An astronaut whose height $h$ is $1.70 \mathrm{~m}$ floats "feet down" in an orbiting space shuttle at distance $r=6.77 \times 10^6 \mathrm{~m}$ away from the center of Earth. What is the difference between the gravitational acceleration at her feet and at her head?
We can approximate Earth as a uniform sphere of mass $M_E$. Then, from Eq. 13-11, the gravitational acceleration at any distance $r$ from the center of Earth is
$$
a_g=\frac{G M_E}{r^2}
$$
We might simply apply this equation twice, first with $r=$ $6.77 \times 10^6 \mathrm{~m}$ for the location of the feet and then with $r=6.77 \times 10^6 \mathrm{~m}+1.70 \mathrm{~m}$ for the location of the head. However, a calculator may give us the same value for $a_g$ twice, and thus a difference of zero, because $h$ is so much smaller than $r$. Here's a more promising approach: Because we have a differential change $d r$ in $r$ between the astronaut's feet and head, we should differentiate Eq. 13-15 with respect to $r$.
Calculations: The differentiation gives us
$$
d a_g=-2 \frac{G M_E}{r^3} d r
$$
where $d a_g$ is the differential change in the gravitational acceleration due to the differential change $d r$ in $r$. For the astronaut, $d r=h$ and $r=6.77 \times 10^6 \mathrm{~m}$. Substituting data into Eq.13-16, we find
$$
\begin{aligned}
d a_g & =-2 \frac{\left(6.67 \times 10^{-11} \mathrm{~m}^3 / \mathrm{kg} \cdot \mathrm{s}^2\right)\left(5.98 \times 10^{24} \mathrm{~kg}\right)}{\left(6.77 \times 10^6 \mathrm{~m}\right)^3}(1.70 \mathrm{~m}) \\
& =-4.37 \times 10^{-6} \mathrm{~m} / \mathrm{s}^2, \quad \text { (Answer) }
\end{aligned}
$$