question_answer

The escape velocity for a satellite is\[\mathbf{11}.\mathbf{2}\text{ }\mathbf{km}{{\mathbf{s}}^{-1}}\]. If the satellite is launched at an angle of \[\mathbf{60}{}^\circ \] with the vertical, what will be the escape velocity?

Answer:

                \[11.2km{{s}^{-1}}\], because escape velocity \[({{\upsilon }_{e}}=\sqrt{2GM/R})\] does not depend on the angle of projection.