Question

Which of the following is correct about the statement “Escape velocity from the surface of moon is less than that from the surface of earth because the moon has no atmosphere while earth have a very dense one”
${\text{A}}{\text{.}}$ True
${\text{B}}{\text{.}}$ False

Answer 1

Hint: Here, we will proceed by defining the term escape velocity. Then, we will write down the mathematical formula for determining escape velocity. Finally, we will find the escape velocities for earth as well as moon.

Formulas Used- ${{\text{v}}_{{\text{Escape}}}} = \sqrt {\dfrac{{2{\text{GM}}}}{{\text{R}}}} $.

Complete Step-by-Step solution:
Every object which is on the planet (or inside the gravitational sphere of attraction) experiences a gravitational pull or force and hence, is not able to escape that planet. But, whenever the object is having sufficient velocity so that it can go out of the gravitational sphere of attraction, it escapes just like a rocket and other celestial bodies. Therefore, escape velocity can be defined as the minimum velocity acquired by an object to escape the gravitational sphere of pull for any planet without undergoing any change in acceleration.
Escape velocity for any object in order to escape another object of mass M and radius R is given by
${{\text{v}}_{{\text{Escape}}}} = \sqrt {\dfrac{{2{\text{GM}}}}{{\text{R}}}} {\text{ }} \to {\text{(1)}}$ where G denotes the universal gravitational constant (G = 6.67$ \times {10^{ - 11}}$ N${{\text{m}}^2}$/${\text{k}}{{\text{g}}^2}$), M denotes the mass of the planet or object from which the given object needs to be escaped and R denotes the radius of the planet or object from which the given object needs to be escaped.
In order to find out the value for escape velocity of the earth, we can substitute the values of mass and radius of earth. Also, in order to find out the value for escape velocity of the moon, we can substitute the values of mass and radius of the moon.
Mass of earth ${{\text{M}}_{{\text{Earth}}}} = 5.972 \times {10^{24}}$ kg
Radius of earth ${{\text{R}}_{{\text{Earth}}}} = 6.4 \times {10^6}$ m
Using the formula given by equation (1), we get
${{\text{v}}_{{\text{Earth}}}} = \sqrt {\dfrac{{2{\text{G}}{{\text{M}}_{{\text{Earth}}}}}}{{{{\text{R}}_{{\text{Earth}}}}}}} = \sqrt {\dfrac{{2\left( {6.67 \times {{10}^{ - 11}}} \right)\left( {5.972 \times {{10}^{24}}} \right)}}{{6.4 \times {{10}^6}}}} = 11.2 \times {10^3}$ m/s
Also, Mass of moon \[{{\text{M}}_{{\text{Moon}}}} = 7.34 \times {10^{22}}\] kg
Radius of moon ${{\text{R}}_{{\text{Moon}}}} = 1.7 \times {10^6}$ m
Using the formula given by equation (1), we get
${{\text{v}}_{{\text{Moon}}}} = \sqrt {\dfrac{{2{\text{G}}{{\text{M}}_{{\text{Moon}}}}}}{{{{\text{R}}_{{\text{Moon}}}}}}} = \sqrt {\dfrac{{2\left( {6.67 \times {{10}^{ - 11}}} \right)\left( {7.34 \times {{10}^{22}}} \right)}}{{1.7 \times {{10}^6}}}} = 2.39 \times {10^3}$ m/s
Clearly, we can see that the escape velocity for moon is less than that for the earth and because of this low escape velocity for moon, all the gas molecules escapes from the moon due to the lesser velocity as compared to the root mean square value of the velocities of these gas molecules.
Therefore, the statement “Escape velocity from the surface of moon is less than that from the surface of earth because the moon has no atmosphere while earth has a very dense one” is true.

Hence, option A is correct.

Note- We must have wondered that when we throw an object into the air it comes back. Then, why don’t the celestial bodies come back like whenever a rocket or a satellite is launched it goes into outer space and leaves earth. This is because escape velocity is being attained by that rocket or satellite.