This can be illustrated with the thought experiment shown in the following figure. Suppose we fire a cannon horizontally from a high mountain; the projectile will eventually fall to earth, as indicated by the shortest trajectory in the figure, because of the gravitational force directed toward the center of the Earth and the associated acceleration. (Remember that an acceleration is a change in velocity and that velocity is a vector, so it has both a magnitude and a direction. Thus, an acceleration occurs if either or both the magnitude and the direction of the velocity change.)
But as we increase the muzzle velocity for our imaginary cannon, the projectile will travel further and further before returning to earth. Finally, Newton reasoned that if the cannon projected the cannon ball with exactly the right velocity, the projectile would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth, which is curving away at the same rate that the projectile falls. That is, the cannon ball would have been put into orbit around the Earth. Newton concluded that the orbit of the Moon was of exactly the same nature: the Moon continuously "fell" in its path around the Earth because of the acceleration due to gravity, thus producing its orbit. By such reasoning, Newton came to the conclusion that any two objects in the Universe exert gravitational attraction on each other, with the force having a universal form:
The constant of proportionality G is known as the universal gravitational constant. It is termed a "universal constant" because it is thought to be the same at all places and all times, and thus universally characterizes the intrinsic strength of the gravitational force.
Consider the diagram shown to the right. We may define a point called the center of mass between two objects through the equations
where R is the total separation between the centers of the two objects. The center of mass is familiar to anyone who has ever played on a see-saw. The fulcrum point at which the see-saw will exactly balance two people sitting on either end is the center of mass for the two persons sitting on the see-saw.
Here is a Center of Mass Calculator that will help you make and visualize
calculations concerning the center of mass.
(Caution: this applet
is written under Java 1.1, which is only supported by the most recent browsers. It
should work on Windows systems under Netscape 4.06 or the most recent version of
Internet Explorer 4.0, but may not yet work on Mac or Unix systems or earlier Windows
where P is the planetary orbital period and the other quantities have the meanings described above, with the Sun as one mass and the planet as the other mass. (As in the earlier discussion of Kepler's 3rd Law, this form of the equation assumes that masses are measured in solar masses, times in Earth years, and distances in astronomical units.) Notice the symmetry of this equation: since the masses are added on the left side and the distances are added on the right side, it doesn't matter whether the Sun is labeled with 1 and the planet with 2, or vice-versa. One obtains the same result in either case.
Now notice what happens in Newton's new equation if one of the masses (either 1 or 2; remember the symmetry) is very large compared with the other. In particular, suppose the Sun is labeled as mass 1, and its mass is much larger than the mass for any of the planets. Then the sum of the two masses is always approximately equal to the mass of the Sun, and if we take ratios of Kepler's 3rd Law for two different planets the masses cancel from the ratio and we are left with the original form of Kepler's 3rd Law:
Thus Kepler's 3rd Law is approximately valid because the Sun is much more massive than any of the planets and therefore Newton's correction is small. The data Kepler had access to were not good enough to show this small effect. However, detailed observations made after Kepler show that Newton's modified form of Kepler's 3rd Law is in better accord with the data than Kepler's original form.
This is the situation in the Solar System: the Sun is so massive compared with any of the planets that the center of mass for a Sun-planet pair is always very near the center of the Sun. Thus, for all practical purposes the Sun IS almost (but not quite) motionless at the center of mass for the system, as Kepler originally thought.
However, now consider the other limiting case where the two masses are equal to each other. Then it is easy to see that the center of mass lies equidistant from the two masses and if they are gravitationally bound to each other, each mass orbits the common center of mass for the system lying midway between them:
This situation occurs commonly with binary stars (two stars bound gravitationally to each other so that they revolve around their common center of mass). In many binary star systems the masses of the two stars are similar and Newton's correction to Kepler's 3rd Law is very large.
Here is a Java applet that implements Newton's modified form of Kepler's 3rd law for two objects (planets or stars) revolving around their common center of mass. By making one mass much larger than the other in this interactive animation you can illustrate the ideas discussed above and recover Kepler's original form of his 3rd Law where a less massive object appears to revolve around a massive object fixed at one focus of an ellipse.
These limiting cases for the location of the center of mass are perhaps familiar from our afore-mentioned playground experience. If persons of equal weight are on a see-saw, the fulcrum must be placed in the middle to balance, but if one person weighs much more than the other person, the fulcrum must be placed close to the heavier person to achieve balance.
Here is a Kepler's Laws Calculator that allows you to make simple calculations for periods, separations, and masses for Keplers' laws as modified by Newton (see subsequent section) to include the effect of the center of mass. (Caution: this applet is written under Java 1.1, which is only supported by the most recent browsers. It should work on Windows systems under Netscape 4.06 or the most recent version of Internet Explorer 4.0, but may not yet work on Mac or Unix systems or earlier Windows browsers.)
Thus, the weight of an object of mass m at the surface of the Earth is obtained by multiplying the mass m by the acceleration due to gravity, g, at the surface of the Earth. The acceleration due to gravity is approximately the product of the universal gravitational constant G and the mass of the Earth M, divided by the radius of the Earth, r, squared. (We assume the Earth to be spherical and neglect the radius of the object relative to the radius of the Earth in this discussion.) The measured gravitational acceleration at the Earth's surface is found to be about 980 cm/second/second.