The mass of the earth is 9 times that of Mars. The radius of the earth is twice that of Mars. The escape velocity on earth is 12km/s. The escape velocity of mars is ________km/sec.
A) $4\sqrt 2 $ km/sec
B) $2\sqrt 2 $km/sec
C) $6\sqrt 2 $km/sec
D) $8\sqrt 2 $km/sec
Answer
268.5k+ views
Hint: Escape velocity is defined as the minimum velocity that is required by a moving body to escape the gravitational field of the celestial body. Here we have to establish a relation between the escape velocity of earth to the escape velocity of mars.
Complete step by step solution:
Find the escape velocity of Earth:
${v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} $;
Here:
${v_e}$= Escape velocity of mars;
$G$= Gravitational Constant;
${M_e}$= Mass of earth;
${R_e}$= Radius of Earth;
Put the given value in the above equation:
Here, it is given that the mass of earth is nine times the mass of mars and it’s radius is twice that of mars. So:
$12 = \sqrt {\dfrac{{2G \times 9 \times {M_m}}}{{2{R_m}}}} $;
$ \Rightarrow 12 = \sqrt {\dfrac{{G \times 9 \times {M_m}}}{{{R_m}}}} $;
Square on both the sides to remove the square root:
$ \Rightarrow {12^2} = \dfrac{{G \times 9 \times {M_m}}}{{{R_m}}}$;
$ \Rightarrow \dfrac{{{{12}^2}}}{{9G}} = \dfrac{{{M_m}}}{{{R_m}}}$;
Now, write the formula for the escape velocity of mars:
\[{v_m} = \sqrt {\dfrac{{2G{M_m}}}{{{R_m}}}} \];
\[ \Rightarrow {v_m} = \sqrt {\left( {2G} \right) \times \left( {\dfrac{{{M_m}}}{{{R_m}}}} \right)} \];
Put the value in the above equation:
\[ \Rightarrow {v_m} = \sqrt {\left( {2G} \right) \times \dfrac{{{{12}^2}}}{{9G}}} \];
\[ \Rightarrow {v_m} = \sqrt {\dfrac{{2 \times 12 \times 12}}{9}} \];
Do the necessary calculation:
\[ \Rightarrow {v_m} = \sqrt {2 \times 4 \times 4} \];
\[ \Rightarrow {v_m} = 4\sqrt 2 \]km/sec;
Option “A” is correct. Therefore, the escape velocity of mars is \[4\sqrt 2 \] km/sec.
Note: First we need to apply the formula of escape velocity of earth and from the formula find the ratio of mass of the earth upon the radius of earth in terms of mars. After finding out the mass to radius ratio write the escape velocity formula for mars and put the mass-radius ratio in the formula for escape velocity of mars and find out the unknown escape velocity of mars.
Complete step by step solution:
Find the escape velocity of Earth:
${v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} $;
Here:
${v_e}$= Escape velocity of mars;
$G$= Gravitational Constant;
${M_e}$= Mass of earth;
${R_e}$= Radius of Earth;
Put the given value in the above equation:
Here, it is given that the mass of earth is nine times the mass of mars and it’s radius is twice that of mars. So:
$12 = \sqrt {\dfrac{{2G \times 9 \times {M_m}}}{{2{R_m}}}} $;
$ \Rightarrow 12 = \sqrt {\dfrac{{G \times 9 \times {M_m}}}{{{R_m}}}} $;
Square on both the sides to remove the square root:
$ \Rightarrow {12^2} = \dfrac{{G \times 9 \times {M_m}}}{{{R_m}}}$;
$ \Rightarrow \dfrac{{{{12}^2}}}{{9G}} = \dfrac{{{M_m}}}{{{R_m}}}$;
Now, write the formula for the escape velocity of mars:
\[{v_m} = \sqrt {\dfrac{{2G{M_m}}}{{{R_m}}}} \];
\[ \Rightarrow {v_m} = \sqrt {\left( {2G} \right) \times \left( {\dfrac{{{M_m}}}{{{R_m}}}} \right)} \];
Put the value in the above equation:
\[ \Rightarrow {v_m} = \sqrt {\left( {2G} \right) \times \dfrac{{{{12}^2}}}{{9G}}} \];
\[ \Rightarrow {v_m} = \sqrt {\dfrac{{2 \times 12 \times 12}}{9}} \];
Do the necessary calculation:
\[ \Rightarrow {v_m} = \sqrt {2 \times 4 \times 4} \];
\[ \Rightarrow {v_m} = 4\sqrt 2 \]km/sec;
Option “A” is correct. Therefore, the escape velocity of mars is \[4\sqrt 2 \] km/sec.
Note: First we need to apply the formula of escape velocity of earth and from the formula find the ratio of mass of the earth upon the radius of earth in terms of mars. After finding out the mass to radius ratio write the escape velocity formula for mars and put the mass-radius ratio in the formula for escape velocity of mars and find out the unknown escape velocity of mars.
Recently Updated Pages
Algebra Made Easy: Step-by-Step Guide for Students

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27

JEE Advanced 2026 Marks vs Rank: Estimate IIT Rank from Your Score

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

