## Essential University Physics: Volume 1 (3rd Edition)

$2.8\times10^5m$
As we know that: $T^2=\frac{4\pi^2r^3}{GM}$ This can be rearranged as: $r=\sqrt[3] {\frac{T^2GM}{4\pi^2}}$ We plug in the known values to obtain: $r=\sqrt[3] {\frac{(112\times 60)^2(6.67\times 10^{-11})(0.642\times 10^{24})}{4\pi^2}}$ $r=3.66\times 10^6m$ We can now find the aircraft's altitude: $h = 3.66\times 10^6m-3.38\times10^6 m=2.8\times10^5m$