
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 V_{o} 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 V_{o} 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 where for a satellite orbiting Earth at distance of one Earth radius R_{E}, the constant k equals k=gR_{E}^{2}. 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 V_{1} be the velocity of such a spacecraft, located at distance R_{E} but with zero energy, i.e. E=0. Then Kepler's Third Law for Earth SatellitesThe 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 (R_{E}/r)^{2}. We then get V^{2}/r = g (R_{E} r)^{2} = g R_{E}^{2}/r^{2} Let T be the orbital period, in seconds. Then (as noted earlier), the distance 2 πr covered in one orbit equals VT 
VT  = 2 π r 
V  = 2 π r/T 
V^{2}  = 4 π^{2}r^{2}/T^{2} 
V^{2}/r  = 4 π^{2} r/T^{2} 
and by the earlier equality 4 π^{2}r/T^{2} = g R_{E}^{2}/r^{2} Get rid of fractions by multiplying both sides by r^{2}T^{2} 4 π^{2}r^{3} = g R_{E}^{2} T^{2} To better see what we have, divide both sides by g R_{E}^{2}, isolating T^{2}: T^{2} = (4 π^{2}/g R_{E}^{2}) r^{3} What's inside the brackets is just a number. The rest tells a simple messageT^{2} is proportional to r^{3}, the orbital period squared is proportional to the distance cubes. This is Kepler's 3rd law, for the special case of circular orbits around Earth. If you are not yet tired of the calculation, you may click here for turning the above into a practical formula. TidbitThe first claim of escape from Earth' gravity goes back to 1634. Marin Mersenne, a friend of Descartes, got involved in an arcane argument: if a cannon barrel were set into the ground precisely vertically and fired, would the cannonball fall back into the barrel or not? Mersenne and a friend of his named Petit, with ties to the military, obtained a cannon and actually tried it out. (My source did not mention how close to the cannon they stood at that time.)The result was somewhat unexpected: the cannon ball was never found again! Frustrated, they consulted Descartes, whose opinion was that the projectile went so far from Earth that it escaped its pull and never came back. (""Retomberatil?", Amer. J. of Physics 22, p.76, 1954) Questions from Users: Spaceflight without escape velocity? *** Does the solar wind have escape velocity? *** Is orbital motion same as free fall? 
Deriving a practical formula from Kepler's 3rd law.
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Author and Curator: Dr. David P. Stern
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