A satellite launched with velocity V\(_E\) just escapes the earth's gravitational attraction. Given that the radius of the earth is R, show that V\(_E\) = \(\sqrt{20R}\) [g = 10ms\(^{-2}\)
Explanation
To show that V\(_E\) = \(\sqrt{20R}\) KE = Gravitational Potential Energy \(\frac{1}{2}\)MV\(_E^2\) = \(\frac{GmM}{R}\) V\(_E^2\) = \(\frac{2GM}{R}\) but GM = gR\(^2\) V\(_E^2\) = \(\frac{2gR^2}{R}\) = 2gR V\(_E^2\) = \(\sqrt{2 \times 10 \times R}\) = \(\sqrt{20R}\) OR
