# Question #5ba10

Apr 9, 2016

#### Explanation:

If we apply Newton's law, the force due to gravity will be
$F = \frac{G M m}{r} ^ 2$

The satellite moves in a circular orbit, and the centripetal force must be provided by the gravitational force. So for a stable orbit
$F = \frac{m {v}^{2}}{r} = \frac{G M m}{r} ^ 2$

So if the velocity is too high, then

$\frac{m {v}^{2}}{r} > \frac{G M m}{r} ^ 2$

The force of gravity is insufficient to maintain the satellite in a circular orbit and the satellite will "fly off into space" unless it can be directed into an orbit of lower radius.

Or seen another way, rearranging the above equations gives:
$G M = r \cdot {v}^{2}$
Since $G M$ is constant, if $v$ is too high, $r$ must be reduced (or the satellite will not maintain its orbit).

If velocity is too low:
$\frac{m {v}^{2}}{r} < \frac{G M m}{r} ^ 2$
The force gravity exceeds the centripetal force required. Unless the orbit can be moved to a larger radius (reducing the force of gravity until the above are in equilibrium), the satellite will spiral in towards the earth.