Question

The escape velocity of a rocket on the earth is $11.2\text{km/sec}$. Its value on a planet where acceleration due to gravity is twice that on the earth and the diameter of the planet is twice that of earth, will be(in $km/\sec $ ):
A. 11.2
B. 5.6
C. 22.4
D. 33.6

Answer 1

Hint: Obtain the expression for the escape velocity of an object from the earth. Express the escape velocity in terms of the acceleration due to gravity of earth. Put the values for the planet on the obtained equation to find the escape velocity of that planet.

Complete step-by-step answer:
The escape velocity of an object from a planet is given by the mathematical expression,
$v=\sqrt{\dfrac{2GM}{R}}$
Where, G is the gravitational constant with value, $G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$ , M is the mass of the planet and R is the radius of the planet.
Again, we can express the acceleration due to gravity g in terms of the gravitational constant as,
$g=\dfrac{GM}{{{R}^{2}}}$
From this we can write that,
$G=\dfrac{g{{R}^{2}}}{M}$
Putting this value on the expression for escape velocity, we get that,
$v=\sqrt{\dfrac{2\dfrac{g{{R}^{2}}}{M}M}{R}}=\sqrt{2gR}$
Given that, the escape velocity of a rocket from earth is $11.2km/\sec $ .
So, we can write that,
$11.2=\sqrt{2{{g}_{e}}{{R}_{e}}}$
Where, ${{g}_{e}}$ is the acceleration due to gravity of earth and ${{R}_{e}}$ is the radius of earth.
On another planet the acceleration due to gravity is twice that of the earth and the radius is also twice that of earth. So,
$g=2{{g}_{e}}$
$R=2{{R}_{e}}$
So, the escape velocity of the rocket in this planet will be,
$\begin{align}
  & v=\sqrt{2gR} \\
 & v=\sqrt{2\times 2{{g}_{e}}\times 2{{R}_{e}}} \\
 & v=\sqrt{8gR} \\
\end{align}$
Dividing the second equation by the first one, we get that,
$\begin{align}
  & \dfrac{v}{11.2}=\dfrac{\sqrt{8{{g}_{e}}{{R}_{e}}}}{\sqrt{2{{g}_{e}}{{R}_{e}}}} \\
 & \dfrac{v}{11.2}=\sqrt{4}=2 \\
 & v=2\times 11.2 \\
 & v=22.4km/\sec \\
\end{align}$
So, the escape velocity of the rocket on the other planet will be $22.4km/\sec $.
The correct option is (C).

Note: Escape velocity of a planet is directly proportional to the mass of the planet and inversely proportional to the radius of the planet. So, the higher the mass of the planet, the higher is the escape velocity and lower the radius of the planet, the higher is the escape velocity of the planet.