Question

The escape velocity of a body from earth is about $11.2\,km/s$. Assuming the mass and radius of the earth to be about 81 and 4 times the mass and radius of the moon, what will be the escape velocity in $km/s$ from the surface of the moon?
A. $0 \cdot 54$
B. $2 \cdot 48$
C. 11
D. $49 \cdot 5$

Answer 1

Hint- We know that escape velocity is the minimum velocity required by a body in order to overcome the gravitational pull provided by a massive object such as planets.
It is given by the equation
$v = \sqrt {\dfrac{{2GM}}{r}}$
Where $G$ is the gravitational constant $M$ is the mass of the planet or moon and $r$ is its radius.
Using this equation we can write the equation for escape velocity of earth and moon and by dividing these equations we can arrive at the final answer.

Step by step solution:
Escape velocity is the minimum velocity required by a body in order to overcome the gravitational pull provided by a massive object such as planets.
It is given by the equation
$v = \sqrt {\dfrac{{2GM}}{r}}$
Where $G$ is the gravitational constant $M$ is the mass of the planet or moon and $r$ is its radius.
Let us use this equation to solve our question.
Given,
The escape velocity of a body from earth is, ${v_e} = 11.2\,km/s$.
Let the mass of the moon be $m$ and radius of the moon be $r$.
The mass of the earth is 81 times the mass of the moon. Therefore, mass of the earth can be written as
${m_e} = 81m$
Radius of the earth is four times that of the moon. Therefore, radius of the earth can be written as
${r_e} = 4r$
We need to find escape velocity on the surface of moon
That is,
${v_m} = \sqrt {\dfrac{{2Gm}}{r}}$ …… (1)
Escape velocity of earth is
${v_e} = \sqrt {\dfrac{{2G{m_e}}}{{{r_e}}}}$
Substitute ${m_e} = 81m$ and ${r_e} = 4r$. Then, we get
${v_e} = \sqrt {\dfrac{{2G \times 81m}}{{4r}}}$
${v_e} = \dfrac{9}{2}\sqrt {\dfrac{{2Gm}}{r}}$ ……. (2)
Now let us divide equation (1) by (2). then we get,
$\dfrac{{{v_m}}}{{{v_e}}} = \dfrac{{\sqrt {\dfrac{{2Gm}}{r}} }}{{\dfrac{9}{2}\sqrt {\dfrac{{2Gm}}{r}} }}$
$\Rightarrow \dfrac{{{v_m}}}{{{v_e}}} = \dfrac{2}{9}$
$\therefore {v_m} = \dfrac{2}{9} \times {v_e}$
Now substitute the escape velocity of a body from earth ${v_e} = 11.2\,km/s$.
Then, we get
${v_m} = \dfrac{2}{9} \times 11.2 = 2.48\,m/s$

So, the answer is option B

Note:
Escape velocity is $v = \sqrt {\dfrac{{2GM}}{r}}$
Where $G$ is the gravitational constant $M$ is the mass of the planet or moon and $r$ is its radius.
Here the mass is the mass of the planet. Remember that escape velocity does not depend upon mass of the body and direction of projection of the body. It depends only on the mass and radius of the massive object providing the gravitational pull.