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to Keep the Planets in their Orbits? |

and the Centrifugal Force on the Planet Due to the Curvature of its Orbit

The gravitational attraction on a planet of mass m due to the mass M of the Sun at a distance R is

where G is the gravitational constant.

The centripetal acceleration a experienced by a body moving with a tangential velocity of v in a circular orbit of radius is

The force required to hold it in that orbit is

If T is the orbit period then the tangential velocity is v=(2πR)/T.

The force holding a planet in orbit is a balance of the centrigugal force by the gravitational force and hence

Thus there are two ways of computing the force. Here are the results of such computations.

The mass of the Sun is 1.989×10^{30} kg. The gravitational constant is 6.67384×10^{-11} m³/(kg*s²)

Data for the Planets and the Computed Level of Force Holding Them in their Orbits | |||||||
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Planet | Mass (kg) | Radius of Orbit (10 ^{9} m) | Orbit Period (Earth days) | Tangential Velocity (m/s) | Centrifugal Force (Newtons) | Gravitational Force (Newtons) | Ratio |

Mercury | 3.30E+23 | 58 | 88 | 47930 | 1.3071E+22 | 1.30217E+22 | 0.996231843 |

Venus | 4.87E+24 | 108 | 224.7 | 34953 | 5.50907E+22 | 5.54233E+22 | 1.006036955 |

Earth | 5.97E+24 | 150 | 365.2 | 29869 | 3.55088E+22 | 3.52211E+22 | 0.991897547 |

Mars | 6.42E+23 | 228 | 687 | 24135 | 1.64017E+21 | 1.63937E+21 | 0.999511945 |

Jupiter | 1.90E+27 | 778 | 4332 | 13060 | 4.1657E+23 | 4.16682E+23 | 1.000269809 |

Saturn | 5.68E+26 | 1,429 | 10760 | 9658 | 3.70755E+22 | 3.69228E+22 | 0.995879774 |

Uranus | 8.68E+25 | 2,871 | 30700 | 6801 | 1.39832E+21 | 1.39786E+21 | 0.99966901 |

Neptune | 1.02E+26 | 4,504 | 60200 | 5441 | 6.70405E+20 | 6.67443E+20 | 0.995581671 |

The computed forces should be equal but due to small differences in the manner of computation they are slightly different

Having computed the force in newtons it is now desirable to express that force as something familiar.
The force on the Earth is about 3.53×10^{22} newtons. A newton is the weight on Earth
of 102 grams of matter. The force of the Earth due to the gravitational attraction of the Sun is equal to the
weight on Earth of about 3.6×10^{21} kilograms of matter, say water.
Water has a density of 1 gram per cubic centimeter, or 1000 kg/m³. Thus 3.6×10^{21}
corresponds to 3.6×10^{18} cubic meters or 3.6×10^{9} km*sup3; of water.
The volume of water in all of Earth's ocean is 5.136×10^{8} km³. Thus the force on the
Earth due to the gravitational pull of the Sun is about 7 times the weight of all the water in Earth's oceans.

In the equation

the mass m of a planet can be cancelled out and v may be replaced by 2πR/T leaving

and hence

This is Kepler's Law. .