Calculate the orbital and escape velocity from earth, mass of earth is \(5.9722 \times {10^{24}}\,kg\) and radius is \(6.378 \times {10^6}\,m\).
Calculate the orbital and escape velocity from earth, mass of earth is \(5.9722 \times {10^{24}}\,kg\) and radius is \(6.378 \times {10^6}\,m\).
Orbital velocity: It is the velocity acquired by an object to orbit around any celestial body. The orbital path is a balance between gravity of the celestial body and inertia to move in a straight line of the object.
Escape velocity: It is the velocity of an object to overcome or escape the gravitational force of a planet and enter into orbit. Orbital velocity is given as
\[{V_{orbital}} = \sqrt {\frac{{G{M_{earth}}}}{r}} \]
Here,
\[G{M_{earth}} = \left( {6.67 \times {{10}^{ – 11}}} \right)\left( {5.9722 \times {{10}^{24}}} \right) = 3.98 \times {10^{14}}\,{m^3}/{s^2}\]
Therefore,
\[{V_{orbital}} = \sqrt {\frac{{3.98 \times {{10}^{14}}}}{{6.378 \times {{10}^6}}}} = 7899.494\,m/s = 7.9\,km/s\]
Escape velocity is \(\sqrt 2 \) orbital velocity,
Therefore,
\[{V_{escape}} = \sqrt 2 {V_{orbital}} = \sqrt 2 \times 7899.494\,m/s = 11171.572\,m/s = 11.2\,km/s\]
