Equation (15.3) gives us
\frac{\Delta T}{T}=\frac{g H}{c^{2}}
The value of H for the clock on the Earth’s surface is r_{e}, and the value for the flying airplane is r_{e}+A, where A is the altitude of the airplane, about 33,000 feet (10,000 m). The value of H is the difference in the height of the two clocks in Earth’s gravitational field g, and H = A. We neglect the change of g over this altitude, and \Delta T / T becomes
\frac{\Delta T}{T}=\frac{\left(9.8 m / s ^{2}\right)(10,000 m )}{\left(3 \times 10^{8} m / s \right)^{2}}=1.09 \times 10^{-12}
The eastward and westward airplane trips took about T = 45 hours flying time. The difference in the two clocks due to the gravitational redshift is
\begin{aligned}\Delta T &=\left(1.09 \times 10^{-12}\right)(45 h )\left(\frac{3600 s }{1 h }\right) \\&=177 \times 10^{-9} s =177 ns\end{aligned}
A clock fixed on Earth will measure a flight time T_{0} of
T_{0}=\frac{4 \times 10^{7} m }{300 m / s }=1.33 \times 10^{5} s
Because a clock in the airplane will run slowly, an observer on Earth will say the time measured on the airplane is T=T_{0} \sqrt{1-\beta^{2}} \text { where } \beta=v / c=(300 m / s ) /\left(3 \times 10^{8} m / s \right)=10^{-6}. The time difference is
\Delta T=T_{0}-T=T_{0}\left(1-\sqrt{1-\beta^{2}}\right)
Because \beta is so small, we can use a power series expansion of the square root and ignore all terms smaller than \beta^{2}. The special relativity effect is
\begin{aligned}\Delta T &=T_{0}\left[1-\left(1-\beta^{2} / 2+\cdots\right)\right]=\frac{\beta^{2} T_{0}}{2} \\&=\frac{1}{2}\left(10^{-6}\right)^{2}\left(1.33 \times 10^{5} s \right)=6.65 \times 10^{-8} s =66.5 ns\end{aligned}
The gravitational time dilation effect of 177 ns is larger than the approximate 66.5-ns special relativity time dilation effect. There is also a third correction due to the rotation of Earth.
