# 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**

_{ }

_{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

The gravitational forces between two particles act along the line
joining them, and form an action-reaction pair (see Figure 14.1).^{-11}N m^{2}/kg^{2}
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

_{ }

_{ }

Figure 14.2. Shell theorem.

and its mass m is equal to_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

_{ }

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.

_{ }

_{ }

_{ }

_{ }

^{-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:**

_{ }

_{e}, the gravitational force is given by

_{ }

_{ }

^{2}and R

_{e}= 6.37 x 10

^{6}m gives

_{ }

^{24}kg. In reality, the situation is more complicated:

**The earth's crust is not uniform**. Precise measurements of the variations of the free-fall acceleration give information about non-uniformaties in the density of the earth. This can suggest the presence of salt domes (which often indicated the presence of oil).**The earth is not a sphere**. The earth is an ellipsoid. It is flattened on the poles and bulging at the equator (difference in radius is 21 km). The free-fall acceleration is larger at the poles than it is at the equator.**The earth is rotating**. The centripetal acceleration will change the free-fall acceleration. To illustrate the effect of the rotation of the earth on the gravitational acceleration, consider a mass m located on a scale at the equator (see Figure 14.4). The mass m will carry out a uniform circular motion with a period T equal to 24 hours. The radius of the circle is equal to the radius of the earth R_{e}. The corresponding centripetal acceleration is given by

_{ }

1. The gravitational force m g_{0}(downwards)

2. The force W exerted by the scale on the mass (upwards)

The net force acting on the mass must be equal to the centripetal force required for the circular motion:

_{ }

Figure 14.4. Mass located at equator.The effective free-fall acceleration, obtained from the measured weight W, is given by

_{ }

g_{0}- g = 0.034 m/s^{2}

## 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

_{ }

_{ }

_{ }

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 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

_{ }

_{ }

_{ }

_{ }

_{ }

**escape speed**. For the earth we obtain

v

_{crit}= 1.1 x 10^{4}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**

_{ }

_{C}:

_{ }

_{ }

_{ }

**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)

_{ }

**eccentricity**of the ellipse and is equal to

_{ }

**perihelion distance**R

_{p}. It is easy to see that this distance is given by

R

_{p}= 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

R

_{a}= 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:

_{ }

- M is the mass of the sun (= 1.99 x 10
^{30}kg) - G is the gravitational constant (= 6.67 x 10
^{-11}Nm^{2}/kg^{2}) - T is the period (= 2.4 x 10
^{9}s).

^{12}m. The perihelion distance R

_{p}is related to the semi-major axis a and the eccentricity e:

R

This equation shows that the eccentricity of the orbit can be
calculated easily:_{p}= a (1 - e)_{ }

_{ }

## 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_{ }

_{ }

Figure 14.10. Angular momentum of planet.

_{ }

_{ }

**" A line joining the planet to the sun sweeps out equal areas in equal time "**

## 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**

_{ }

_{ }

_{ }

_{ }

## Comments

## Post a Comment