Selasa, 22 November 2011

Universal Gravitation

Chapter 13

Universal Gravitation

13.1 Newton’s Law of Universal Gravitation

You may have heard the legend that Newton was struck on the head by a falling apple while napping under a tree. This alleged accident supposedly prompted him to imagine that perhaps all objects in the Universe were attracted to each other in the same way the apple was attracted to the Earth. Newton analyzed astronomical data on the motion of the Moon around the Earth. From that analysis, he made the bold assertion that the force law governing the motion of planets was the same as the force law that attracted a falling apple to the Earth. This was the first time that “earthly” and “heavenly”
motions were unified. We shall look at the mathematical details of Newton’s analysis in this section.
In 1687 Newton published his work on the law of gravity in his treatise Mathematical
Principles of Natural Philosophy. Newton’s law of universal gravitation states that

Every particle in the universe attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them.

If the particles have masses m1 and m2 and are separated by a distance r, the magnitude of this gravitational force is

Fg=G (m_1 m_2)/r2
(13.1)
with the statement:
F is greater than the gravitational force between two point masses
G is the gravitational constant = 6.67 x 10-11 Nm2 / kg2
m1 is the mass of the first point
m2 is the mass of the second point
r is the distance between the two mass points
Where G is a constant called the universal gravitational constant, which has been measured experimentally. Its value in SI units are:

G=6,673 x 10-11N.m2 / kg2
( 13.2)

The form of the force law given by Equation 13.1 is often referred to as an inverse square law because the magnitude of the force varies as the inverse square of the separation of the particles.
the magnitude of the force
exerted by the Earth on a particle of mass m near the Earth’s surface is
Fg=G (M_E m)/(R_E^2 )
(13.3)
where ME is the Earth’s mass and RE its radius. This force is directed toward the center of the Earth

Quick Quiz 13.1 The Moon remains in its orbit around the Earth rather
than falling to the Earth because
(a) it is outside of the gravitational influence of the
Earth
(b) it is in balance with the gravitational forces from the Sun and other planets
(c) the net force on the Moon is zero
(d) none of these
(e) all of these.
Answer :
(d). The gravitational force exerted by the Earth on the
Moon provides a net force that causes the Moon’s centripetal
acceleration.

13.2 Measuring the gravitational constan
experimentally that the force is attractive. The universal gravitational constant G was measured in an important experiment by Henry Cavendish (1731–1810) in 1798. The Cavendish apparatus consists of two small spheres, each of mass m, fixed to the ends of a light horizontal rod suspended by a fine fiber or thin metal wire. The experiment is carefully repeated with different masses at various separations.
For get the value of G, the results show e, proportional to the product mM, and inversely proportional to the square of the distance r.

Quick Quiz 13.2 A planet has two moons of equal mass. Moon 1 is in a circular orbit of radius r. Moon 2 is in a circular orbit of radius 2r. The magnitude of the gravitational force exerted by the planet on moon 2 is
(a) four times as large as that on moon 1
(b) twice as large as that on moon 1
(c) equal to that on moon 1
(d) half as large as that on moon 1
(e) one fourth as large as that on moon 1.

Answer

(e). The gravitational force follows an inverse-square behavior,
so doubling the distance causes the force to be
one fourth as large.




13.3 Free Fall Acceleration and the Gravitational Force
Because the magnitude of the force acting on a freely falling object of mass m near the Earth’s surface is given by Equation 13.4, we can equate mg to this force to obtain
mg=G (M_E m)/(R_E^2 )
g=G (M_E )/(R_E^2 )
Now consider an object of mass m located a distance h above the Earth’s surface or
a distance r from the Earth’s center, where r=RE + h. The magnitude of the gravitational force acting on this object is
Fg=G (M_E m)/r^2 =G 〖GM〗_E/(R_E+ h )2
The magnitude of the gravitational force acting on the object at this position is also
Fg = mg, where g is the value of the free-fall acceleration at the altitude h. Substituting this expression for Fg into the last equation shows that g is
g=(〖GM〗_E )/r^2 =〖GM〗_E/(R_E+ h )2

Thus, it follows that g decreases with increasing altitude. Because the weight of an object is
mg, we see that as r ∞ its weight approaches zero.

Quick Quiz 13.3 Superman stands on top of a very tall mountain and throws a baseball horizontally with a speed such that the baseball goes into a circular orbit around the Earth. While the baseball is in orbit, the acceleration of the ball
(a) depends on how fast the baseball is thrown
(b) is zero because the ball does not fall to the ground
(c) is slightly less than 9.80 m/s2
(d) is equal to 9.80 m/s2.

Answer
(c). An object in orbit is simply falling while it moves
around the Earth. The acceleration of the object is that
due to gravity. Because the object was launched from a
very tall mountain, the value for g is slightly less than that
at the surface.

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