Chapter 12 - Kinematics of a Particle - Section 12.7 - Curvilinear Motion: Normal and Tangential Components - Problems - Page 66: 119

$h=5.99$ Mn

$v=20$ Mm/h $=\frac{20(10^6)}{3600}=5.56(10^3)$ m/s Since $a_t=\frac{dv}{dt}=0$, then, $a=a_n=2.5=\frac{v^2}{\rho}$ $\rho=\frac{(5.56(10^3))^2}{2.5}=12.35(10^6)$ m The radius of the earth is $\frac{12713(10^3)}{2}=6.36(10^6)$ m Hence $h=12.35(10^6)-6.36(10^6)=5.99(10^6)$m $=5.99$ Mm