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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?

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    $\begingroup$ 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. $\endgroup$ – Organic Marble Jun 11 at 3:41
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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.

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  • $\begingroup$ Thank you for responding. $\endgroup$ – BernieP Jun 11 at 20:36

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