Finding Orbit Type with Energy Equation

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SUMMARY

The discussion focuses on using the energy equation to determine the type of orbit for celestial bodies, specifically using the formula $$E = \frac{v^2}{2} - \frac{\mu}{r}$$ where $\mu = G(m_1+m_2)$. Participants clarify that an energy value of $E=0$ indicates a parabolic orbit, $E>0$ indicates a hyperbolic orbit, and $E<0$ indicates an elliptic orbit. Additionally, a condition where $v^{2}r=\mu$ signifies a circular orbit. The confusion regarding the absence of the mass term in the kinetic energy calculation was addressed, leading to a resolution of the issue.

PREREQUISITES
  • Understanding of gravitational constant (G) and its application in orbital mechanics
  • Familiarity with the concepts of kinetic and potential energy in celestial mechanics
  • Knowledge of vector representation in three-dimensional space
  • Basic understanding of orbital types: circular, elliptical, parabolic, and hyperbolic
NEXT STEPS
  • Study the derivation of the gravitational parameter $\mu = G(m_1+m_2)$
  • Learn how to calculate orbital energy and its implications for different types of orbits
  • Explore the application of the energy equation in real-world orbital mechanics problems
  • Investigate the relationship between velocity, radius, and orbital type in celestial dynamics
USEFUL FOR

Astronomy students, astrophysicists, and engineers involved in orbital mechanics or satellite design will benefit from this 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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