This is the velocity required by the satellite to stay in its orbit ("1" in the drawing). Any slower and it loses altitude and hits the Earth ("2"), any faster and it rises to greater distance ("3"). For comparison, a jetliner flies at about 250 m/sec, a rifle bullet at about 600 m/sec.
We again need a notation for square root. Since the HTML language does not provide one, we use the notation SQRT found in some computer languages. The square root of 2, for instance, can be written
SQRT(2) = 1. 41412. . .
If the speed V of our satellite is only moderately greater than Vo curve "3" will be part of a Keplerian ellipse and will ultimately turn back towards Earth. If however V is greater than 1. 4142. . .times Vo the satellite has attained escape velocity and will never come back: this comes to about 11.2 km/sec.
The value of 11.2 km/s was already derived in the section on Kepler's 2nd law, where the expression for the energy of Keplerian motion was given (without proof) as
E = 1/2 mv2 – k m/r
where for a satellite orbiting Earth at distance of one Earth radius RE, the constant k equals k=gRE2. The energy is negative for any spacecraft captured by Earth's gravity, positive for any not held captive, and zero for one just escaping. Let V1 be the velocity of such a spacecraft, located at distance RE but with zero energy, i.e. E=0. Then
(1/2) m V12 = mgRE
V12 = 2gRE
which gives, as claimed,
Kepler's Third Law for Earth Satellites
The velocity for a circular Earth orbit at any other distance r is similarly calculated, but one must take into account that the force of gravity is weaker at greater distances, by a factor (RE/r)2. We then get
V2/r = g (RE r)2 = g RE2/r2
Let T be the orbital period, in seconds. Then (as noted earlier), the distance 2
πr covered in one orbit equals VT