Question

An astronaut, orbiting in a spaceship round the earth, has a centripetal acceleration of $6.67m/s^2$. Find the height of the spaceship above the surface of the earth.
($G=6.67\times {10}^{11}N{m}^2/k{g}^2$, radius of the earth = 6400 km)

Answer 1

$\begin{array}{c}\Rightarrow g^{{}^{\prime }}=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}\\ \Rightarrow {\left(1+\frac{h}{R}\right)}^2=\frac{g}{g^{{}^{\prime }}}=\frac{9.8}{6.67}\\ \Rightarrow \left(1-\frac{h}{R}\right)=\sqrt{\frac{9.8}{6.67}}=1.212\\ \Rightarrow \frac{h}{R}=0.212\\ \Rightarrow h=0.212R\\ =0.212\times 6400km\\ Hence,\\ h=1357.64Km\end{array}$