Q. A spaceship is launched into a circular orbit close to earth’s surface. The additional velocity that should be imparted to the spaceship in the orbit to overcome the gravitational pull is (Radius of earth = 6400 km and $g = 9.8\ ms^{- 2}$)
A) $11.2\ kms^{- 1}$
B) $8\ kms^{- 1}$
C) $3.2\ kms^{- 1}$
D) $1.5\ kms^{- 1}$
Explanation
(c) For orbiting the earth close to its surface
$= \frac{mv^{2}}{R} = \frac{GMm}{R^{2}},ie,\ v_{0} = \sqrt{\frac{GM}{R}} = \sqrt{gR}$
$\therefore v_{0} = \sqrt{(9.8 \times 6.4 \times 10^{6})} = 8\ kms^{- 1}$
For escaping from close to the surface of earth,
$\frac{GMm}{R} = \frac{1}{2}mv_{t}^{2},v_{e} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$
$v_{e} = \sqrt{2} \times v_{0} = 1.41 \times 8kms^{- 1} = 11.2\ kms^{- 1}$
$\therefore$ the additional velocity to be imparted to the orbiting satellite for escaping is
11.2 $- 8 = 3.2\ kms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.