what force causes an apple to fall from a tree

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Sir Isaac Newton: The
Universal Law of Gravitation

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At that place is a popular story that Newton was sitting nether an apple tree, an apple tree brutal on his head, and he suddenly thought of the Universal Law of Gravitation. As in all such legends, this is well-nigh certainly not true in its details, but the story contains elements of what actually happened.

What Really Happened with the Apple tree?

Probably the more right version of the story is that Newton, upon observing an apple fall from a tree, began to remember along the following lines: The apple tree is accelerated, since its velocity changes from zero as it is hanging on the tree and moves toward the basis. Thus, by Newton'southward 2nd Law there must be a force that acts on the apple to crusade this acceleration. Let'southward call this force "gravity", and the associated acceleration the "dispatch due to gravity". So imagine the apple tree tree is twice every bit high. Again, we expect the apple to be accelerated toward the footing, so this suggests that this force that we call gravity reaches to the superlative of the tallest apple tree.

Sir Isaac's Most Fantabulous Idea

Now came Newton'south truly bright insight: if the force of gravity reaches to the acme of the highest tree, might it not reach even further; in particular, might it not accomplish all the way to the orbit of the Moon! Then, the orbit of the Moon about the Earth could exist a consequence of the gravitational strength, considering the acceleration due to gravity could modify the velocity of the Moon in just such a fashion that it followed an orbit effectually the world.

This can be illustrated with the idea experiment shown in the following figure. Suppose nosotros burn down a cannon horizontally from a high mountain; the projectile will eventually fall to earth, equally 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 dispatch is a modify in velocity and that velocity is a vector, so it has both a magnitude and a direction. Thus, an dispatch occurs if either or both the magnitude and the direction of the velocity alter.)

But as nosotros increase the cage velocity for our imaginary cannon, the projectile volition travel further and further before returning to globe. Finally, Newton reasoned that if the cannon projected the cannon ball with exactly the right velocity, the projectile would travel completely around the Globe, always falling in the gravitational field but never reaching the Earth, which is curving abroad at the same rate that the projectile falls. That is, the cannon ball would take been put into orbit around the World. Newton concluded that the orbit of the Moon was of exactly the aforementioned nature: the Moon continuously "fell" in its path around the World because of the acceleration due to gravity, thus producing its orbit.

Past such reasoning, Newton came to the conclusion that whatsoever ii objects in the Universe exert gravitational allure on each other, with the force having a universal grade:

The constant of proportionality G is known as the universal gravitational constant. It is termed a "universal constant" because information technology is thought to be the same at all places and all times, and thus universally characterizes the intrinsic strength of the gravitational strength.

The Center of Mass for a Binary System

If yous recall about it a moment, it may seem a little strange that in Kepler's Laws the Sun is stock-still at a betoken in space and the planet revolves around it. Why is the Sun privileged? Kepler had rather mystical ideas about the Sun, endowing information technology with nearly god-similar qualities that justified its special place. However Newton, largely equally a corollary of his 3rd Law, demonstrated that the state of affairs actually was more symmetrical than Kepler imagined and that the Sun does non occupy a privileged position; in the process he modified Kepler's tertiary Law.

Consider the diagram shown to the correct. We may define a betoken called the center of mass between two objects through the equations

where R is the total separation between the centers of the ii objects. The center of mass is familiar to anyone who has ever played on a see-saw. The fulcrum bespeak at which the come across-saw volition exactly balance ii people sitting on either end is the center of mass for the two persons sitting on the come across-saw.

Two Limiting Cases

We tin can gain further insight past considering the position of the middle of mass in two limits. First consider the example just addressed, where one mass is much larger than the other. And so, we run across that the center of mass for the organisation essentially coincides with the eye of the massive object:

This is the state of affairs in the Solar System: the Sun is so massive compared with whatever of the planets that the center of mass for a Sun-planet pair is always very near the center of the Sunday. Thus, for all practical purposes the Sun IS about (but not quite) motionless at the centre of mass for the system, as Kepler originally thought.

However, now consider the other limiting instance where the two masses are equal to each other. Then it is piece of cake to run across that the center of mass lies equidistant from the two masses and if they are gravitationally spring to each other, each mass orbits the mutual heart of mass for the system lying midway between them:

This situation occurs usually with binary stars (two stars spring 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 feel. If persons of equal weight are on a see-saw, the fulcrum must be placed in the middle to residue, merely if one person weighs much more than than the other person, the fulcrum must be placed shut to the heavier person to achieve residual.

Circular Velocity and Geosynchronous Orbit

How fast do we have to travel in a rocket to stay in circular orbit around a much more massive body such as the Earth? I will present the answer without derivation here, merely information technology follows from Newton's law of gravitation.

If nosotros presume that the mass of the rocket is small compared to the Earth, then the CIRCULAR VELOCITY is given past the square root of GM/r, that is

V_circ = (GM/R)^1/two

where Grand 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 non 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 ii objects of dissimilar mass dropped from the same pinnacle fall to Earth in the aforementioned time.)

We can use this formula to summate how fast the moon moves in its orbit around the Earth. Plugging in the Globe's mass of One thousand=6 ten 10²⁴ kg, the radius of the moon'south orbit of R=3.84 10 ten⁸ meters, and gravitational constant G= 6.67 ten ten^-xi Newton meter² / kgⁱⁱ, the magnitude of the moon'due south velocity is so 1020 meter/due south. This is about 2278 miles per hour.

(Hey and so if its moving so fast, 5 times faster than jet airplanes, why practice jets seem to motility faster on the sky than the moon? You lot'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 aforementioned calculation for R=6578 km above the center of the Earth tells us that a satellite must motion at speeds of 17,400 miles per hour (=7790 m/southward). Thus rockets have to motility incredibly fast. The rocket has to get up, and and so turn to point in a circular orbit at the right speed. Merely 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. Well-nigh the World the orbital menses is most one.five hours. If 1 goes out to most 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'due south equator in geosynchronous orbit: and then the satellite will stay over the aforementioned spot on the World at all times in its orbit.

Open and Closed Orbits

Orbits which close on themselves like circular or elliptical orbits are called CLOSED ORBITS. The object in such orbits always return to the aforementioned 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 so slows as the object proceeds to the farther role 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 come across this Imagine a cannon shooting off of a mountain on the surface of the Earth (fig 5-13). There are 5 qualitatively dissimilar possibilities depending on how fast the cannonball travels with respect to the round velocity initially.

If it travels at the circular velocity, the orbit will exist circular.

If information technology travels much less than circular velocity the cannon brawl will drop to the surface.

If information technology travels a little flake less than circular velocity, information technology will form an elliptical orbit with the cannon at APOGEE (the farthest point in the orbit).

If the ball travels a petty bit faster than circular velocity, so it will form an elliptical orbit with the cannon at the PERIGEE (the nearest bespeak in the orbit).

If the ball travels at the ESCAPE VELOCITY (see below), and so the orbit will be OPEN and will be a parabola. If the ball travels faster than the ESCAPE VELOCITY, the orbit with exist Open and exist a hyperbola.

Escape Velocity

The ESCAPE VELOCITY from an object like the Earth is given by

V_es = (2GM/R)^i/ii

where R is the radius of the launch point for the object. When the launch point is on the Earth'southward surface, then R would be the radius of the Globe. This escape velocity is the critical velocity an object must have to afterwards coast to infinity when shot up from within a gravitational field. That is, if a rocket is shot from the Earth and consumes all of its fuel to accelerate to this velocity, then even later on the rocket is no longer called-for fuel, it will coast to infinity and the Earth's gravity cannot pull the rocket dorsum to Earth.

Plugging in for the Globe's mass, radius, and for Thou, we obtain 11.2 km/s for the escape velocity for an object launched from the Earth's surface. This is most 25,000 miles per hour! I cartel you to try to escape. (Then much for Aristotle'due south worries nigh birds flying off the Earth's surface-- no chance in the presence of gravity.)

Weight and the Gravitational Force

We have seen that in the Universal Law of Gravitation the crucial quantity is mass. In popular linguistic communication mass and weight are oftentimes used to hateful the same thing; in reality they are related just quite different things. What we ordinarily call weight is really but the gravitational force exerted on an object of a sure mass. We can illustrate by choosing the Earth as one of the two masses in the previous illustration of the Law of Gravitation:

Thus, the weight of an object of mass m at the surface of the Globe is obtained by multiplying the mass m by the acceleration due to gravity, grand, 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 World M, divided by the radius of the Earth, r, squared. (Nosotros 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 Globe's surface is found to be about 9.8 m/2d².

Mass and Weight

Mass is a measure out of how much material is in an object, but weight is a measure of the gravitational force exerted on that fabric in a gravitational field; thus, mass and weight are proportional to each other, with the acceleration due to gravity equally the proportionality constant. Information technology follows that mass is abiding for an object (really this is non quite true, only we will save that surprise for our later discussion of the Relativity Theory), simply weight depends on the location of the object. For example, if we transported the preceding object of mass thousand to the surface of the moon, the gravitational acceleration would modify because the radius and mass of the Moon both differ from those of the Earth. Thus, our object has mass k both on the surface of the Earth and on the surface of the Moon, simply it will counterbalance much less on the surface of the Moon because the gravitational dispatch at that place is a factor of half-dozen less than at the surface of the World.

Newton'due south Derivation of Kepler's Laws

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:

V_circ = (GM/R)^1/ii = (2 pi R)/ P

where pi=3.1415 and P is the orbital period. The latter equality follows from merely noting that the magnitude of velocity is a measurement of distance traveled per unit time. The distance traveled is the circumference of the orbit (= 2 pi R), and the menstruation P is the time it takes to travel this distance. Merely expect: now magically nosotros have recovered Kepler's third law!: Rearranging the latter equality we have

(GM) P² = 4 pi² R^three

the bespeak being that P² is proportional to R^three. That is Kepler'south 3rd police force, now falling direct from Newton'due south theory.

Newton's Estimation of Kepler's Laws

Because for every action in that location is an equal and opposite reaction, Newton realized that in the planet-Sunday arrangement the planet does not orbit around a stationary Lord's day. Instead, Newton proposed that both the planet and the Sun orbited effectually the common center of mass for the planet-Dominicus system. He then modified Kepler's 3rd Law so that the mass M used is at present the sum of the mass of the Lord's day plus the planet. Instead of using M₁ and M_ii as to a higher place, let us apply M_{due south} and G_p. Then we have M= Chiliad_s + Yard_p and so

Grand(M_{due south} + G_p ) P ² = 4 pi² R³

But notice what happens in Newton's new equation if the mass of the sunday is much larger than the mass for whatsoever of the planets (which is e'er the instance). So the sum of the two masses is always approximately equal to the mass of the Sun and so are back to

M Thou_s P² = 4 piⁱⁱ R³

for planet-Sun systems. If nosotros accept ratios of Kepler's 3rd Police force for two different planets the Sun mass so cancels from the ratio and we are left with the original class of Kepler'southward 3rd Police force:

Thus Kepler's 3rd Law is approximately valid because the Sunday is much more than massive than any of the planets and therefore Newton's correction is small. The information Kepler had access to were not skillful enough to show this small-scale effect. However, detailed observations fabricated after Kepler evidence that Newton's modified form of Kepler's third Police is in better accord with the data than Kepler's original form.

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Source: http://www.pas.rochester.edu/~blackman/ast104/newtongrav.html

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