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.
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.
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.
If we assume that the mass of the rocket is small compared to the Earth, then the CIRCULAR VELOCITY is given by the square root of GM/r, that is
where G is the gravitational constant, R is the radius of the orbit, M is the mass of the larger object, like the Earth, around which the smaller object orbits. Note that the formula does not depend on the mass of the smaller object. (This last fact follows from the Newtonian theory, and is related to the experimental inference by Galileo that two objects of different mass dropped from the same height fall to Earth in the same time.)
We can use this formula to calculate how fast the moon moves in its orbit around the Earth. Plugging in the Earth's mass of M=6 x 1024 kg, the radius of the moon's orbit of R=3.84 x 108 meters, and gravitational constant G= 6.67 x 10-11 Newton meter2 / kg2, the magnitude of the moon's velocity is then 1020 meter/s. This is about 2278 miles per hour.
(Hey so if its moving so fast, 5 times faster than jet airplanes, why do jets seem to move faster on the sky than the moon? You'll have to answer this question on your own..)
Since the circular velocity varies inversely with the square root of R, an object in a smaller orbit has faster speed because the gravity is stronger. The same calculation for R=6578 km above the center of the Earth tells us that a satellite must move at speeds of 17,400 miles per hour (=7790 m/s). Thus rockets have to move incredibly fast. The rocket has to get up, and then turn to point in a circular orbit at the correct speed. But once it is at that speed, it will stay in orbit without subsequent rocket propulsion.
Since the orbital velocity of a satellite depends on its distance from the center of the Earth, the farther out, the longer the period of the orbit. Near the Earth the orbital period is about 1.5 hours. If one goes out to about 42,000 km (26,000) miles, the orbital period is 24 hours. Thus the satellite would be in GEOSYNCHRONOUS ORBIT. Imagine launching a satellite eastward above the Earth's equator in geosynchronous orbit: then the satellite will stay over the same spot on the Earth at all times in its orbit.
Orbits which close on themselves like circular or elliptical orbits are called CLOSED ORBITS. The object in such orbits always return to the same place in the orbit periodically. In circular orbits, the speed of the object remains the same everywhere in the orbit. In elliptical orbits the speed is faster as the object moves to the closer part of its orbit and then slows as the object proceeds to the farther part of the orbit.
There are also orbits called ESCAPE ORBITS or OPEN ORBITS. In these orbits the object never returns and goes off to large distances.
To see this Imagine a cannon shooting off of a mountain on the surface of the Earth (fig 5-13). There are 5 qualitatively different possibilities depending on how fast the cannonball travels with respect to the circular velocity initially.
Plugging in for the Earth's mass, radius, and for G, we obtain 11.2 km/s for the escape velocity for an object launched from the Earth's surface. This is about 25,000 miles per hour! I dare you to try to escape. (So much for Aristotle's worries about birds flying off the Earth's surface-- no chance in the presence of gravity.)
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 9.8 m/second2.
Notice that magnitude of the circular velocity of an object in orbit is also equal to the the circumference of an orbit divided by the period of the orbit. That is:
But notice what happens in Newton's new equation if the mass of the sun is much larger than the mass for any of the planets (which is always the case). Then the sum of the two masses is always approximately equal to the mass of the Sun and so are back to
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.