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Next: Changing Orbits Up: Orbital Mechanics Previous: Orbital Mechanics

Orbital Speed

The force and the acceleration are both centripetal (i.e. back towards the centre of the Earth, so we can just talk about the magnitudes of $\vec{\mbox{\boldmath$\space F $\unboldmath }}$ and $\vec{\mbox{\boldmath$\space a $\unboldmath }}$:

\begin{displaymath}F = {GmM_E \over r^2} \qquad \hbox{\rm and} \qquad a = {v^2 \over r}. \end{displaymath}


\begin{displaymath}\hbox{\rm but} \qquad F = ma, \qquad \hbox{\rm so} \end{displaymath}


\begin{displaymath}{GmM_E \over r^2} = {m v^2 \over r}
\Longrightarrow {G \, M_E \over r} = v^2. \end{displaymath}

We can ``solve'' this equation for v in terms of r,

 \begin{displaymath}v = \sqrt{G \, M_E \over r} ,
\end{displaymath} (10.6)

or for r in terms of v:

 \begin{displaymath}r = {G \, M_E \over v^2} .
\end{displaymath} (10.7)

You can try your hand with these equations. See if you can show that the orbital velocity at the Earth's surface (i.e. the speed required for a frictionless train moving through an Equatorial tunnel to be in free fall all the way around the Earth) is 7.905 km/s. For a more practical example, try calculating the radius and velocity of a geosynchronous satellite - i.e. a signal-relaying satellite in an Equatorial orbit with a period of exactly one day, so that it appears to stay at exactly the same place in the sky all the time.10.10


next up previous
Next: Changing Orbits Up: Orbital Mechanics Previous: Orbital Mechanics
Jess H. Brewer
1998-10-08