Question

A rocket is launched straight up from the surface of the earth. When its altitude is one fourth of the radius of the earth, its fuel runs out and therefore it coasts. The minimum velocity which the rocket should have when it starts to coast if it is to escape from the gravitational pull of the earth is approximately:
$A)\text{ }1km.{{s}^{-1}}$
$B)\text{ 6}km.{{s}^{-1}}$
$C)\text{ }10km.{{s}^{-1}}$
$D)\text{ }15km.{{s}^{-1}}$

Answer 1

This problem can be solved by finding out the potential and kinetic energy of the rocket when it starts to coast and at the final position when it escapes the gravitational pull of the earth. The sum of the kinetic energy and the potential energy of the rocket, that is the mechanical energy will remain conserved throughout.

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

Complete answer:
We will write the sum of the kinetic and potential energy of the rocket when it starts to coast and equate it to the sum of the potential and kinetic energy of the rocket when it escapes the gravitational pull of the earth since the sum of the energies, that is, the mechanical energy of the rocket will remain conserved throughout.
The gravitational potential energy $PE$ of a rocket of mass $m$ at a height $h$ above the surface of the earth is given by
$PE=-\dfrac{GMm}{\left( R+h \right)}$ --(1)
Where $R,M$ is the radius and mass of the earth respectively and $G=6.67\times {{10}^{-11}}N.{{m}^{2}}k{{g}^{-2}}$ is the universal gravitational constant.
The kinetic energy $KE$ of a body of mass $m$ and moving with speed $v$ is given by
$KE=\dfrac{1}{2}m{{v}^{2}}$ --(2)
Now, let us analyze the question.
The height of the rocket in the initial position (when its fuel becomes zero) is $h=\dfrac{R}{4}$, where $R$ is the radius of the earth.
Let the velocity of the rocket in the initial position be $u$.
Let the mass of the rocket be $m$.
Let the initial kinetic energy of the rocket be $K{{E}_{i}}$ and the initial potential energy be $P{{E}_{i}}$.
Let the final kinetic energy of the rocket (when it escapes the earth’s gravitational pull) be $K{{E}_{f}}$. In this case, the final potential energy will be zero as there is no gravitational pull on the rocket. Therefore,
$P{{E}_{f}}=0$ --(3)
Let the final velocity of the rocket be $v$.
Using (1), we get
$P{{E}_{i}}=-\dfrac{GMm}{\left( R+\dfrac{R}{4} \right)}=-\dfrac{4GMm}{4R+R}=-\dfrac{4GMm}{5R}$ --(4)
Using (2), we get
$K{{E}_{i}}=\dfrac{1}{2}m{{u}^{2}}$ --(5)
$K{{E}_{f}}=\dfrac{1}{2}m{{v}^{2}}$ --(6)
Now, since, the mechanical energy will remain conserved,
$K{{E}_{i}}+P{{E}_{i}}=K{{E}_{f}}+P{{E}_{f}}$
Using (3), (4), (5) and (6) in the above equation, we get
$\dfrac{1}{2}m{{u}^{2}}-\dfrac{4GMm}{5R}=\dfrac{1}{2}m{{v}^{2}}+0$
$\therefore \dfrac{1}{2}m{{u}^{2}}=\dfrac{1}{2}m{{v}^{2}}+\dfrac{4GMm}{5R}$
Now, to get the minimum initial velocity, $\dfrac{1}{2}m{{v}^{2}}=0$, that is, the final kinetic energy of the rocket will be zero and it will have no speed.
$\Rightarrow \dfrac{1}{2}m{{u}^{2}}=0+\dfrac{4GMm}{5R}=\dfrac{4GMm}{5R}$
$\Rightarrow \dfrac{1}{2}{{u}^{2}}=\dfrac{4GM}{5R}$
$\Rightarrow {{u}^{2}}=\dfrac{8GM}{5R}$
Square rooting both sides, we get
$\sqrt{{{u}^{2}}}=\sqrt{\dfrac{8GM}{5R}}$
$\Rightarrow u=\sqrt{\dfrac{8GM}{5R}}=\sqrt{\dfrac{8}{5}\dfrac{GM}{{{R}^{2}}}R}=\sqrt{\dfrac{8}{5}gR}$ $\left( \because g=\dfrac{GM}{{{R}^{2}}} \right)$ is the acceleration due to gravity on the surface of the earth.
$\Rightarrow u=\sqrt{\dfrac{8}{5}\times 9.8\times 6.4\times {{10}^{6}}}=10017.58m/s=10.017km/s\approx 10km/s$ $\left( \because g=9.8m/{{s}^{2}},R=6.4\times {{10}^{6}}m,1m/s={{10}^{-3}}km/s \right)$
Therefore, the required minimum speed of the rocket should be $10km/s$.

Therefore, the correct answer is $C)\text{ }10km.{{s}^{-1}}$.

Note:
Students must note that saying that the gravitational pull of the earth on the rocket becomes zero means that the potential energy of the rocket becomes zero because the gravitational pull of the earth on the rocket essentially becomes zero when the rocket is at an infinite distance away from the center of the earth and then putting this value in the formula for the gravitational potential energy of the rocket, we will get that the potential energy tends to zero as the radial distance (in the denominator of the fraction) tends to infinity.