When deriving the eccentricity vector you first start of with the equation of motion for the two body problem (we are concerned about the motion of the smaller body around a much larger mass) the equations follows :

$$\mathbf{\ddot{r}} = - \frac{\mu}{r^3} \mathbf{r}$$

where mu is the multiplication of the larger body and the gravitational constant.

Though later in the textbook I am referring to (fundamentals of astrodynamics) the author cross products both sides by angular momentum h.

$$\mathbf{\ddot{r}} \times \mathbf{h} = \frac{\mu}{r^3}$$

Though I am not too sure why the author does this, it does say that now the equation can be integrated though I am not too sure why that is the case, like could this have been done with any old vector? Is it possible that someone can explain this to me?

Edit : https://www.google.co.uk/books/edition/Fundamentals_of_Astrodynamics/UEC9DwAAQBAJ?hl=en&gbpv=1&dq=fundamentals+of+astrodynamics+second+edition&printsec=frontcover

Page 21


Page 16

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    $\begingroup$ It's very odd that the second equation crosses two three-dimensional vector quantities and outputs a scalar. $\endgroup$
    – notovny
    Commented Dec 24, 2021 at 2:04
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    $\begingroup$ @notovny This is the old version of the book. The one im using the second edition which has much less errors has the equation as : acceleration X angular momentum = mu/r^3 * (angular momentum X r) Though I am not too sure what happened to negative sign. $\endgroup$
    – John
    Commented Dec 24, 2021 at 2:26
  • $\begingroup$ @John reversed the order of the cross product, so a cross b = -b cross a $\endgroup$ Commented Dec 24, 2021 at 3:10
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    $\begingroup$ Consider adding a few more steps of the derivation into the question. Or a link to the relevant page of this book, if preview is available online (Google books). $\endgroup$
    – AJN
    Commented Dec 24, 2021 at 12:26
  • $\begingroup$ Upon reading a bit, It seems that you may get better answers if you got this migrated to Mathematics, as the cross product seems to be mathematical trick to make the system easy to solve. $\endgroup$
    – AJN
    Commented Dec 26, 2021 at 15:13

1 Answer 1


This question may be more suited for Math SE. The method of using $(h\times\cdot)$ to integrate the system is not limited to the text book you mentioned. For example, it is mentioned briefly in this Math.SE answer.

Easiest expressions to integrate are expressions of the form $\frac{d}{dt}(f(t))$. To integrate them, we simply need to "cancel" the $\int dt$ and the $\frac{d}{dt}$. Expressions of this form appear to be called Exact differentials.

In case, the equation to be solved doesn't contain such easy-to-solve expressions, we can multiply them with a carefully chosen[1] Integrating factor (Wolfram article) to transform it into the easy-to-solve form.

Here, the integrating factor is the cross product operation with $\vec h$. By multiplying both LHS and RHS of the differential equation to be solved ($\ddot{\vec r} = \frac{-\mu \vec r}{r^3}$) with the integrating factor, both LHS and RHS are converted into the easy-to-integrate form.

Chapter 3 of this pdf introduces the concept of integrating factor. Its appendix shows an example of integrating factor which uses dot product with a vector to help convert the expression to be integrated into the $\frac{d}{dt}(\cdot)$ form.

[1] If we choose zero as the integrating factor, then we don't really solve anything. So clearly, the integrating factor should satisfy some properties, and can't be arbitrary expressions.


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