Q 11-07-031NEETJEE MainEasy
The escape velocity from the Earth is about $11.2$ km/s. The orbital speed of a satellite very close to the Earth's surface is about
Answer: (C) $7.9$ km/s
Near the surface, $v_o = \sqrt{gR}$ and $v_e = \sqrt{2gR}$, so $v_o = \dfrac{v_e}{\sqrt{2}} = \dfrac{11.2}{1.414} \approx 7.9$ km/s.
Solution by Sreeraj P, M.Sc Physics