Question

With what velocity should a particle be projected so that its height becomes equal to radius of earth?
A. ${\left( {\dfrac{{GM}}{R}} \right)^{1/2}}$
B. ${\left( {\dfrac{{8GM}}{R}} \right)^{1/2}}$
C. ${\left( {\dfrac{{2GM}}{R}} \right)^{1/2}}$
D. ${\left( {\dfrac{{4GM}}{R}} \right)^{1/2}}$

Answer 1

Use the formula for kinetic energy of an object and potential energy of the object on the surface of the Earth. Use the law of conservation of energy and apply this law of conservation of energy when the particle is on the surface of the Earth and when it is at a height equal to radius of the Earth.

Formulae used:
The kinetic energy $K$ of an object is
$K = \dfrac{1}{2}m{v^2}$ …… (1)
Here, $m$ is the mass of the object and $v$ is the velocity of the object.
The potential energy $U$ of an object on the surface of the Earth is
$U = - \dfrac{{GMm}}{R}$ ….. (2)
Here, $G$ is a universal gravitational constant, $M$ is mass of the Earth, $m$ is mass of the object and $R$ is radius of the Earth.

Complete step by step answer:
Let $m$ be the mass of the particle and $v$ be the velocity of the particle on the surface of the Earth. The kinetic energy ${K_i}$ of the particle on the surface of Earth is
${K_i} = \dfrac{1}{2}m{v^2}$
The potential energy ${U_i}$ of the particle on the surface of Earth is
${U_i} = - \dfrac{{GMm}}{R}$
The kinetic energy ${K_f}$ of the particle at a height equal to radius of the Earth from the surface of Earth is zero.
${K_f} = 0$
The potential energy ${U_f}$ of the particle on the surface of Earth is
${U_f} = - \dfrac{{GMm}}{{R + h}}$
Substitute $R$ for $h$ in the above equation.
${U_f} = - \dfrac{{GMm}}{{R + R}}$
$\Rightarrow {U_f} = - \dfrac{{GMm}}{{2R}}$

According to law of conservation of energy, the sum of kinetic energy ${K_i}$ and potential energy ${U_i}$ of the particle on the surface of the Earth is equal to the sum of the kinetic energy ${K_f}$ and potential energy ${U_f}$ at the height equal to radius of the Earth.
${K_i} + {U_i} = {K_f} + {U_f}$

Substitute $\dfrac{1}{2}m{v^2}$ for ${K_i}$, $- \dfrac{{GMm}}{R}$ for ${U_i}$, $0$ for ${K_f}$ and $- \dfrac{{GMm}}{{2R}}$ for ${U_f}$ in the above equation.
$\dfrac{1}{2}m{v^2} - \dfrac{{GMm}}{R} = 0 - \dfrac{{GMm}}{{2R}}$
$\Rightarrow \dfrac{1}{2}{v^2} = - \dfrac{{GM}}{{2R}} + \dfrac{{GM}}{R}$
$\Rightarrow {v^2} = - \dfrac{{GM}}{R} + \dfrac{{2GM}}{R}$
$\therefore v = \sqrt {\dfrac{{GM}}{R}}$
Therefore, the particle should be projected with velocity ${\left( {\dfrac{{GM}}{R}} \right)^{1/2}}$.

Hence, the correct option is A.

Note: The students should be careful while determining the values of various energies of the particle on the surface of the Earth because if these values are not taken correctly, the final answer for the velocity of the particle will also be incorrect. Also, the students should not forget to take the kinetic energy of the particle at given height as zero and to add the radius of the Earth in the denominator for the potential energy of the particle at given height.