[SOLVED]energy equation

dwsmith

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

dwsmith

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