Finding Orbit Type with Energy Equation

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    Energy Orbit Type
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Discussion Overview

The discussion revolves around using the energy equation to classify the type of orbit (elliptic, parabolic, hyperbolic) based on specific parameters. Participants explore the implications of the energy equation in the context of orbital mechanics, addressing both theoretical and practical aspects.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that for the energy equation, $E=0$ corresponds to a parabolic orbit, $E>0$ to a hyperbolic orbit, and $E<0$ to an elliptic orbit, with a specific condition for circular orbits.
  • Another participant expresses confusion regarding the absence of a mass term in the kinetic energy component of the equation.
  • A participant questions whether the definition of $\mu$ is something that should be derived rather than assumed.
  • One participant mentions that they were later informed about the value of $\mu$ for their specific problem, which helped them proceed.

Areas of Agreement / Disagreement

There is no clear consensus on the handling of the mass term in the kinetic energy equation or the derivation of $\mu$. Multiple viewpoints exist regarding the interpretation and application of the energy equation in determining orbit types.

Contextual Notes

Participants highlight potential missing assumptions, such as the necessity of including mass in the kinetic energy term and the derivation of $\mu$, which may affect the clarity of the discussion.

Dustinsfl
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How does one use the energy equation to determine the type of orbit?
$$
E = \frac{v^2}{2} - \frac{\mu}{r}
$$
where $\mu = G(m_1+m_2)$ and
$$
\mathbf{r} = \begin{pmatrix}
-4069.503\\
2861.786\\
4483.608
\end{pmatrix}\text{km},\quad
\mathbf{v} = \begin{pmatrix}
-5.114\\
-5.691\\
-1.000
\end{pmatrix}\text{km/sec}
$$
 
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As I understand it, $E=0$ is parabolic, $E>0$ is hyperbolic, and $E<0$ is elliptic. If $v^{2}r=\mu$, then it's circular.
 
Ackbach said:
As I understand it, $E=0$ is parabolic, $E>0$ is hyperbolic, and $E<0$ is elliptic. If $v^{2}r=\mu$, then it's circular.

My issue was I didn't have a mu term.
 
dwsmith said:
How does one use the energy equation to determine the type of orbit?
$$
E = \frac{v^2}{2} - \frac{\mu}{r}
$$
where $\mu = G(m_1+m_2)$ and
$$
\mathbf{r} = \begin{pmatrix}
-4069.503\\
2861.786\\
4483.608
\end{pmatrix}\text{km},\quad
\mathbf{v} = \begin{pmatrix}
-5.114\\
-5.691\\
-1.000
\end{pmatrix}\text{km/sec}
$$
I'm confused about two things:
1. You are missing an "m" from the kinetic energy term.

2. You defined \mu in your original post. Is this a result you are supposed to derive perhaps?

-Dan
 
topsquark said:
I'm confused about two things:
1. You are missing an "m" from the kinetic energy term.

2. You defined \mu in your original post. Is this a result you are supposed to derive perhaps?

-Dan

Later on I was told what mu is for this problem so I was able to do it.
 

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