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Next: The hollow earth Up: Gravitational Potential Energy Previous: solution

Energy of orbits

Let's think a bit about the total energy of orbiting objects. Suppose an object with mass $m$ doing a circular orbit around a much heavier object with mass $M$. Now we know its potential energy. It's

\begin{displaymath}
U ~=~ - {GMm\over R}
\end{displaymath} (1.32)

How about it's kinetic energy? From eqn. 1.11 and the fact that $v ~=~ \omega R$ have

\begin{displaymath}
v^2 ~=~ \omega^2 R^2 ~=~ G {M\over R}
\end{displaymath} (1.33)

so that

\begin{displaymath}
K ~=~ {1 \over 2}mv^2 ~=~ {1 \over 2}{GMm\over R}
\end{displaymath} (1.34)

Notice that $K ~=~ -U/2$ and that
\begin{displaymath}
E ~=~ K+U ~=~ U/2 ~=~ - {GMm\over 2R}
\end{displaymath} (1.35)

So the total energy is always negative. In the same way that electrons in an atom are bound to their nucleus, we can say that a planet is bound to the sun. It's energy is negative so it doesn't have enough energy to escape to infinity.

But what if the energy were positive? In that case the trajectories are no longer elliptical, and instead you get hyperbolic orbits! The object comes in from inter-stellar space (as they say on star trek) almost going in a straight line, and then cruises around the sun and is finally deflected in a straight line off into never-never land, never to be seen by us again!

\begin{figure}\centerline{\psfig{file=hyperbolic.eps,width=4in}}\end{figure}


next up previous
Next: The hollow earth Up: Gravitational Potential Energy Previous: solution
Joshua Deutsch 2003-03-05