## (21) Kepler's Third LawBesides the Moon, Earth now has many artificial satellites, put up by us earthlings for a variety of purposes. The calculation applied by Newton to the Moon can also be used for them. ## Orbital velocitySuppose the Earth were a perfect sphere of radius 1 R If the satellite is in a stable circular orbit and its velocity is
Dividing both sides by
Multiplication of both sides by R
V = 7905. 66 m/sec = 7.90566 km/s = V 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
If the speed ## 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
Let |

VT | = 2 p r |

V | = 2 p r/ T |

V^{2} | = 4 p^{2}r^{2}/ T^{2} |

V^{2}/ r | = 4 p^{2}r/ T^{2} |

and by the earlier equality
Get rid of fractions by multiplying both sides by
To better see what we have, divide both sides by T:^{2}
What's inside the brackets is just a number. The rest tells a simple message-- , the orbital period squared is proportional to the distance cubes. r^{3}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. |

**Deriving a practical formula from Kepler's 3rd law.**

**Next Regular Stop: Frames of Reference: The Basics**

**
**

Author and curator: David P. Stern, u5dps@lepvax.gsfc.nasa.gov

Last updated 24 August 1998