# Determining orbital elements from r and v vectors - Fundamentals of Astrodynamics Example Problem

I am fighting my way through Fundamentals of Astrodynamics 1st edition for the fun of it. I have made it up to section 2.4 and have successfully completed all the example problems and chapter exercises.

However, I am stuck on the Example Problem on page 68. Specifically, I do not understand the solution for e.

1) It seems to me that the second term in the brackets should evaluate to zero since r dot v is equal to: (rv cos theta). Since r has only an i value and v has only a J value, these vectors are perpendicular, theta = 90 degrees, cos 90 = 0

2) if the above is true you have only first team left and there is no way I can see to get it to be a value that being divided by the gravitational constant nu would result in the 1i solution?

• Consider posting more information about the problem. Someone who does not have that exact book may know how to answer the question, but not with what you have supplied. Jun 11, 2020 at 3:41

I assume you are referring to "Fundamentals of Astrodynamics" by Bate, R. et al. (1971).

The author is using non-dimensional units in this example problem.

Therefore, as explained on p. 41,

$$\mu = 1~DU^3/TU^2$$

And, in this example problem,

$$\mathbf{r} = 2~\mathbf{I} DU, \quad \mathbf{v} = 1~\mathbf{J} DU/TU$$

With these, the equation for the eccentriciy vector is

$$\mathbf{e} = \dfrac{1}{\mu} [ (v^2-\dfrac{\mu}{r})\mathbf{r} - (\mathbf{r} \cdot \mathbf{v})v ] = 1\mathbf{I}$$

The second term inside the brackets that consists of a dot product is indeed zero because position and velocity are orthogonal in this case. The first term becomes 1 after replacing the variables with their values.

• Thank you for responding. Jun 11, 2020 at 20:36