There is one small problem with the admonition that you must use Newton’s laws in an inertial reference frame: A reference frame attached to the ground isn’t truly inertial because of the earth’s rotation. Fortunately, we can make a simple correction that allows us to continue using Newton’s laws on the earth’s surface.

FIGURE 8.13 shows an object being weighed by a spring scale on the earth’s equator. An observer hovering above the north pole sees two forces on the object: the gravitational force $\vec{F}{}_{M\text{on }m}$, given by Newton’s law of gravity, and the outward spring force $\vec{F}{}_{\text{Sp}}$. The object moves in a circle as the earth rotates, so Newton’s second law is

where $\omega$ is the angular speed of the rotating earth. The spring-scale reading $F_{\text{Sp}}=F_{M\text{on }m}-m{\omega }^{2}R$ is less than it would be on a nonrotating earth.

FIGURE 8.13 The earth’s rotation affects the measured value of g.

The blow-up in Figure 8.13 shows how we see things in a noninertial, flat-earth reference frame. For us the object is at rest, in equilibrium, hence the upward spring force must be exactly balanced by a downward gravitational force $\vec{F}{}_{\text{G}}$. Thus $F_{\text{Sp}}=F_{\text{G}}$.

Now both we and the hovering, inertial observer see the same reading on the scale. If $F_{\text{Sp}}$ is the same for both of us, then

In other words, force $\vec{F}{}_{\text{G}}$, the effective gravitational force of Chapter 6, is slightly less than the true gravitational force $\vec{F}{}_{M\text{on }m}$ because of the earth’s rotation. In essence, $m{\omega }^{2}R$ is the centrifugal force, a fictitious force trying—from our perspective in a noninertial reference frame—to “throw” us off the rotating platform. There really is no such force, but—this is the important point—we can continue to use Newton’s laws in our rotating reference frame if we pretend there is.

Because $F_{\text{G}}=mg$ for an object at rest, the effect of the centrifugal term in Equation 8.14 is to make g a little smaller than it would be on a nonrotating earth:

We calculated $g_{\text{earth}}=9.82\text{m}\text{/}{\text{s}}^2$ in Chapter 6. Using $\omega =1\text{rev}\text{/}\text{day}$ (which must be converted to SI units) and $R=6370\text{km,}$ we find ${\omega }^{2}R=0.033{\text{m/s}}^2$ at the equator and $\approx 0.02\text{m}\text{/}{\text{s}}^2$ at midlatitudes where the distance from the rotation axis is smaller. Thus the free-fall acceleration—what we actually measure in our rotating reference frame—is about $9.80{\text{m/s}}^2$, exactly what we measure in the laboratory.

Things are a little more complicated at other latitudes, but the bottom line is that we can safely use Newton’s laws in our rotating, noninertial reference frame on the earth’s surface if we calculate the gravitational force—as we’ve been doing—as $F_{\text{G}}=mg$ with g the measured free-fall value, a value that compensates for our rotation, rather than the purely gravitational $g_{\text{earth}}$.