Specific Mechanical Energy

The energy of an orbit is given by the kinetic energy of the reduced mass \(m\) and the gravitational potential energy,

\[ E = \frac{1}{2}mv^2 - \frac{Gm_1m_2}{r} = \frac{1}{2}mv^2 - \frac{Gm(m_1 + m_2)}{r} \]

A more useful vector for orbital mechanics is the specific mechanical energy (SME) \(\epsilon \equiv E / m\),

\[ \epsilon = \frac{v^2}{2} - \frac{\mu}{r} \]

This SME for the two-body problem is called the energy integral

In terms of the semimajor axis \(a\),

\[ \epsilon = -\frac{\mu}{2a} \]

Constant SME for the Two Body Problem : \(\epsilon\) is constant which can be shown by first taking the dot product of the two body equation with the velocity vector of the satellite \(\vec v\) (note that \(\vec v \neq \dot{\vec r}\))

$$
\ddot{\vec r} \cdot \vec v = -\frac{\mu}{r^2} (\hat r \cdot \vec v)
$$

Note that the dot product is the radial component of the velocity vector $v_r = \hat r \cdot \vec v$. Equivalently its the magnitude of the velocity of $r$

$$
\dot{r} = v_r = \vec v \cdot \hat r
$$

$$
\ddot{\vec r} \cdot \vec v = -\frac{\mu}{r^2} \dot{r}
$$

Notice that the acceleration vector of the tangential and radial velocity vector is the same $\ddot{\vec r} = \dot{\vec v}$

$$
\dot{\vec v} \cdot \vec v = -\frac{\mu}{r^2} \dot{r}
$$

The two sides are derivatives of the kinetic energy and potential energy respectively thus we get the energy equivalence,

$$
\frac{\rm d}{\rm d t} \left(\frac{v^2}{2}\right) = \frac{\rm d}{\rm d t} \left(\frac{\mu}{r}\right)\\
\frac{\rm d}{\rm d t} \left(\frac{v^2}{2}\right) - \frac{\rm d}{\rm d t} \left(\frac{\mu}{r}\right) = 0\\
\epsilon = \frac{v^2}{2} - \frac{\mu}{r} = \text{constant}
$$