Equation 13.7 gives $g_{\text{surface}}$ at the surface of a planet. More generally, imagine an object of mass m at distance $r>R$ from the center of a planet. Further, suppose that gravity from the planet is the only force acting on the object. Then its acceleration, the free-fall acceleration, is given by Newton’s second law:

This more general result agrees with Equation 13.7 if $r=R$, but it allows us to determine the “local” free-fall acceleration at distances $r>R$. Equation 13.8 expresses Newton’s discovery, with regard to the moon, that g decreases inversely with the square of the distance.

FIGURE 13.7 shows a satellite orbiting at height h above the earth’s surface. Its distance from the center of the earth is $r=R_{\text{e}}+h$. Most people have a mental image that satellites orbit “far” from the earth, but in reality h is typically $200\text{miles}\approx$ $3\times {10}^5\text{m,}$ while $R_{\text{e}}=6.37\times {10}^6\text{m}$. Thus the satellite is barely “skimming” the earth at a height only about 5% of the earth’s radius!

FIGURE 13.7 A satellite orbits the earth at height h.

The value of g at height h above the earth (i.e., above sea level) is

where $g_{\text{earth}}=9.83\text{m}\text{/}{\text{s}}^2$ is the value calculated from Equation 13.7 for $h=0$ on a nonrotating earth. TABLE 13.1 shows the value of g evaluated at several values of h.