Question

A spaceship is launched into circular orbit close to earth’s surface. The additional velocity that should be imparted to the spaceship in orbit to overcome the gravitational pull is: (Radius of earth =6400km and$g = 9 \cdot 8m{s^{ - 1}}$).
A) $9 \cdot 8km{s^{ - 1}}$.
B) $8km{s^{ - 1}}$
C) $3 \cdot 2km{s^{ - 1}}$
D) $1 \cdot 5km{s^{ - 1}}$

Answer 1

Hint:The escape velocity is the velocity of the body which if attained then the body can leave the atmosphere of the earth and enter the space. The escape velocity does not depend upon the mass of the body therefore everybody has to achieve the same velocity in order to escape to space.

Formula used:The formula of the escape velocity is given by ${v_e} = \sqrt {2{R_e}g} $ where g is acceleration of gravity and ${R_e}$ is the radius of earth. The orbital velocity of the spaceship is given by ${v_o} = \sqrt {g \cdot R} $ where $R$ is the radius of orbital of the spaceship and g is acceleration due to gravity.

Complete step by step answer:
It is given that a spaceship launched into earth’s orbit close to the earth’s surface and we need to calculate the additional velocity that should be imparted so as to overcome the gravitational pull.
The velocity of the spaceship orbiting in the space is given by,
${v_o} = \sqrt {g \cdot R} $
Where $R$ is the radius of orbital of the spaceship and g is acceleration due to gravity.
The velocity of the spaceship orbiting in the space is with radius $R = {R_e}$ is given by,
$ \Rightarrow {v_o} = \sqrt {g \cdot {R_e}} $………eq. (1)
The escape velocity of the spaceship will be,
$ \Rightarrow {v_e} = \sqrt {2{R_e}g} $………eq. (2)
Where g is acceleration of gravity and ${R_e}$ is the radius of earth.
The additional velocity of that needs to be imparted is given by,
$ \Rightarrow {v_e} - {v_o}$
Replace the value of ${v_e}$ and ${v_o}$ from equation (1) and equation (2).
$ \Rightarrow {v_e} - {v_o}$
$ \Rightarrow {v_e} - {v_o} = \sqrt {2{R_e}g} - \sqrt {g \cdot {R_e}} $
$ \Rightarrow {v_e} - {v_o} = 11 \cdot 2 - 7 \cdot 9$
$ \Rightarrow {v_e} - {v_o} = 3 \cdot 2km{s^{ - 1}}$
The additional velocity is equal to${v_e} - {v_o} = 3 \cdot 2km{s^{ - 1}}$.

The correct answer for this problem is option C.

Note:It does not matter if the body is small or the body is very big the escape velocity is independent of mass of the body which means that anybody that wants to leave the atmosphere of the earth should attain the escape velocity otherwise the body will return to the earth surface.