Question

(d) The escape velocity from the surface of planet of radius 347 km and g of 9 ms-is
(1) 2.4 km/sec (ii) 2.5 km/sec (iii) V24 km/sec (iv) v2.4 km/sec​

Answer 1

The escape velocity from the surface of a planet of radius 347 km and g of 9 m/s is 2.5 km/s

         Given,

                 Radius of the planet, R = 347 km

                 Acceleration due to gravity of planet, g = 9 m/s = 9 x 10^-3 km/s

                Escape velocity of a planet is given by,

                      $v_{e}=\sqrt{2gR}$

                      $v_{e}=\sqrt{2(9)(10^{-3})347}$

                      $v_{e}=3\sqrt{(694)(10^{-3})}$

                      $v_{e}=3\sqrt{0.694}$

                      $v_{e}=3(0.83)$

                      $v_{e}=2.49$

                      $v_{e}=2.5km/s$

       ∴ The correct option is (ii) 2.5 km/s