14.1. The Gravitational Force
Gravity is the weakest force we know, but it is the force of gravity that
controls the evolution of the universe. Every body in the universe attracts
every other body. Newton proposed that the magnitude of this force is given
by
where m
1 and m
2 are the masses of the particles,
r is the distance between them and G is a universal constant whose value is
G = 6.67 x 10-11 N m2/kg2
The gravitational forces between two particles act along the line
joining them, and form an action-reaction pair (see Figure 14.1).
Figure 14.1. The gravitational force.
In real life we are not dealing with point particles; instead we are
dealing with extended objects. To evaluate the gravitational force between
extended objects, the
shell theorem can be used:
"A uniform shell of matter attracts an external particle as if all the shell's
mass were concentrated at its center"
Proof:
Figure 14.2 shows a shell located a distance r from a particle with mass m.
The radius of the shell is R and its mass is M. The mass density of the shell
is given by
All points on the small hoop indicated in Figure 14.2 have the same
distance to the particle m. The magnitude of the gravitational attraction
between any of these points and the mass m is therefore the same. The net
force between the hoop and mass m acts along the axis connecting the center of
the shell and mass m. The area of the hoop is given by
Figure 14.2. Shell theorem.
and its mass m is equal to
The net force is equal to
The angles [theta] and a can be eliminated by using the following
relations:
and
Differentiating the first of these two equations with respect to
[theta] we obtain
or
Further more we see that
The total force acting on mass m can now be obtained easily
The shell theorem immediately shows that a sphere of uniform density
(and mass M) attracts an external particle as if all the mass of the sphere is
concentrated in its center.
In a similar fashion we can proof that a uniform shell of matter exerts no
gravitational force on a particle located inside it .
14.2. The gravitational constant G
The strength of the gravitational force depends on the value of G. The value
of the gravitational constant can be determined using the Cavendish apparatus.
Two small lead spheres of mass m are connected to the end of a rod of length L
which is suspended from it midpoint by a fine fiber, forming a torsion balance.
Two large lead spheres, each of mass M, are placed in the location indicated in
Figure 14.3. The lead spheres will attract each other, exerting a torque on
the rod. In the equilibrium position the gravitational torque is just balanced
by the torque exerted by the twisted fiber. The torque exerted by the twisted
wire is given by
Figure 14.3. The Cavendish Apparatus.
The torque exerted by the gravitational force is given by
where R is the equilibrium distance between the center of the large and
the small spheres. If the system is in equilibrium, the net torque acting on
the rod is zero. Thus
All of a sudden the large spheres are rotated to a new position
(position B in Figure 14.3). The net torque acting on the twisted fiber is now
not equal to zero, and the system will start to oscillate. The period of
oscillation is related to the rotational inertia and the torsion constant
[kappa]
The angle between the two equilibrium positions is measured to be
2[theta]. This, combined with the measured torsion constant, is sufficient to
determine the torque [tau] acting on the torsion balance due to the
gravitational force. Measurements show that G = 6.67 x 10
-11
Nm
2/kg
2.
14.3. Free-fall Acceleration
If the mass density of the earth depends only on the distance from the center
of the earth (homogeneous shells), we can easily calculate the net
gravitational force acting on a particle of mass m, located at an external
point, a distance r from the center of the earth:
where M is the mass of the earth. For a particle on the earth surface,
r = R
e, the gravitational force is given by
We conclude that the free-fall acceleration depends on the mass of the
earth and its radius:
The measured value of g = 9.8 m/s
2 and R
e = 6.37
x 10
6 m gives
which is in good agreement with the accepted value of 5.98 x
10
24 kg. In reality, the situation is more complicated:
14.4. Gravitational Potential Energy
In chapter 8 we have discussed the relation between the force and the
potential energy. Consider two particles of masses
m
1 and m
2, separated by a distance r. In the
gravitational field it is convenient to define the zero potential
energy configuration to be one in which the two particles are separated by a
large distance (infinity). Suppose the two masses are brought together
(distance r) from infinity, along the path connecting the centers of the two
masses. The work done by the gravitational force can be calculated as
follows
(note that the force F and the position vector r are pointed in an
opposite direction, and the angle between them is 180deg.). The potential
energy U(r) is now given by
The potential energy is always negative and is a property of the two
masses together rather than of either mass alone. We can verify our
calculation by using U(r) to calculate the gravitational force
which is of course equal to Newton's law of gravity.
The work done by the gravitational force depends only on its initial
and its final position, and not on the actual path followed. For example, a
baseball travels from point A to point B (see Figure 14.5). The work done by
the gravitational force on the baseball along the arcs is zero since the force
and displacement are perpendicular. The only segments that contribute to the
work done are those segments along the radial direction. The work done is
negative if the force and the displacement are pointing in the opposite
direction; if the force and the displacement are pointing in the same direction
the work is positive Therefore the net work done if we travel along the
radial direction back-and-forth (initial and final points coincide) is zero.
We can now easily show that the net work done by the gravitational force on the
baseball is just determined by its initial radial position and its final radial
position.
Figure 14.5. Work done by the gravitational force.
Figure 14.6. A system of three particles.
If the system contains more than two particles, the
principle of
superposition applies. In this case we consider each pair and the total
potential energy is equal to the sum of the potential energies of each pair.
This is illustrated in Figure 14.6 for a system consisting of 3 particles. In
calculating the total potential energy of a system of particles
one should
take great care not to double count the interactions. The total potential
energy of the system shown in Figure 14.6 can be easily calculated:
The total potential energy of a system of particles is sometimes
called
the binding energy of the system. The total potential
energy is the amount of work that needs to be done to separate the individual
parts of the system and bring them to infinity.
Example
The gravitational potential can be used to calculate the minimum initial speed
that a projectile must have to escape from the earth. Suppose a projectile of
mass m has a speed v. Its initial kinetic energy if given by
The initial potential energy of the projectile is given by
and its initial mechanical energy is equal to
In deep space the potential energy of the projectile will be zero, and
its minimum kinetic energy will also be equal to zero. We conclude that the
minimum mechanical energy of the projectile must be zero. Therefore
and
This initial speed is called the
escape speed. For the
earth we obtain
vcrit = 1.1 x 104 m/s
14.5. Motion of planets( Kepler's Laws)
Suppose a planet with mass m is in a circular orbit around the sun, whose mass
is M. The radius of the orbit is r. The gravitational force between the sun
and the planet is given by
This is the force that keeps the planet in its circular orbit and its
magnitude should therefore be equal to the centripetal force F
C:
This implies that
or
This shows that for circular orbits, the square of the period of any
planet is proportional to the cube of the radius of the orbit (
law
of periods). The constant depends only on the mass of the sun (M) and the
gravitational constant (G).
In reality none of the planets carry out a circular orbit; their orbits are
elliptical. The general equation of an ellipse is given by (see
Figure 14.7)
The parameter a is called the semi-major axis of the ellipse
(if a > b). It corresponds to the longest distance between the center of
the ellipse (x=0,y=0) and the trajectory. The parameter b is called the
semi-minor axis of the ellipse (if a > b). It corresponds to the
shortest distance between the center of the ellipse (x=0,y=0) and the
trajectory. An ellipse has two focuses (see Figure 14.7): each focus
is located on the x-axis, a distance (e a) away from the center of the ellipse.
The parameter e is called the
eccentricity of the ellipse and
is equal to
We see that for a circle the eccentricity is equal to zero, and the
semi-major axis is equal to the radius of the circle. The shortest distance
between the focus and the ellipse is called the
perihelion
distance R
p. It is easy to see that this distance is given by
Rp = a (1 - e)
Figure 14.7. The ellipse.
The largest distance between the focus and the ellipse is called the
aphelion distance R
a which is given by
Ra = a (1 + e)
Figure 14.8. Trajectory of planet around sun.
The planets move about the sun in an elliptic path with the focus at
the position of the sun (see Figure 14.8). The elliptical shape of the
trajectory of the planet is a result of the 1/r
2 nature of the
gravitational force and the initial conditions. Under certain conditions the
trajectory will be hyperbolic and the planet will approach the sun only once in
its lifetime. Examples of hyperbolic trajectories are the trajectories of
satellites that use the gravitational fields of the planets to change
direction. The law of periods, previously derived for the special case of
circular orbits, also holds for elliptical orbits, provided we replace r by a,
the semi-major axis of the ellipse.
Sample Problem 14-8
Comet Halley has a period of 76 years and, in 1986, has a distance of closest
approach to the sun of 8.9 x 10
10 m. (a) What is the aphelion
distance ? (b) What is the eccentricity of the orbit of Comet Halley ?
The semi-major axis of the orbit of Comet Halley can be found using the law of
periods:
where
- M is the mass of the sun (= 1.99 x 1030 kg)
- G is the gravitational constant (= 6.67 x 10-11 Nm2/kg2)
- T is the period (= 2.4 x 109 s).
Substituting these numbers we obtain a = 2.7 x 10
12 m. The
perihelion distance R
p is related to the semi-major axis a and the
eccentricity e:
Rp = a (1 - e)
This equation shows that the eccentricity of the orbit can be
calculated easily:
The aphelion distance can now be calculated
14.6. The Law of Areas
The trajectory of a planet about the sun is described by an ellipse
with the sun in one of its focuses. Figure 14.9 shows the position of the
planet at two instances (t and t + [Delta]t). The shaded wedge shows the area
swept out in the time [Delta]t. The area, [Delta]A, is approximately one-half
of its base, [Delta]w, times its height r. The width of the wedge is related
to r and [Delta][theta]:
Figure 14.9. Area swept out by planet during a time [Delta]t.
[Delta]w = r [Delta][theta]
We conclude that the area [Delta]A is given by
If the time interval [Delta]t approaches zero, the expression for
[Delta]A becomes more exact. The instantaneous rate at which the area is being
swept out is
The rate at which the area is being swept out depends on the velocity
of the planet and is also related to its angular momentum L. Figure 14.10
shows how to calculate the angular momentum of the planet. The angular
momentum of the planet can be calculated as follows
Figure 14.10. Angular momentum of planet.
Substituting this in the expression obtained for dA/dt we conclude
that
Since no external torques are acting on the sun-planet system, the
angular momentum of the system is constant. This immediately indicates that
dA/dt also remains constant. We conclude that
" A line joining the planet to the sun sweeps out equal areas in equal time
"
This shows that the velocity of the planet will be highest when the distance
between the sun and planet is smallest. The slowest velocity of the planet
will occur when the distance between the sun and the planet is largest.
14.7. Orbits and Energy
Suppose a satellite of mass m is in orbit around the earth (mass M). The
radius of the orbit is given to be r. The kinetic and potential energy of the
satellite can be easily expressed in terms of r. The potential energy of the
satellite is given by
The velocity of the satellite can be found by requiring that the
magnitude of the gravitational force is equal to the centripetal force:
The kinetic energy can therefore given by
The total mechanical energy can now be calculated
