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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 $\unboldmath }}$:

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

$$\hbox{\rm but} \qquad F = ma, \qquad \hbox{\rm so}$$

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

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

$$v = \sqrt{G \, M_E \over r} ,$$ (10.6)

or for r in terms of v:

$$r = {G \, M_E \over v^2} .$$ (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