### Gravity between planets

We can now use Newton's Law to derive some results concerning planets in circular orbits. Although we know from Kepler's Laws that the orbits are not circular, in most cases approximating the orbit by a circle gives satisfactory results. When two massive bodies exert a gravitational force on one another, we shall see (in the SparkNote on Orbits) that planets describe circular or elliptical paths around their common center of mass. In the case of a planet orbiting the sun, however, the sun's mass is so much greater than the planets, that the center of mass lies well within the sun, and in fact very close to its center. For this reason it is a good approximation to assume that the sun stays fixed (say at the origin) and the planets move around it. The force is then given by:

From the central force acting on the planet is exerting a centripetal force. We know that a centripetal motion has acceleration = and thus = . We can therefore write (note that in what follows

*r*, without the vector arrow denote the magnitude of

*r*--that is

*r*= ||):

= |

Rearranging we have that:

v^{2} = |

Thus we have derived an expression for the speed of the planet orbiting the sun. However, we can also express the speed as the distance around the orbit divided by the time taken

*T*(the period):

v = |

Squaring this and equating this with the result from above:

= âá’T^{2} = |

Thus we have derived Kepler's Third Law for circular orbits from the Universal Law of Gravitation.

### Gravity near the earth

We can apply the Universal Law of Gravitation to objects near the earth
also. For an object at or near the surface of the earth, the force due to
gravity acts (for reasons that will become clearer in the section on Newton's
Shell Theory) toward the center of the earth. That is, it acts
downwards because every particle in the earth is attracting the object. The
magnitude of the force on an object of mass *m* is given by:

F = |

where

*r*

_{e}

^{2}is the radius of the earth. Let us calculate the constant :

= 9.74 |

This is the acceleration due to gravity on the earth (the figure is usually given as

9.8 m/sec^{2}

*g*, then we have the familiar equation

*F*=

*mg*which determines all free-fall motion near the earth.

We can also calculate the value of *g* that an astronaut in a space shuttle
would feel orbiting at a height of 200 kilometers above the earth:

g_{1} | = | ||

= | (6.67×10^{-11})(5.98×10^{24})(6.4×10^{6} +2×10^{5})^{-2} | ||

= | 9.16 |

This small reduction in

*g*is not sufficient to explain why the astronauts feel "weightless." In fact, this is caused by the fact that the shuttle's orbit is in fact a constant free-fall around the earth. An orbit is essentially a perpetual "falling" around a planet--since an orbiting shuttle and its occupant astronauts are falling with the same acceleration as the gravitational field, they feel no gravitational force.

### Determining G

Because the gravitational force between everyday-sized objects is very small,
the gravitational constant, *G*, is extremely difficult to measure accurately.
Henry Cavendish (1731-1810) devised a clever apparatus for measuring the
gravitational constant. A fiber is attached to the center of the beam to
which *m* and *m'* are attached, as shown in . This is
allowed to reach an equilibrium, untwisted state before, the two larger masses
*M* and *M'* are lowered next to them. The gravitational force between the two
pairs of masses causes the string to twist such that the amount of twisting is
just balanced by the gravitational force. By appropriate calibration (knowing
how much force causes how much twisting), the gravitational force may be
measured. Since the masses and the distances between them may also be measured,
only *G* remains unknown in the Universal Law of Gravitation. Thus *G* can
be calculated from the measured quantities. Accurate measurements of *G* now
place the value at 6.673×10^{-11} N.m^{2}/kg^{2}.