第13章万有引力

Equation 13.2 is a scalar equation, giving the magnitude of the gravitational acceleration as a function of the distance from the center of the mass that causes the acceleration. But we could have retained the vector form for the force of gravity in Equation 13.1, and written the acceleration in vector form as

We identify the vector field represented by $\overrightarrow{g}$ as the gravitational field caused by mass $M$. We can picture the field as shown Figure 13.9. The lines are directed radially inward and are symmetrically distributed about the mass.

The direction of $\overrightarrow{g}$ is parallel to the field lines at any point. The strength of $\overrightarrow{g}$ at any point is inversely proportional to the line spacing. Another way to state this is that the magnitude of the field in any region is proportional to the number of lines that pass through a unit surface area, effectively a density of lines. Since the lines are equally spaced in all directions, the number of lines per unit surface area at a distance r from the mass is the total number of lines divided by the surface area of a sphere of radius r, which is proportional to $r^2$. Hence, this picture perfectly represents the inverse square law, in addition to indicating the direction of the field. In the field picture, we say that a mass m interacts with the gravitational field of mass M. We will use the concept of fields to great advantage in the later chapters on electromagnetism.

As we saw in Applications of Newton’s Laws, objects moving at constant speed in a circle have a centripetal acceleration directed toward the center of the circle, which means that there must be a net force directed toward the center of that circle. Since all objects on the surface of Earth move through a circle every 24 hours, there must be a net centripetal force on each object directed toward the center of that circle.

Let’s first consider an object of mass m located at the equator, suspended from a scale (Figure 13.10). The scale exerts an upward force ${\overrightarrow{F}}_{\text{SE}}$ away from Earth’s center. This is the reading on the scale, and hence it is the apparent weight of the object. The weight (mg) points toward Earth’s center. If Earth were not rotating, the acceleration would be zero and, consequently, the net force would be zero, resulting in $F_{\text{s}}=mg$. This would be the true reading of the weight.

With rotation, the sum of these forces must provide the centripetal acceleration, $a_{\text{c}}$. Using Newton’s second law, we have

Note that $a_{\text{c}}$ points in the same direction as the weight; hence, it is negative. The tangential speed v is the speed at the equator and r is $R_{\text{E}}$. We can calculate the speed simply by noting that objects on the equator travel the circumference of Earth in 24 hours. Instead, let’s use the alternative expression for $a_{\text{c}}$ from Motion in Two and Three Dimensions. Recall that the tangential speed is related to the angular speed $\left(\omega \right)$ by $v=r\omega$. Hence, we have $a_c=\text{−}r{\omega }^2$. By rearranging Equation 13.3 and substituting $r=R_{\text{E}}$, the apparent weight at the equator is

The angular speed of Earth everywhere is

Substituting for the values or $R_{\text{E}}$ and $\omega$, we have $R_{\text{E}}{\omega }^2=0.0337{\text{m/s}}^2$. This is only 0.34% of the value of gravity, so it is clearly a small correction.