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**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.

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