Question

Calculate the escape velocity of a body from the surface of the earth. [Average density of earth $=5.5\times {10}^{3}kg/m^3,G=6.67\times {10}^{-11}N{m}^2/k{g}^2,$ radius of earth $R=6.4\times {10}^{6}m$].

Answer 1

$\begin{array}{c}\rho =5.5\times {10}^{3}Kg/m^3\\ G=6.67\times {10}^{-11}N{m}^2/k{g}^2\\ R=6.4\times {10}^{6}m\\ V=volume=\frac{4}{3}\pi R^3=\frac{4}{3}\pi {\left(6.4\times {10}^6\right)}^3\\ V=\left(\frac{4}{3}\pi \times 262.144\right)\times {10}^{18}\\ Mass=\rho \times V=\left(5.5\times {10}^3\times \frac{4}{3}\pi \times 262.144\times {10}^{18}\right)\\ V_{esc}=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times {10}^{-11}\times 5.5\times {10}^3\times \frac{4}{3}\pi \times 262.144\times {10}^{18}}{6.4\times {10}^6}}\\ V_{esc}=11219.75925m/s\\ V_{esc}=11.2197Km/s\end{array}$