Conservation of Quantities in Orbital Motion

Energy

The potential energy associated with any central force, eg gravitational force, is given by the following:

\[U = -\int_{\infty}^{r} \vec F \cdot d \vec r\]

This effectively means the amount of central force exerted by an object to bring a faraway object to the current position.

This may seem like a daunting formula, but we do remember that we can disassociate many variables in the description.

For instance, \(d \vec r\) is simply equivalent to \(d(r \vec r)\) which can by expressed as \(r \cdot d\vec r + \vec r \cdot r\), or more clearly:

\[\begin{align*}d\vec{r} &= d(r\hat{r}) \\ &= r \cdot d\hat{r} + dr \cdot \hat{r}\end{align*}\]

Noting that \(d \hat{r}\) is an absolutely unnecessary term to consider, since it isn't affected when considering the movement of an object infinitely away moving radially to a specific location. Hence, we can simplify the above expression to:

\[d\vec{r} = \hat{r} \cdot dr\]

From here, we also note that we represent \(\vec F\) as shown below:

\[\vec F_g = - G \frac{Mm}{r^2} \hat r\]

This helps us piece the final step, as shown below:

\[\begin{align*}U &= -\int_\infty^r \vec F_g \cdot d \vec r \\ &= -\int_\infty^r - G \frac{Mm}{r^2} \hat r \cdot \hat{r} \cdot dr \\ &= \int_\infty^r G \frac{Mm}{r^2} dr \\ &= GMm \int_\infty^r r^{-2} dr \\ &= GMm [-r^{-1}]^r_\infty \\ &= GMm \left(\lim_{n \to \infty} \frac{1}{n} - \frac{1}{r} \right) \\ U &= -G \frac{Mm}{r}\end{align*}\]

And there you have it! It is pretty clear that the potential energy is negative, since the force is effectively pulling inwards and thus exerting a force antiparallel to the radial axis on the object, which of course, by the definition of dot product, implies the introduction of \(\cos\pi = -1\) into the equation.

One thing we can still note is that, for two isolated point masses \(m\) and \(M\), the total mechanical energy \(E\) of the mass is conserved. Thus,

\[E = K + U = \frac{1}{2} mv^2 - \frac{GMm}{r}\]

Angular Momentum

We note that since the force only acts radially, there is also zero torque, \(\vec \tau\) that acts on the object. This is, of course, in an isolated binary system, where other forces do not exist and hence \(\vec \tau = 0 \cdot \hat{\mathbf{k}}\). In this case, we note that \(\vec \tau = \dot{\vec{L}} = 0 \cdot \hat{\mathbf{k}}\), the angular momentum, \(\vec L\) is conserved.

Now, we define \(\vec L\):

\[\vec L = \vec r \times \vec p = \vec r \times m \vec v\]

But what is the velocity, \(\vec v\)?

Find out, in the next page.