Question

The kinetic energy of a satellite in its orbit around the earth is $E$ . What should be the minimum kinetic energy of the satellite so as to enable it to escape from the gravitational pull of the earth?
(A) $4E$
(B) $2E$
(C) $\sqrt 2 E$
(D) $E$

Answer 1

Hint: The satellite is launched with the help of the rockets. It is set like that it has to move upward with the velocity called escape velocity to overcome the earth’s gravitational force. After that it is not necessary to provide the force to move against the earth’s gravitational force.
Useful formula:
(1) The formula of the kinetic energy is given by
$KE = \dfrac{1}{2}m{v^2}$
Where $KE$ is the kinetic energy of a satellite, $m$ is the mass of the satellite and $v$ is the velocity of the satellite.
(2) The formula of the escape velocity is given by
${v_e} = \sqrt 2 {v_0}$
Where ${v_e}$ is the escape velocity of the satellite and ${v_0}$ is the orbital velocity.

Complete step by step solution:
It is given that the energy of the satellite is $E$ .
Let us consider the formula of the kinetic energy of the satellite,
$KE = \dfrac{1}{2}m{v^2}$
Substituting the orbital velocity in it, we get
$KE = \dfrac{1}{2}m{v_0}^2$ ……………….(1)
Substituting the escaping velocity in the formula, we get
$KE = \dfrac{1}{2}m{v_e}^2$
Substitute the formula of the escape velocity in the above step,
$
  KE = \dfrac{1}{2}m{\left( {\sqrt 2 {v_0}} \right)^2} \\
  KE = \dfrac{1}{2}m{v_0}^2 \times 2 \\
 $
Substituting the equation (1) in the above step, we get
$KE = 2E$
Hence the kinetic energy of the satellite to escape from the earth’s gravitational force is $2E$ .
Thus the option (B) is correct.

Note: Most probably, on the earth’s surface, the velocity needed for the escape of the earth’s gravitational force is $11.2\,km{s^{ - 1}}$. The velocity less than this cannot make the escape. This speed is $33$ times greater than that of the speed of the sound.