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[SOLVED] energy equation

dwsmith

Well-known member
Feb 1, 2012
1,673
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
$$
\mathbf{r} = \begin{pmatrix}
-4069.503\\
2861.786\\
4483.608
\end{pmatrix}\text{km},\quad
\mathbf{v} = \begin{pmatrix}
-5.114\\
-5.691\\
-1.000
\end{pmatrix}\text{km/sec}
$$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
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 [tex]\mu[/tex] in your original post. Is this a result you are supposed to derive perhaps?

-Dan
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I'm confused about two things:
1. You are missing an "m" from the kinetic energy term.

2. You defined [tex]\mu[/tex] 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.