Astrodynamics Question: Derivation of Sp. Orbital Energy?

[AdrianGriff]

So this should be coming easily, but for some reason I can't seem to grasp why or how this is being done:

So say we have equation:
0 = a + (μ/r³) r , where μ = G(M+m) or ≈ GM and M >> m.

According to this book, the first step to finding ξ, the Specific Mechanical/Orbital Energy they dot multiplied the vector v through the equation above like so:

v ⋅ a + v ⋅ (μ/r³) r = 0

And just below that, there is:

va + (μ/r³) rv = 0

So basically, my question is:
Why do all the vectors turn into scalars, considering that the dot product cannot actually be performed because we do not know the components of those variable vectors?

Should it not just stay as the equation with vectors?
Or why does a ⋅ v not equal a₁v₁ + a₂v₂?

Thank you!
- Adrian

 

[mfb]

The sign in the last equation is wrong.

a and r point in opposite directions. You know $\vec v \cdot \vec a = v a \cos(theta)$ where $\theta$ is the angle between them. For v and r the corresponding angle is $\pi - \theta$, and $\cos(\pi-\theta) = -\cos(\theta)$.

 

[AdrianGriff]

The sign in the last equation is wrong.

a and r point in opposite directions. You know $\vec v \cdot \vec a = v a \cos(theta)$ where $\theta$ is the angle between them. For v and r the corresponding angle is $\pi - \theta$, and $\cos(\pi-\theta) = -\cos(\theta)$.

I assure you that my equations are true to the text, perhaps the text is incorrect? It happens.

Nevertheless, if $\vec v \cdot \vec a = v a \cos(\theta)$, does it not follow that $\cos(\theta) = 0$ because $\theta = \pi/2$, where the vectors are orthogonal as velocity is tangent, and acceleration is towards the center of the orbit? And thus $||a||||b||$ will conclude to zero as well?

Also the same result arises with $\pi - \theta$ where $\theta = \pi/2$ and $\pi - \pi/2 = \pi/2$ again.

 

[mfb]

The angle is pi/2 only for circular orbits. The formulas here are more general.
In this special case the last equation is true, of course, but adding zero twice is not very interesting.