《物理双语教学课件》Chapter 8 Gravitation 万有引力

8.1 Newton’s Law of Gravitation

1. In 1665, the 23-year-old Isaac Newton made a basic contribution to physics when he showed that the force that holds Moon in its orbit is the same force that makes an apple fall. Newton concluded that not only does Earth attract an apple and the Moon but every body in the universe attracts every other body; this tendency of bodies to move toward each other is called gravitation. 2. Quantitatively, Newton proposed a force law that we call Newton’s law of gravitation: every particle attracts any other particle with a gravitational force whose magnitude is given by

m2FGm1m2r2. Here

m1

and

are the masses of the particles,

G

r is the distance

between them, and is the gravitational constant, whose

G6.671011Nm2/kg2. The directions

value is known to be

of the gravitation are shown in above figure.

3. Although Newton’s law of gravitation applies strictly to particles, we can also apply it to real object as long as the sizes of the objects are small compared to the distance

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between them.

4. For the apple-Earth problem, Newton proved an important theorem called the shell theorem: A uniform spherical shell of matter attracts a particle that is outside the shell as if all the shell’s mass were concentrated at its center. Earth can be thought of as a nest of such shells. One within another, and each attracting a particle outside Earth’s surface as if the mass of that shell were at the center of the shell. Thus, from the apple’s point of view, Earth does behave like a particle, located at the center of Earth and having a mass equal to that of the planet.

8.2 Gravitation and the Principle of Superposition

1. Given a group of particles, we find the net (or resultant) gravitational force exerted on any one of them by using the principle of superposition. This is a general principle that says a net effect is the sum of the individual effects. Here, the principle means that we first compute the gravitational force that acts on our selected particle due to each of the other particles, in turn. We then find the net force by adding these forces vectorially, as usual.

2. For n interacting particles, we can write the principle of

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superposition for gravitational forces as Here

F1

nF1F12F13F14F15F1nF1ii2.

F1i

is the net force on particle 1 and is the force

exerted on particle 1 by the ith particle.

3. To an extended object, we can divide the object into differential units of mass differential force

dF

dm, each of which exerts only a

on the particle. To get the gravitational

F1dF.

force on the particle exerted by the extended force, we take an integral over the entire extended object

8.3 Gravitation near Earth’s Surface

1. Let us ignore the rotation of Earth for a moment and assume the planet to be a non-rotating uniform sphere of mass M. The magnitude of the gravitational force acting on a particle of mass m, located outside Earth a distance r from Earth’s center, is then given by

FGMm 2r2. If the particle is released, it will fall toward the center of Earth, as a result of the gravitational force F, with an acceleration that we shall call the gravitational acceleration

ag. Newton’s second law tells us that F and ag

are related by

agGMr2.

3. The gravitational acceleration compared with above equation

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is not the same as the free-fall acceleration g that we would measure for the falling particle. The two accelerations differ for three reasons: (1). Earth is not uniform, (2) it’s not a perfect sphere, and (3). It rotates. 4. To see how Earth’s rotation causes the free-fall acceleration g to differ from the gravitational acceleration

ag, we analyze a simple situation in

which a crate of mass m rests on a

scale at the equator, as shown in the figure. From Newton’s second law, we have FmagNmagmgma. So we get the relation:

agga2R(22)R0.034m/s2. Therefore the Tfree-fall acceleration g measured on the equator of the real rotating planet is slightly less than the gravitational acceleration

8.4 Gravitation Inside Earth

1. Newton’s shell theorem can also applied to a situation in which a particle is located inside a uniform shell, to show the following: A uniform shell of matter exerts no net gravitational force on a particle located inside it.

2. For a uniform Earth, the force on the particle steadily

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ag due strictly to the gravitational force.

decreases to zero as the particle approach the center of Earth. However, for the real (non-uniform) Earth, the force on the particle actually increases a maximum at a certain depth; only then does it begin to decrease as particle descends further.

8.5 Gravitational Potential Energy

1. Work done by gravitational force: We suppose the particle m exerted the gravitational force by the particle M moves from a to b. Then the work done by the gravitational force is:

MmMmdWFdlG2r0dlG2drrr bb1MmMmMmMmWdWGMmd()GG(G)(G)aarrbrarbraThe work done by the gravitational force is independent of the path, so the gravitational force is a conservative force. Thus we can introduce the gravitational potential energy. 2. (1). The change of the gravitational potential energy is equal to minus of the work done by the gravitational force:

MmMmUUfUiW(G)(G)

rfri(2). If we choose these two particles are separated far away (infinite) as a reference configuration with U equal to zero. With these facts in mind, we take the gravitational potential

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energy of the two-particle system to be (3). Note that

U(r)

UGMm. r

approaches zero as r approaches infinity

U(r)

and that for any finite value of r, the value of negative.

is

3. If the system contains more than two particles, we consider each pair of particles in turn, calculate the gravitational potential energy of pair with above equation as if the other particles were not there, and then algebraically sum the results.

4. Potential energy and force: We also can derive the force function from the potential energy function:

FdUdGMmMm()G2 drdrrr This is just Newton’s law of gravitation. 5. Escape speed

(1). If you fire a projectile upward, usually it will slow, stop momentarily, and return to Earth. There is, however, a certain minimum initial speed that will cause it move upward forever, theoretically coming to rest only at infinity. This initial speed is called the escape speed.

(2). Consider a projectile of mass m, leaving the surface of a planet (or some other astronomical body or system) with

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escape speed v. Its mechanical energy can be expressed as

EKU12GMmmv(), in which M is the mass of the planet 2Rand R its radius.

(3). When the projectile reaches infinity and it stops, both its kinetic energy and gravitational potential energy will be zero. (4). From the principle of conservation of energy, we will have

8.6 Planets and Satellites: Kepler’s Laws:

The motions of the planets, as they seemingly wander against the background of the stars, have been a puzzle since the dawn of history. Johannes Kepler (1571-1630), after a lifetime of study, worked out the empirical laws that govern these motions. Tycho Brahe (16-1601), the last of the great astronomers to make observations without the help of a telescope, compiled the extensive data from which Kepler was able to derive the three laws of planetary motion that now bear his name. Later, Newton (12-1727) showed that his law of gravitation lead to Kepler’s empirical laws.

We discuss each of Kepler’s law in turn. Although we apply the laws here to planets orbiting the Sun, they hold equally well for satellites, either natural or artificial, orbiting Earth or any other

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E12GMmmv()0. This yields v2r2GMR.

massive central body. 1. The law of orbit

(1). All planets move in elliptical orbit, with the Sun at one focus. (2). The figure shows a planet, of mass m, moving in such an orbit around the Sun, whose mass is M. We assume that

Mm, so that the

center of mass of the planet-Sun system is virtually at the center of the Sun.

(3). The orbit in the figure is described by giving its semi-major axis a and its eccentricity e, the later defined so that ea is the distance from the center of the ellipse to either focus F or F’. An eccentricity of zero corresponds to a circle, in which the two foci merge to a single central point. The eccentricities of the planetary orbits are not large, so-sketched on paper-the orbits look circular. 2. The law of areas

(1). A line that connects a planet to the Sun sweeps out equal areas in equal time.

(2). Quantitatively, this second law tells us that the planet will move most slowly when it is farthest from the Sun and most rapidly when it is nearest to the Sun.

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(3). As it turns out, Kepler’s second law is totally equivalent to the law of conservation of angular momentum. (4). The area swept out in time

t

by a line connecting the Sun

Ar2/2and the planet can be expressed as . The

instantaneous rate at which area is being swept out is then

dAr2dr2dt2dt2, in which  is the angular speed of the rotating

line connecting Sun and planet.

(5). The magnitude of the angular momentum of the planet about the Sun can be written as

dAL. If dA/dt dt2mLrp(r)(mv)(r)(mr)mr2.

(6). Comparing above two equations, we come to the conclusion that

is constant, as Kepler said it, then L must

also be constant-angular momentum is conserved. 3. The law of periods

(1). The square of the period of any planet is proportional to the cube of the semi-major axis of its orbit.

(2). Considering a circular orbit with radius r, and applying Newton’s second law, we yield

GMm(m)(2r). If we replace  2rwith the period of the motion, we will obtain Kepler’s third law

423T()r.

GM2

(3). The above equation holds for elliptical orbits, provided we replace r with a, the semi-major axis of the ellipse.

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(4). This law predicts that the ratio same value for every planetary orbit around a given massive body. The table shows how well it holds for the orbit of the planets of the solar system.

8.7 Satellite: Orbits and Energy

T2/a3 has essentially the

As a satellite orbits Earth on its elliptical path, both its speed and its distance from the center of Earth fluctuate with fixed periods. However, the mechanical energy of the satellite remains constant.

1. The potential energy is given by UGMm.

r2. The kinetic energy of a satellite in a circular orbit can be

der

v2/r

GMmv2m2rr, where

is the centripetal acceleration of the satellite. So can get

K12GMmUmv22r2the kinetic energy as .

3. The total mechanical energy of the orbiting satellite is

EKUGMmGMmGMmK. 2rr2r 10

4. For a satellite in an elliptical orbit of semi-major axis a, the mechanical energy can be written as energy of an orbiting satellite depends only on the semi-major axis of its orbit and not on its eccentricity e.

EGMm. So the total 2a 11