretty good approximation for [latex]\vec{R}_{\text{site}}[/latex] except for two facts.
- The earth is not spherically symmetrical
- The site may not be at sea level.
For example, Schriever Space Force Base is at an altitude of 1894 m (6214 ft) and Air Force’s Maui Ground Tracking Station is at a whopping 3058 meters (10032 ft) (Pike, Schriever).
We will start by addressing the first point: how we can account for the earth’s oblateness.
EARTH’S OBLATENESS
We could simply use a spherical earth model, but in order to be more accurate we will need to consider a non-spherical earth. The first Vanguard satellite showed the earth to be slightly pear-shaped – this is an effect we will consider in more detail later. However, a reference ellipsoid is a good approximation to “mean sea level” (Bate)
The earth can be modeled fairly accurately as a type of ellipsoid, or a stretched sphere. In the figure below, two types of ellipsoids are shown (Math).
![The image illustrates two types of ellipsoids, which are 3D geometric figures with elliptical shapes. An ellipsoid is akin to a stretched sphere, and any plane cutting through it forms an ellipse. In the figure, the ellipsoid on the left represents an oblate spheroid, characterized by a flattened shape, while the one on the right is a prolate spheroid, exhibiting an elongated form. Examples of ellipsoids in real life include an egg or a blimp. The Earth itself can be accurately modeled as a type of ellipsoid, reflecting its slightly flattened shape at the poles (oblate spheroid).](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/Ellipsoids-300x105-1.jpg)
The ellipsoid on the left is an oblate spheroid; the one on the right is a prolate spheroid. The earth takes the shape of an oblate spheroid due to the rotation of the earth, which causes the earth to swell more at the equator than at the poles. When the earth rotates, there is a strong outward force on earth matter that is near the equator. This means the radius of the earth at the equator is approximately 6,378.145 km, while the distance from the center of the earth to the poles is slightly less, at about 6,356.785 km. The earth’s eccentricity is 0.08182. This results in the geometry in the figure below.
![The figure depicts geographic coordinates expressed as geodetic latitude and longitude, referencing a specific ellipsoid with a defined shape and size. Geodetic latitude and longitude differ slightly from geocentric coordinates. In the illustration, the exaggerated difference between the equatorial and polar axes is shown in the right figure. This emphasizes the deviation from a perfect sphere and highlights the ellipsoidal nature of the Earth's shape. The coordinates are crucial for accurate location representation on the Earth's surface.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/geodetic-and-geocentric.-300x135-1.jpg)
Where:
L = Geodetic latitude, measured from the perpendicular to the earth’s surface to the point where the line crossed the equatorial plane.
L’ = Geocentric latitude
What we are accustomed to using on earth is the Geodetic latitude, L.
Consider what would happen in real life if a person dropped a plumb bob and it pointed toward the center of the earth!
![The image shows a plumb bob pointed toward the Earth's center, representing a geocentric perspective. A plumb bob is a weight suspended by a string, and its direction points directly towards the center of the Earth due to the force of gravity. This serves as a visual representation of the geocentric reference, aligning with the Earth's gravitational pull towards its center.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/Plumb-bob-geodetic-300x230-1.jpg)
![The image illustrates a plumb bob pointed perpendicular to the Earth's surface, representing a geodetic perspective. In this context, geodetic coordinates take into account the actual shape of the Earth, which is not a perfect sphere but rather an ellipsoid. The plumb bob aligns with the local gravity vector, providing a reference perpendicular to the Earth's surface according to the geodetic coordinate system.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/Plumb-bob-geodetic-2-300x237-1.jpg)
In this case, using Geodetic Latitude, more accurate expressions are obtained for x and z.
[latex]x = \left[ \frac{a_e}{\sqrt{1 - e^2 \sin^2 L}} + H \right] \cos L[/latex]
[latex]z = \left[ \frac{a_e (1 - e^2)}{\sqrt{1 - e^2 \sin^2 L}} + H \right] \sin L[/latex]
where
ae = Rearth = 6378.137 km
e = eearth = 0.08182
____________________________________________________
Example 1
Given a radar tracking site at:
L = Latitude = 42 degrees
H = altitude above sea level = 77 meters
LST = local sidereal time = 256 degrees
Find
[latex]\vec{R}_{\text{site}}[/latex]
___________________________________________________
POSITION VECTOR
But we must remember that [latex]\vec{\rho}[/latex] is referenced to the SEZ coordinate frame.
In other words:
[latex](\vec{R})_{IJK} = (\vec{R}_{\text{site}})_{IJK} + (\vec{\rho})_{SEZ}[/latex]
The above equation will not work because, in order to add vectors, all must be referenced to the same coordinate frame!
![The image represents the SEZ (Solar Ecliptic Zenith) coordinate frame. This coordinate frame is defined with respect to the satellite's orbital plane and is commonly used in satellite-based applications, especially for Earth observation satellites. The axes are defined as follows: S (Sun-Pointing): Points from the satellite to the Sun. E (Orbital Velocity): Points along the direction of the satellite's velocity vector in its orbital plane. Z (Zenith): Points from the satellite towards the center of the Earth. Understanding the orientation of these axes is crucial for analyzing and interpreting the satellite's orientation in orbit.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/SEZ-1-286x300-1.jpg)
Now considering the yellow triangle
[latex]\rho_{SE} = \rho \cos(\text{El})[/latex]
[latex]\rho_{Z} = \rho \cos(\text{El})[/latex]
Then, considering the purple box
[latex]\vec{\rho}_S = \rho_{SE} \cos(180 - \text{Az}) \hat{S}[/latex]
so
[latex]\vec{\rho}_S = -\rho_{SE} \cos(\text{Az}) \hat{S}[/latex]
[latex]\vec{\rho}_E = \rho_{SE} \sin(180 - \text{Az}) \hat{E}[/latex]
so
[latex]\vec{\rho}_E = \rho_{SE} \sin(\text{Az}) \hat{E}[/latex]
and
[latex]\vec{\rho}_Z = \rho \sin(\text{El}) \hat{Z}[/latex]
Now we end up with a final set of equations:
[latex]\rho_S = -\rho \cos(\text{El}) \cos(\text{Az})[/latex]
[latex]\rho_E = \rho \cos(\text{El}) \sin(\text{Az})[/latex]
[latex]\rho_Z = \rho \sin(\text{El})[/latex]
But it still does not work! We will need a coordinate transformation between the IJK and SEZ coordinate frames.
[latex]\vec{\rho}_{IJK} = \begin{bmatrix} \sin L \cos \text{LST} & -\sin L \sin \text{LST} & \cos L \\ \sin \sin \text{LST} & \cos L \sin \text{LST} & 0 \\ -\cos L & 0 & \sin L \end{bmatrix} \vec{\rho}_{SEZ}[/latex]
Where we will define:
[latex]\vec{\rho}_{IJK} = \text{ROT} \vec{\rho}_{SEZ}[/latex]
See Bate for more information.
Now we can add the vectors since they are all referenced to the IJK coordinate frame.
[latex]\vec{R} = \vec{R}_{\text{site}} + \vec{\rho}[/latex]
________________________________________________
Example 2
Given satellite parameters as spotted by the radar tracking station in Example
[latex]{\rho}[/latex] = 7000 km
Az = 40 degrees
El = 45 degrees
Find the position of the satellite, [latex]\vec{R}[/latex] , in the IJK coordinate frame.
________________________________________________
VELOCITY VECTOR
Now all that is needed is the velocity vector, [latex]\vec{V}[/latex] , to find all of the satellite’s COEs. Recall,
[latex]\vec{R} = \vec{R}_{\text{site}} + \vec{\rho}[/latex]
So to find the velocity vector, [latex]\vec{V}[/latex] , it will be:
[latex]\vec{V} = \frac{d}{dt} \vec{R}_{IJK}[/latex]
[latex]\vec{V} = \frac{d}{dt} (\vec{R}_{\text{site}} + \vec{\rho})[/latex]
Thus,
[latex]\vec{V} = \frac{d}{dt} \vec{R}_{\text{site}} + \frac{d}{dt} \vec{\rho}[/latex]
and since the range site is not moving around (typically!) on the surface of the earth, then
[latex]\vec{V} = \frac{d}{dt} \vec{R}_{\text{site}} = 0[/latex]
So,
[latex]\vec{V} = \frac{d}{dt} \vec{\rho}[/latex]
However, since [latex]\vec{\rho}[/latex] is attached to the earth’s surface, which is rotating, transport theorem from dynamics must be used. Recall, to take the derivative of a vector [latex]\vec{A}[/latex] in a moving frame:
[latex]\dot{\vec{A}} = \dot{\vec{R}} + \vec{\omega} \times \vec{A}[/latex]
Where:
[latex]\dot{\vec{A}}[/latex] = Derivative of vector A relative to the inertial frame, I
[latex]\dot{\vec{R}}[/latex] = Derivative of vector A relative to the moving frame, R
[latex]{\omega}[/latex] = Angular velocity of frame R relative to the intertial frame, I
[latex]\vec{V}_{IJK} = \frac{d}{dt} \vec{R}_{IJK} = \vec{\rho}_{IJK} + \vec{\omega} \times \vec{R}_{IJK}[/latex]
Notice:
[latex]\vec{\rho}_{IJK} = \text{ROT} \vec{\rho}_{SEZ}[/latex]
where [latex]\dot{\vec{\rho}}_{SEZ}[/latex] is determined from taking the derivative of [latex]\dot{\vec{\rho}}_{SEZ}[/latex] and yields:
[latex]\dot{\vec{\rho}}_{SEZ} = \begin{bmatrix} -\dot{\rho} \cos(El) \cos(Az) + \rho \dot{El} \sin(El) \cos(Az) + \rho \dot{Az} \cos(El) \sin(Az) \\ \dot{\rho} \cos(El) \sin(Az) - \rho \dot{El} \sin(El) \sin(Az) + \rho \dot{Az} \cos(El) \cos(Az) \\ \dot{\rho} \sin(El) + \rho \dot{El} \cos(El) \end{bmatrix}[/latex]
Where the angular velocity [latex]\vec{\omega}[/latex] of the SEZ frame relative to the inertial, IJK frame,
[latex]\vec{\omega} = \begin{bmatrix} 0 \\ 0 \\ 15.041^\circ / \text{hr} \end{bmatrix} \times \begin{bmatrix} \frac{0}{\text{hr}} \\ \frac{0}{3600 \text{ secs}} \\ \frac{\pi}{180^\circ} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 7.292082578 \times 10^{-5} \text{ rad/s} \end{bmatrix}[/latex]
This type of Preliminary Orbit Determination is called COMFIX (complete fix).
Given: Find:
![The image depicts a COMFIX (complete fix) in the context of Preliminary Orbit Determination. COMFIX is a method used for obtaining a complete set of observational data to determine the initial orbit of a satellite or celestial body. It involves collecting various types of observations or fixes to precisely determine the satellite's position and velocity parameters.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/comfix-300x95-1.jpg)
OTHER TYPES OF ORBIT DETERMINATION
OTHER TYPES OF ORBIT DETERMINATION
Comfix is one type of an orbit determination problem and requires the range site data. Alternative methods of determining a satellite’s orbit is possible if other parameters are available. Below are a few other methods and are briefly mentioned and described. For more information, see the references.
Gibbsian Method
The Gibbsian orbit determination method determines an orbit from three position vectors. It is attributed to Joshua Gibbs (1839 – 1903), an engineer who was believed to be the first American contributor to celestial mechanics. Gibbs is more well-known for his contributions to thermodynamics, but his work in orbit determination is the basis for many current orbit determination methods.
When three position vectors, , are known over a relatively long time, you can determine the satellite’s orbit. As shown in the figure below:
![The image illustrates the Gibbs method for orbit determination. The Gibbs method is a technique used in astrodynamics to determine the orbit of a satellite or celestial object based on three known position vectors observed over a significant period. By utilizing these three position vectors, the Gibbs method allows for the calculation of the satellite's orbital elements, providing valuable information about its trajectory in space.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/gibbs-300x137-1.jpg)
Given: Find:
![The image demonstrates the Gibbs method applied to the determination of orbital elements. The Gibbs method is a mathematical approach used in astrodynamics to ascertain the orbital elements of a satellite or celestial body by analyzing three known position vectors observed over a certain duration. This method aids in deriving crucial orbital parameters that define the satellite's path in space, contributing to a comprehensive understanding of its orbital dynamics.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/Gibbscoes-300x90-1.jpg)
Where (Vallado):
[latex]\vec{R}_1 \text{ to } \vec{R}_3 > 5^\circ[/latex]
Note that you do not need to know anything about the rate of change of the vectors.
Herrick-Gibbs Method
The Herrick-Gibbs orbit determination method also determines an orbit from three position vectors, but requires that they be gathered over a short amount of time. It is not as robust as the Gibbs method, but three closely related space positions may be the only information available due to ground radar site limitations. It is attributed to Sam Herrick (1911 – 1974), who corresponded regularly with Robert Goddard. Herrick said that Goddard’s encouragement led him to anticipate the basic problems of space navigation (Samuel Herrick).
When three position vectors, , are known over a relatively short time, you can determine the satellite’s orbit. As shown in the figure below:
![The figure illustrates the Herrick-Gibbs method, a technique used in orbital mechanics for preliminary orbit determination when three position vectors of a satellite are known over a considerable time span. This method is employed to calculate orbital elements and trajectory information based on observed position vectors, aiding in the analysis and prediction of a satellite's orbital path.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/herrick-300x126-1.jpg)
Given: Find:
![The image demonstrates the Herrick-Gibbs method applied to the determination of orbital elements. The Herrick-Gibbs method is a mathematical approach used in astrodynamics to ascertain the orbital elements of a satellite or celestial body by analyzing three known position vectors observed over a certain duration. This method aids in deriving crucial orbital parameters that define the satellite's path in space, contributing to a comprehensive understanding of its orbital dynamics.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/Gibbscoes-300x90-1.jpg)
Where (Vallado):
[latex]\vec{R}_1 \text{ to } \vec{R}_3 < 1^\circ[/latex]
Note that you do not need to know anything about the rate of change of the vectors.
Gauss-Lambert-Euler
The Gauss-Lambert-Euler orbit determination method determines an orbit from two position vectors and a time of flight between the two. As Lambert first formed the solution, it is better known as Lambert’s problem. It was first posed in the 18th century by Johann Lambert (1728 – 1777), who is known for proving that is an irrational number. The problem was formally solved with mathematical proof by Joseph Lagrange.
When two position vectors, [latex]\vec{R}[/latex] , are known and a time of flight is known between the two, the orbit between the two positions may be found. This can be seen in the figure below:
![The figure illustrates the Lambert's problem, a mathematical approach used in orbital mechanics when two position vectors of a satellite are known along with the time of flight between them. Lambert's problem is concerned with finding the orbit that connects two given positions in space over a specified time interval. This method is valuable for mission planning and trajectory optimization in space missions.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/lambert-300x149-1.jpg)
Given: Find:
![The figure depicts the process of determining orbital elements when two position vectors and the time of flight between them are known. This involves solving Lambert's problem, a mathematical challenge in orbital mechanics. Lambert's problem helps find the orbit that connects two specified positions in space over a given time interval, providing crucial information for space mission planning and trajectory optimization.](https://colorado.pressbooks.pub/app/uploads/sites/14/2023/11/11.jpg)
where:
[latex]\vec{R}_1 \text{ to } \vec{R}_2 > 60^\circ[/latex]
Lambert’s problem is often used for interplanetary trajectories.
Complete this quick quiz in order to see how well you understand the concepts from this section.
In the next section, Local Sidereal Time (LST) will be introduced as well as ground tracks.
SOLUTIONS TO EXAMPLES:
***Example 1 Solution***
Given a radar tracking site at:
L = Latitude = 42 degrees
H = altitude above sea level = 77 meters
LST = local sidereal time = 256 degrees
Find
[latex]\vec{R}_{\text{site}}[/latex]
Solution:
[latex]\vec{R}_{\text{site}} = x \cos(\text{LST}) \hat{i} + x \sin(\text{LST}) \hat{j} + z \hat{k}[/latex]
Where:
[latex]x = \left[ \frac{a_e}{\sqrt{1 - e^2 \sin^2 L}} + H \right] \cos L[/latex]
[latex]x = \left[ \frac{6378.137}{\sqrt{1 - 0.08182^2 \sin^2 42}} + 0.077 \right] \cos 42[/latex]
x = 4747.06 km
and
[latex]z = \left[ \frac{a_e(1 - e^2)}{\sqrt{1 - e^2 \sin^2 L}} + H \right] \sin L[/latex]
[latex]z = \left[ \frac{6378.137(1 - 0.08182^2)}{\sqrt{1 - 0.08182^2 \sin^2 42}} + 0.077 \right] \sin 42[/latex]
z = 4245.65 km
[latex]\vec{R}_{\text{site}} = x \cos(\text{LST}) \hat{i} + x \sin(\text{LST}) \hat{j} + z \hat{k}[/latex]
[latex]\vec{R}_{\text{site}} = 4747.06 \cos 25.6 \hat{i} + 4747.06 \sin 25.6 \hat{j} + 4239.15 \hat{k}[/latex]
so
[latex]\vec{R}_{\text{site}} = -1148.42 \hat{i} - 4606.05 \hat{j} + 4245.65 \hat{k}[/latex]
***Example 2 Solution***
Given satellite parameters as spotted by the radar tracking station in Example 1:
[latex]\rho[/latex] = 7000 km
Az = 40 degrees
El = 45 degrees
Find the position of the satellite, [latex]\vec{R}[/latex] , in the IJK coordinate frame.
Solution:
[latex]\vec{\rho}_{SEZ} = \rho_S \hat{S} + \rho_E \hat{E} + \rho_Z \hat{Z}[/latex]
Where:
[latex]\begin{align*} \rho_S &= -\rho \cos(\text{El}) \cos(\text{Az}) \\ \rho_E &= \rho \cos(\text{El}) \sin(\text{Az}) \\ \rho_Z &= \rho \sin(\text{El}) \end{align*}[/latex]
Plugging in the given parameters yields:
[latex]\rho_S = -7000 \cos(45) \cos(40) = -3791.73 \text{ km}[/latex]
[latex]\rho_E = 7000 \cos(45) \sin(40) = 3181.64 \text{ km}[/latex]
[latex]\rho_Z = 7000 \sin(45) = 4949.75 \text{ km}[/latex]
[latex]\vec{\rho}_{SEZ} = -3791.73 \hat{S} + 3181.64 \hat{E} + 4949.75 \hat{Z}[/latex]
Now applying the coordinate transformation between the SEZ and IJK coordinate frames:
[latex]\vec{\rho}_{IJK} = \begin{bmatrix} \sin L \cos \text{LST} & -\sin \text{LST} & \cos L \cos \text{LST} \\ \sin L \sin \text{LST} & \cos \text{LST} & \cos L \sin \text{LST} \\ -\cos L & 0 & \sin L \end{bmatrix} \vec{\rho}_{SEZ}[/latex]
[latex]\vec{\rho}_{IJK} = \begin{bmatrix} \sin(42) \cos(256) & -\sin(256) & \cos(42) \cos(256) \\ \sin(42) \sin(256) & \cos(256) & \cos(42) \sin(256) \\ -\cos(42) & 0 & \sin(42) \end{bmatrix} \vec{\rho}_{SEZ}[/latex]
hence,
[latex]\vec{\rho}_{IJK} = \begin{bmatrix} -0.16188 & 0.97030 & -0.17978 \\ -0.64925 & -0.24192 & -0.72107 \\ -0.74314 & 0 & 0.66913 \end{bmatrix} \begin{bmatrix} -3791.73 \\ 3181.64 \\ 4949.75 \end{bmatrix}[/latex]
[latex]\vec{\rho}_{IJK} = 2811.05 \hat{i} - 1877.03 \hat{j} + 6129.83 \hat{k} \text{ km}[/latex]
so now
[latex]\vec{R} = \vec{R}_{\text{site}} + \vec{\rho}[/latex]
[latex]\vec{R} = (-1148.42 \hat{i} - 4606.05 \hat{j} + 4245.65 \hat{k}) + (2811.05 \hat{i} - 1877.03 \hat{j} + 6129.83 \hat{k})[/latex]
[latex]\vec{R} = 1662.63 \hat{i} - 6483.08 \hat{j} + 10375.48 \hat{k}[/latex]
REFERENCES
Bate, R. R., Mueller, D. D., White, J. E., & Saylor, W. W. (1971). Fundamentals of astrodynamics. Mineola (N.Y.): Dover Publications.
Math.com. Ellipsoid. (2021). www.math.net/ellipsoid.
NASA Exploration and Space Communications. (n.d.). Space Network. https://esc.gsfc.nasa.gov/projects/SN.
Pike, J. (2021). Maui Space Surveillance Site (MSSS). https://www.globalsecurity.org/space/systems/msss.htm
Samuel Herrick, engineering; Astronomy: Los Angeles. University of California: In Memoriam, March 1976. (2011). https://oac.cdlib.org/view?docId=hb9k4009c7&chunk.id=div00024&brand=calisphere&doc.view=entire_text
Schriever Space Force Base > Home. (2021). https://www.schriever.spaceforce.mil/
Space News, (2015, March 23). Editorial: The Breakup of DMSP F13. SpaceNews. https://spacenews.com/editorial-the-breakup-of-dmsp-f13/.
Vallado, D. A., & McClain, W. D. (2013). Fundamentals of Astrodynamics and Applications. Microcosm Press.