Astrodynamics Question: Derivation of Sp. Orbital Energy?

In summary, the conversation discusses the equation 0 = a + (μ/r3) r and how it is used to find ξ, the Specific Mechanical/Orbital Energy. The book suggests using the dot product of v and a to solve this equation, but there is confusion about how the vectors are being transformed into scalars. The confusion is resolved by understanding that the vectors point in opposite directions, resulting in a negative sign. The conversation also mentions that the equations are general and apply to more than just circular orbits.
  • #1
AdrianGriff
21
0
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
 
Last edited:
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  • #2
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)##.
 
  • #3
mfb said:
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.
 
  • #4
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.
 

Related to Astrodynamics Question: Derivation of Sp. Orbital Energy?

1. What is astrodynamics?

Astrodynamics is the study of the motion of objects in space, particularly the motion of man-made objects such as spacecraft, satellites, and probes.

2. What is specific orbital energy?

Specific orbital energy is the energy required to move an object from one point to another in space, taking into account its position, velocity, and the gravitational forces acting on it.

3. How is specific orbital energy derived?

Specific orbital energy is derived from the Law of Conservation of Energy, which states that energy cannot be created or destroyed, only transferred from one form to another. In astrodynamics, specific orbital energy is derived by considering the potential and kinetic energy of an object in orbit.

4. Why is specific orbital energy important?

Specific orbital energy is important because it helps us understand and calculate the behavior of objects in space. It is used in trajectory analysis, orbital maneuvering, and predicting the behavior of objects in orbit.

5. How is specific orbital energy related to orbital mechanics?

Specific orbital energy is a fundamental concept in orbital mechanics, as it is used to calculate and analyze the motion of objects in space. It is closely related to other important parameters such as specific orbital angular momentum and specific orbital velocity.

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