Question

A particle is fired vertically upwards from the surface of Earth and reaches a height $6400\,{\text{km}}$. The initial velocity of the particle, if $R = 6400\,{\text{km}}$ and of at the surface of Earth is $10\,{\text{m/}}{{\text{s}}^2}$.
A. $8\,{\text{km/s}}$
B. $4\,{\text{km/s}}$
C. $11.2\,{\text{km/s}}$
D. $2\,{\text{km/s}}$

Answer 1

Use the law of conservation of energy for the particle at the surface of the Earth and at the height h from the surface of the Earth. Hence, the sum of kinetic and potential energies at the surface of the Earth and at height h is the same. The kinetic energy of the particle at height h is zero.

Formulae used:
The kinetic energy $K$ of an object is
$K = \dfrac{1}{2}m{v^2}$
Here, $m$ is the mass of the object and $v$ is the velocity of the object.
The potential energy $U$ of an object at the surface of the Earth is
$U = - \dfrac{{GMm}}{R}$
Here, $G$ is the 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:
We have given that the height of the particle from the surface of the Earth is $6400\,{\text{km}}$.
$h = 6400\,{\text{km}}$
The radius of the Earth is $6400\,{\text{km}}$.
$R = 6400\,{\text{km}}$

According to the 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 kinetic energy ${K_f}$ and potential energy ${U_f}$ of the particle at height h.
${K_i} + {U_i} = {K_f} + {U_f}$
The kinetic energy of the particle at height h is zero.

Substitute $\dfrac{1}{2}m{v^2}$ for ${K_i}$, $- \dfrac{{GMm}}{R}$ for ${U_i}$, 0 for ${K_f}$ and $- \dfrac{{GMm}}{{R + h}}$ for ${U_f}$ in the above equation.
$\dfrac{1}{2}m{v^2} - \dfrac{{GMm}}{R} = 0 - \dfrac{{GMm}}{{R + h}}$
$\Rightarrow \dfrac{1}{2}{v^2} - \dfrac{{GM}}{R} = - \dfrac{{GM}}{{R + h}}$
$\Rightarrow v = \sqrt {2\left[ {\dfrac{{GM}}{R} - \dfrac{{GM}}{{R + h}}} \right]}$
$\Rightarrow v = \sqrt {2\left[ {GM\left( {\dfrac{{R + h - R}}{{R\left( {R + h} \right)}}} \right)} \right]}$
$\Rightarrow v = \sqrt {2GM\left( {\dfrac{h}{{R\left( {R + h} \right)}}} \right)}$
Substitute $6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^{\text{2}}}$ for $G$, $6 \times {10^{24}}\,{\text{kg}}$ for $M$, $6400\,{\text{km}}$ for $h$ and $6400\,{\text{km}}$ for $R$ in the above equation.
$\Rightarrow v = \sqrt {2\left( {6.67 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^{\text{2}}}} \right)\left( {6 \times {{10}^{24}}\,{\text{kg}}} \right)\left( {\dfrac{{6400\,{\text{km}}}}{{\left( {6400\,{\text{km}}} \right)\left( {6400\,{\text{km}} + 6400\,{\text{km}}} \right)}}} \right)}$
$\Rightarrow v = \sqrt {2\left( {6.67 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^{\text{2}}}} \right)\left( {6 \times {{10}^{24}}\,{\text{kg}}} \right)\left( {\dfrac{1}{{12800\,{\text{km}}}}} \right)}$
$\Rightarrow v = \sqrt {2\left( {6.67 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}/{\text{k}}{{\text{g}}^{\text{2}}}} \right)\left( {6 \times {{10}^{24}}\,{\text{kg}}} \right)\left( {\dfrac{1}{{12.8 \times {{10}^6}\,{\text{m}}}}} \right)}$
$\Rightarrow v = 7907\,{\text{m/s}}$
$\therefore v \approx 8\,{\text{km/s}}$

Therefore, the initial speed of the particle is $8\,{\text{km/s}}$. Hence, the correct option is A.

Note: The students should not forget to convert the units of radius of the Earth and the height of the particle from the surface of the Earth in the SI system of units. One may directly conclude that this initial velocity is $11.2\,{\text{km/s}}$. But $11.2\,{\text{km/s}}$ is the escape velocity of the particle required to escape the particle from the influence of Earth’s gravitation which is not asked in the present question.