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\(_E\) = \...
Question 1 Report
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\(^{-2}\)
Answer Details
Escape velocity. A body just escapes the earth gravitational field when its total mechanical energy is zero, i.e. its initial kinetic energy equals the work needed to move it from the earth surface to infinity (the gravitational potential energy magnitude).
\[\tfrac{1}{2}mV_E^{2}=\dfrac{GMm}{R}.\]
At the earth surface the acceleration due to gravity is \(g=\dfrac{GM}{R^{2}}\), so \(GM=gR^{2}\). Substituting:
\[\tfrac{1}{2}mV_E^{2}=\dfrac{gR^{2}m}{R}=mgR\Rightarrow V_E^{2}=2gR.\]
Therefore \(V_E=\sqrt{2gR}\). With \(g=10\,\text{m s}^{-2}\):
\[V_E=\sqrt{2\times10\times R}=\sqrt{20R}.\]
Hence \(V_E=\sqrt{20R}\) as required.
Download The App On Google Playstore
Everything you need to excel in your exams
