Gauss's method

Orbiting body direction cosine vector

The orbiting body direction cosine vector can be determined from the right ascension and declination (from Topocentric Equatorial Coordinate System) of the orbiting body from the observation points via:

$${\displaystyle {\hat{\mathit{\rho }}}_{\mathbf{n}}=\cos {\delta }_n\cos {\alpha }_n\hat{\mathbf{I}}+\cos {\delta }_n\sin {\alpha }_n\hat{\mathbf{J}}+\sin {\delta }_n\hat{\mathbf{K}}}$$ where,

• ${\displaystyle {\hat{\mathit{\rho }}}_{\mathbf{n}}}$ is the respective unit vector in the direction of the position vector ${\displaystyle \rho }$ (from observation point to orbiting body in Topocentric Equatorial Coordinate System)
• ${\displaystyle {\delta }_n}$ is the respective declination
• ${\displaystyle {\alpha }_n}$ is the respective right ascension

Algorithm

The initial derivation begins with vector addition to determine the orbiting body's position vector. Then based on the conservation of angular momentum and Keplerian orbit principles (which states that an orbit lies in a two dimensional plane in three dimensional space), a linear combination of said position vectors is established. Also, the relation between a body's position and velocity vector by Lagrange coefficients is used which results in the use of said coefficients. Then with vector manipulation and algebra, the following equations were derived. For detailed derivation, refer to Curtis.

Gauss's method provides a preliminary orbit determination. The approximation of the Lagrange coefficients and the limitations of the required observation conditions (i.e., insignificant curvature in the arc between observations) cause inaccuracies. Gauss's method can be improved, however, by increasing the accuracy of sub-components, such as solving Kepler's equation. Another way to increase the accuracy is through more observations.

Step 1

Calculate time intervals, subtract the times between observations: $${\displaystyle \begin{array}{cc}{\tau }_1& =t_1-t_2\\ {\tau }_3& =t_3-t_2\\ \tau & =t_3-t_1\end{array}}$$ where

• ${\displaystyle {\tau }_n}$ is the time interval
• ${\displaystyle t_n}$ is the respective observation time

Step 2

Calculate cross products, take the cross products of the observational unit direction (order matters): $${\displaystyle \begin{array}{cc}{\mathbf{p}}_{\mathbf{1}}& ={\hat{\mathit{\rho }}}_{\mathbf{2}}\times {\hat{\mathit{\rho }}}_{\mathbf{3}}\\ {\mathbf{p}}_{\mathbf{2}}& ={\hat{\mathit{\rho }}}_{\mathbf{1}}\times {\hat{\mathit{\rho }}}_{\mathbf{3}}\\ {\mathbf{p}}_{\mathbf{3}}& ={\hat{\mathit{\rho }}}_{\mathbf{1}}\times {\hat{\mathit{\rho }}}_{\mathbf{2}}\end{array}}$$ where

• ${\displaystyle \mathbf{a}\times \mathbf{b}}$ is the cross product of vectors ${\displaystyle \mathbf{a}\text{ and }\mathbf{b}}$
• ${\displaystyle {\mathbf{p}}_{\mathbf{n}}}$ is the respective cross product vector
• ${\displaystyle {\hat{\mathit{\rho }}}_{\mathbf{n}}}$ is the respective unit vector

Step 3

Calculate common scalar quantity (scalar triple product), take the dot product of the first observational unit vector with the cross product of the second and third observational unit vector:

$${\displaystyle D_0={\hat{\mathit{\rho }}}_{\mathbf{1}}\cdot {\mathbf{p}}_{\mathbf{1}}={\hat{\mathit{\rho }}}_{\mathbf{1}}\cdot ({\hat{\mathit{\rho }}}_{\mathbf{2}}\times {\hat{\mathit{\rho }}}_{\mathbf{3}})}$$ where

• ${\displaystyle \mathbf{a}\cdot \mathbf{b}}$ is the dot product of vectors ${\displaystyle \mathbf{a}}$ and ${\displaystyle \mathbf{b}}$
• ${\displaystyle D_0}$ is the common scalar triple product
• ${\displaystyle {\mathbf{p}}_{\mathbf{n}}}$ is the respective cross product vector
• ${\displaystyle {\hat{\mathit{\rho }}}_{\mathbf{n}}}$ is the respective unit vector

Step 4

Calculate nine scalar quantities (similar to step 3): $${\displaystyle \begin{array}{cccccc}{D}_{11}& ={\mathbf{R}}_{\mathbf{1}}\cdot {\mathbf{p}}_{\mathbf{1}}& D_{12}& ={\mathbf{R}}_{\mathbf{1}}\cdot {\mathbf{p}}_{\mathbf{2}}& D_{13}& ={\mathbf{R}}_{\mathbf{1}}\cdot {\mathbf{p}}_{\mathbf{3}}\\ D_{21}& ={\mathbf{R}}_{\mathbf{2}}\cdot {\mathbf{p}}_{\mathbf{1}}& D_{22}& ={\mathbf{R}}_{\mathbf{2}}\cdot {\mathbf{p}}_{\mathbf{2}}& D_{23}& ={\mathbf{R}}_{\mathbf{2}}\cdot {\mathbf{p}}_{\mathbf{3}}\\ D_{31}& ={\mathbf{R}}_{\mathbf{3}}\cdot {\mathbf{p}}_{\mathbf{1}}& D_{32}& ={\mathbf{R}}_{\mathbf{3}}\cdot {\mathbf{p}}_{\mathbf{2}}& D_{33}& ={\mathbf{R}}_{\mathbf{3}}\cdot {\mathbf{p}}_{\mathbf{3}}\end{array}}$$ where

• ${\displaystyle D_{mn}}$ is the respective scalar quantity
• ${\displaystyle {\mathbf{R}}_{\mathbf{m}}}$ is the respective observer position vector
• ${\displaystyle {\mathbf{p}}_{\mathbf{n}}}$ is the respective cross product vector

Step 5

Calculate scalar position coefficients:

$${\displaystyle \begin{array}{cc}A& =\frac{1}{D_0}\left(-D_{12}\frac{{\tau }_3}{\tau }+D_{22}+D_{32}\frac{{\tau }_1}{\tau }\right)\\ B& =\frac{1}{6D_0}\left[D_{12}\left({\tau }_3^2-{\tau }^2\right)\frac{{\tau }_3}{\tau }+D_{32}\left({\tau }^2-{\tau }_1^2\right)\frac{{\tau }_1}{\tau }\right]\\ E& ={\mathbf{R}}_{\mathbf{2}}\cdot {\hat{\mathit{\rho }}}_{\mathbf{2}}\end{array}}$$ where

• ${\displaystyle A}$, ${\displaystyle B}$, and ${\displaystyle E}$ are scalar position coefficients
• ${\displaystyle D_0}$ is the common scalar quantity
• ${\displaystyle D_{mn}}$ is the respective scalar quantities
• ${\displaystyle {\tau }_n}$ is the time interval
• ${\displaystyle R_n}$ is the respective observer position vector
• ${\displaystyle {\hat{\mathit{\rho }}}_{\mathbf{n}}}$ is the respective unit vector

Step 6

Calculate the squared scalar distance of the second observation, by taking the dot product of the position vector of the second observation: $${\displaystyle {R_2}^2={\mathbf{R}}_{\mathbf{2}}\cdot {\mathbf{R}}_{\mathbf{2}}}$$ where

• ${\displaystyle {R_2}^2}$ is the squared distance of the second observation
• ${\displaystyle {\mathbf{R}}_{\mathbf{2}}}$ is the position vector of the second observation

Step 7

Calculate the coefficients of the scalar distance polynomial for the second observation of the orbiting body: $${\displaystyle \begin{array}{cc}a& =-\left(A^2+2AE+{R_2}^2\right)\\ b& =-2\mu B(A+E)\\ c& =-{\mu }^2{B}^2\end{array}}$$ where

• ${\displaystyle a\text{, }b\text{, and }c}$ are coefficients of the scalar distance polynomial for the second observation of the orbiting body
• ${\displaystyle A\text{, }B\text{, and }E}$ are scalar position coefficients
• ${\displaystyle \mu }$ is the gravitational parameter of the focal body of the orbiting body

Step 8

Find the root of the scalar distance polynomial for the second observation of the orbiting body: $${\displaystyle {r_2}^8+a{r_2}^6+b{r_2}^3+c=0}$$ where

• ${\displaystyle r_2}$ is the scalar distance for the second observation of the orbiting body (it and its vector, r₂, are in the Equatorial Coordinate System)
• ${\displaystyle a\text{, }b\text{, and }c}$ are coefficients as previously stated

Various methods can be used to find the root, a suggested method is the Newton–Raphson method. The root must be physically possible (i.e., not negative nor complex) and if multiple roots are suitable, each must be evaluated and compared to any available data to confirm their validity.

Step 9

Calculate the slant range, the distance from the observer point to the orbiting body at their respective time: $${\displaystyle \begin{array}{cc}{\rho }_1& =\frac{1}{D_0}\left[\frac{6\left(D_{31}{\displaystyle \frac{{\tau }_1}{{\tau }_3}}+D_{21}{\displaystyle \frac{\tau }{{\tau }_3}}\right){r_2}^3+\mu D_{31}\left({\tau }^2-{{\tau }_1}^2\right){\displaystyle \frac{{\tau }_1}{{\tau }_3}}}{6{r_2}^3+\mu \left({\tau }^2-{{\tau }_3}^2\right)}-D_{11}\right]\\ {\rho }_2& =A+\frac{\mu B}{{r_2}^3}\\ {\rho }_3& =\frac{1}{D_0}\left[\frac{6\left(D_{13}{\displaystyle \frac{{\tau }_3}{{\tau }_1}}-D_{23}{\displaystyle \frac{\tau }{{\tau }_1}}\right){r_2}^3+\mu D_{13}\left({\tau }^2-{{\tau }_3}^2\right){\displaystyle \frac{{\tau }_3}{{\tau }_1}}}{6{r_2}^3+\mu \left({\tau }^2-{{\tau }_1}^2\right)}-D_{33}\right]\end{array}}$$ where

• ${\displaystyle {\rho }_n}$ is the respective slant range (it and its vector, ${\displaystyle {\rho }_{\mathbf{n}}}$, are in the Topocentric Equatorial Coordinate System)
• ${\displaystyle D_0}$ is the common scalar quantity
• ${\displaystyle D_{mn}}$ is the respective scalar quantities
• ${\displaystyle {\tau }_{(n)}}$ is the time interval.
• ${\displaystyle r_2}$ is the scalar distance for the second observation of the orbiting body
• ${\displaystyle \mu }$ is the gravitational parameter of the focal body of the orbiting body

Step 10

Calculate the orbiting body position vectors, by adding the observer position vector to the slant direction vector (which is the slant distance multiplied by the slant direction vector):

$${\displaystyle \begin{array}{cc}{\mathbf{r}}_{\mathbf{1}}& ={\mathbf{R}}_{\mathbf{1}}+{\rho }_1{\hat{\mathit{\rho }}}_{\mathbf{1}}\\ {\mathbf{r}}_{\mathbf{2}}& ={\mathbf{R}}_{\mathbf{2}}+{\rho }_2{\hat{\mathit{\rho }}}_{\mathbf{2}}\\ {\mathbf{r}}_{\mathbf{3}}& ={\mathbf{R}}_{\mathbf{3}}+{\rho }_3{\hat{\mathit{\rho }}}_{\mathbf{3}}\end{array}}$$ where

• ${\displaystyle {\mathbf{r}}_{\mathbf{n}}}$ is the respective orbiting body position vector (in Equatorial Coordinate System)
• ${\displaystyle {\mathbf{R}}_{\mathbf{n}}}$ is the respective observer position vector
• ${\displaystyle {\rho }_n}$ is the respective slant range
• ${\displaystyle {\hat{\mathit{\rho }}}_{\mathbf{n}}}$ is the respective unit vector

Step 11

Calculate the Lagrange coefficients: $${\displaystyle \begin{array}{cc}{f}_1& \approx 1-\frac{1}{2}\frac{\mu }{{r_2}^3}{{\tau }_1}^2\\ f_3& \approx 1-\frac{1}{2}\frac{\mu }{{r_2}^3}{{\tau }_3}^2\\ g_1& \approx {\tau }_1-\frac{1}{6}\frac{\mu }{{r_2}^3}{{\tau }_1}^3\\ g_3& \approx {\tau }_3-\frac{1}{6}\frac{\mu }{{r_2}^3}{{\tau }_3}^3\end{array}}$$ where,

• ${\displaystyle f_1}$, ${\displaystyle f_3}$, ${\displaystyle g_1}$ and ${\displaystyle g_3}$ are the Lagrange coefficients (these are just the first two terms of the series expression based on the assumption of small time interval)
• ${\displaystyle \mu }$ is the gravitational parameter of the focal body of the orbiting body
• ${\displaystyle r_2}$ is the scalar distance for the second observation of the orbiting body
• ${\displaystyle {\tau }_{(n)}}$ is the time interval

Step 12

Calculate the velocity vector for the second observation of the orbiting body:

$${\displaystyle {\mathbf{v}}_{\mathbf{2}}=\frac{1}{f_1{g}_3-f_3{g}_1}\left(-f_3{\mathbf{r}}_{\mathbf{1}}+f_1{\mathbf{r}}_{\mathbf{3}}\right)}$$ where

• ${\displaystyle {\mathbf{v}}_{\mathbf{2}}}$ is the velocity vector for the second observation of the orbiting body (in Equatorial Coordinate System)
• ${\displaystyle f_1}$, ${\displaystyle f_3}$, ${\displaystyle g_1}$ and ${\displaystyle g_3}$ are the Lagrange coefficients
• ${\displaystyle {\mathbf{r}}_{\mathbf{n}}}$ is the respective orbiting body position vector

Step 13

The orbital state vectors have now been found, the position (r₂) and velocity (v₂) vector for the second observation of the orbiting body. With these two vectors, the orbital elements can be found and the orbit determined.