Question

A rocket is launched with a velocity of $10km.{{s}^{-1}}$. If the radius of the earth is $R$, then the maximum height attained by it will be
$A)\text{ }2R$
$B)\text{ 3}R$
$C)\text{ 4}R$
$D)\text{ 5}R$

Answer 1

This problem can be solved by realizing that the total mechanical energy of the rocket remains conserved. Hence, the sum of the kinetic and potential energy of the rocket on the surface of the earth will be equal to the sum of the kinetic and potential energy of the rocket at its maximum height.

Formula used:
$KE=\dfrac{1}{2}m{{v}^{2}}$
$PE=-\dfrac{GMm}{\left( R+h \right)}$

Complete answer:
The mechanical energy of the rocket will remain conserved throughout its journey. Therefore, the sum of the kinetic and potential energy of the rocket on the surface of the earth will be equal to the sum of the kinetic and potential energy of the rocket at its maximum height.
Now, let us write the formula for the gravitational potential energy.
The gravitational potential energy $PE$ of a body of mass $m$ at a height $h$ above the surface of a planet of mass $M$ and radius $R$ is given by
$PE=-\dfrac{GMm}{\left( R+h \right)}$ --(1)
The kinetic energy $KE$ of a body of mass $m$ moving at a speed $v$ is given by
$KE=\dfrac{1}{2}m{{v}^{2}}$ --(2)
Now, let us analyze the question.
Let the mass of the rocket be $m$ and the mass of earth be $M$.
Let the radius of earth be $R$.
Let the maximum height attained be $h$.
The initial speed of the rocket is $u=10km.{{s}^{-1}}={{10}^{4}}m.{{s}^{-1}}$ $\left( \because 1km={{10}^{3}}m \right)$

Let the initial kinetic energy of the rocket at the surface of the earth be $K{{E}_{i}}$ and the initial potential energy be $P{{E}_{i}}$.
Let the final kinetic energy of the rocket at the maximum height be $K{{E}_{f}}$ and the final potential energy be $P{{E}_{f}}$.
From (1), we get
$P{{E}_{i}}=-\dfrac{GMm}{\left( R+0 \right)}=-\dfrac{GMm}{R}$
$P{{E}_{f}}=-\dfrac{GMm}{\left( R+h \right)}$
From (2), we get
$K{{E}_{i}}=\dfrac{1}{2}m{{u}^{2}}$

Now, for maximum height to be gained, all the initial kinetic energy must be converted to increase the potential energy. Therefore, at the maximum height
$K{{E}_{f}}=0$
Now, since, the mechanical energy will remain conserved

$P{{E}_{i}}+K{{E}_{i}}=P{{E}_{f}}+K{{E}_{f}}$
$\Rightarrow -\dfrac{GMm}{R}+\dfrac{1}{2}m{{u}^{2}}=-\dfrac{GMm}{\left( R+h \right)}+0=-\dfrac{GMm}{\left( R+h \right)}$
$\Rightarrow -\dfrac{GM}{{{R}^{2}}}R+\dfrac{1}{2}{{u}^{2}}=-\dfrac{GM}{\left( R+h \right){{R}^{2}}}{{R}^{2}}$

Putting $R+h=x$ , we get

$\Rightarrow \dfrac{1}{2}{{u}^{2}}=-\dfrac{GM{{R}^{2}}}{\left( R+h \right){{R}^{2}}}+\dfrac{GM}{{{R}^{2}}}R=-\dfrac{g{{R}^{2}}}{R+h}+gR$ $\dfrac{GM}{{{R}^{2}}}=g=9.8m/{{s}^{2}}$
$\Rightarrow \dfrac{1}{2}{{\left( {{10}^{4}} \right)}^{2}}=$ $gR-\dfrac{g{{R}^{2}}}{x}$
$\dfrac{1}{2}\times {{10}^{8}}=9.8\times 6.4\times {{10}^{6}}-\dfrac{9.8\times {{\left( 6.4\times {{10}^{6}} \right)}^{2}}}{x}$ $\left( \because R=6.4\times {{10}^{6}}m \right)$
$\dfrac{1}{2}\times {{10}^{8}}=62.72\times {{10}^{6}}-\dfrac{401.408\times {{10}^{12}}}{x}$
$\Rightarrow \dfrac{1}{2}\times 100=62.72-\dfrac{401.408\times {{10}^{6}}}{x}$
$\Rightarrow \dfrac{401.48\times {{10}^{6}}}{x}=62.72-\left( \dfrac{1}{2}\times 100 \right)=62.72-50=12.72$

$\Rightarrow x=\dfrac{401.408\times {{10}^{6}}}{12.72}=31.56\times {{10}^{6}}\approx 32\times {{10}^{6}}m=5R$
$\Rightarrow R+h=5R$
$\Rightarrow h=5R-R=4R$
Therefore, the maximum height attained by the rocket will be $4R$.

So, the correct answer is “Option C”.

Note:
Students should note that the final kinetic energy is zero in this case only because we were required to find out the maximum height of the rocket. Generally, the final kinetic energy of a rocket is not zero at it would defy the purpose of the rocket. A similar approach is also used to calculate the escape velocity of a planet. In this case both the final kinetic energy and potential energy of the body will be zero.