Question

If a rocket is fired with a velocity, $V=2\sqrt{gR}$ near the earth’s surface and goes upwards, its speed in the inter-stellar space is

& A)4\sqrt{gR} \\\ & B)\sqrt{2gR} \\\ & C)\sqrt{gR} \\\ & D)\sqrt{4gR} \\\ \end{aligned}$$

Answer 1

We will use the law of conservation of mechanical energy to solve this question. That is the total energy on the earth's surface will be equal to total energy in the interstellar space. The total energy will be the sum of kinetic and potential energies and the potential energy in the interstellar space is zero. By equating the two total energies we will find the velocity of the rocket at interstellar space.

Formula used:

& K.E=\dfrac{1}{2}m{{v}^{2}} \\\ & P.E=-\dfrac{GMm}{R} \\\ \end{aligned}$$ **Complete step by step answer:** We know that according to the law of conservation of energy, total energy at earth surface and interstellar space will be equal. i.e. $$T.{{E}_{\text{surface}}}=T.{{E}_{\text{space}}}$$ The total energy is given as the sum of kinetic and potential energies. $${{\left( K.E+P.E \right)}_{\text{surface}}}={{\left( K.E+P.E \right)}_{\text{space}}}$$ Now, the kinetic energy of the rocket is given by $$K.E=\dfrac{1}{2}m{{v}^{2}}$$. Where, $$m$$ is the mass of the rocket and $$v$$ is the velocity of the rocket. Potential energy is given by, $$P.E=-\dfrac{GMm}{R}$$ Where, $$G$$ is the universal gravitational constant. $$M$$ is the mass of earth. $$m$$ is the mass of the rocket. $$R$$ is the radius of earth. But, we can change this expression as, $$P.E=-\dfrac{GMm}{R}=-mgR$$ Where, $$g$$ is the acceleration due to gravity. Because, $$\begin{aligned} & g=\dfrac{GM}{{{R}^{2}}} \\\ & \Rightarrow gR=\dfrac{GM}{R} \\\ \end{aligned}$$ We must know that potential energy in interstellar space is zero. Therefore, $$\begin{aligned} & {{\left( K.E+P.E \right)}_{\text{surface}}}={{\left( K.E+P.E \right)}_{\text{space}}} \\\ & \Rightarrow \dfrac{1}{2}m{{v}_{\text{surface}}}^{2}+\left( -mgR \right)=\dfrac{1}{2}m{{v}_{\text{space}}}^{2}+0 \\\ \end{aligned}$$ Now, $$V=2\sqrt{gR}$$ is the velocity of the rocket at the earth’s surface. So, velocity at the interstellar space is, $$\begin{aligned} & \dfrac{1}{2}m{{v}_{\text{surface}}}^{2}-mgR=\dfrac{1}{2}m{{v}_{\text{space}}}^{2} \\\ & \Rightarrow \dfrac{1}{2}m{{\left( 2\sqrt{gR} \right)}^{2}}-mgR=\dfrac{1}{2}m{{v}_{\text{space}}}^{2} \\\ & \Rightarrow \dfrac{1}{2}m\times 4gR-mgR=\dfrac{1}{2}m{{v}_{\text{space}}}^{2} \\\ & \Rightarrow mgR=\dfrac{1}{2}m{{v}_{\text{space}}}^{2} \\\ & \Rightarrow {{v}_{\text{space}}}^{2}=2gR \\\ & \Rightarrow {{v}_{\text{space}}}=\sqrt{2gR} \\\ \end{aligned}$$ Therefore, the velocity of the rocket at the interstellar space is $$\sqrt{2gR}$$. **So, the correct answer is “Option B”.** **Note:** Here, we took potential energy as negative on the earth surface. This is because the work is done against the gravitational field as we are taking the rocket away from the field. Also, we must know that the gravitational potential energy at interstellar space is taken as zero because the acceleration due to gravity is taken as zero.