Question

If a rocket is fired with a speed of $v = 2\sqrt {gR} $ near the earth’s surface and coasts upwards, its speed in the interstellar space is.
A. $4\sqrt {gR} $
B. $\sqrt {2gR} $
C. $\sqrt {gR} $
D. $\sqrt {4gR} $

Answer 1

Hint: The gravitational potential energy in the interstellar space is assumed to be zero. But the rocket will have some kinetic energy by virtue of its motion. Also, the energy of the system will be conserved. Use energy conservation and find the required speed.

Complete step by step answer:
Lets understand the energies the rocket will have in both the cases:
First when the rocket is near the Earth’s surface, the rocket will have gravitational potential energy as well as kinetic energy.
In the second case, when the rocket is in the interstellar space. It will only have kinetic energy as the rocket is in motion and moving with certain velocity.
Now, as energy of the system must be conserved. Hence, the energies in first case and second case must be equal. Therefore, we will have:
$\dfrac{1}{2}m{v^2} + ( - \dfrac{{GMm}}{R}) = \dfrac{1}{2}m{v_i}^2 + 0$
It must be noted that the gravitational potential energy is taken as negative, as the body is taken in the opposite direction to that of the gravitational force.
Where $m$ is the mass of the rocket
$v$ is the speed of the rocket when it is fired
$G$ is the universal gravitational constant
$M$ is the mass of the Earth
$R$ is the radius of the Earth
${v_i}$ is the speed of a rocket in interstellar space.
Substituting the value of the initial speed of rocket and solving further we will have:
$\dfrac{1}{2}m{(2\sqrt {gR} )^2} - \dfrac{{GMm}}{R} = \dfrac{1}{2}m{v_i}^2$
$\Rightarrow \dfrac{1}{2}m(4gR) - \dfrac{{(g{R^2})m}}{R} = \dfrac{1}{2}m{v_i}^2$ as $GM = g{R^2}$
$ \therefore {v_i} = \sqrt {2gR} $

So, the correct answer is “Option B”.

Note:
In these types of questions, remember that energy is always conserved. So, try to solve using energy conservation. Do remember that the gravitational potential energy in the interstellar space is taken as zero though practically it is not zero but for calculation it is taken as zero. Also, remember that the gravitational potential energy is taken as negative.