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