Answer

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**Hint:**The escape velocity is the velocity with which an object can just get out of the gravitational pull. Here, in this scenario you can think that an object can get away from the gravitational pull of the earth with escape velocity ${v_e}$ and that same object can get away from the gravitational pull of the other planet with escape velocity ${v_p}$. The key point to approach the problem is to understand that the potential due to gravitation should be overcome by the obtained kinetic energy to escape the gravitational pull.

**Formula Used:**

If the mass of the object is $m$ and the mass of the planet is $M$ and the distance between their centers is $R$, then the gravitational potential $U$ is

$U = \dfrac{{GMm}}{R}$

**Complete step by step answer:**

Given:

The escape velocity of the earth is ${v_e}$.

The escape velocity of the other planet is ${v_p}$.

The radius of the planet is twice that of the earth.

The mean density of the planet is twice that of the earth.

To get: The ratio of escape velocity at earth $\left( {{v_e}} \right)$ to the escape velocity at the planet $\left( {{v_p}} \right)$, ${v_e}$: ${v_p}$.

Step 1:

Let an object of mass $m$ escape the earth and the other planet with escape velocity ${v_e}$ and ${v_p}$ respectively.

${K_e} = \dfrac{1}{2}m{v_e}^2$The kinetic energy of the object with escape velocity at the earth is

The kinetic energy of the object with escape velocity at the planet is ${K_p} =\dfrac{1}{2}m{v_p}^2 \\$

Step 2:

Let the mass of the earth is ${M_e}$ and the mass of the planet is ${M_p}$

The potential from eq (1) is to be equal to the value of the kinetic energy acquired from escaping.

So, calculate the escape velocity for the object at earth ${v_e}$

\[

{U_e} = {K_e} \\

\Rightarrow \dfrac{{G{M_e}m}}{{{R_e}}} = \dfrac{1}{2}m{v_e}^2 \\

\Rightarrow \dfrac{{G{M_e}}}{{{R_e}}} = \dfrac{1}{2}{v_e}^2 \\

\Rightarrow {v_e}^2 = \dfrac{{2G{M_e}}}{{{R_e}}} \\

\Rightarrow {v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} \\

\]

Similarly, calculate the escape velocity for the object at the other planet ${v_p}$

\[

{U_p} = {K_p} \\

\Rightarrow \dfrac{{G{M_p}m}}{{{R_p}}} = \dfrac{1}{2}m{v_p}^2 \\

\Rightarrow \dfrac{{G{M_p}}}{{{R_p}}} = \dfrac{1}{2}{v_p}^2 \\

\Rightarrow {v_p}^2 = \dfrac{{2G{M_p}}}{{{R_p}}} \\

\Rightarrow {v_p} = \sqrt {\dfrac{{2G{M_p}}}{{{R_p}}}} \\

\]

Step 3:

Let the mean density of the earth is ${\rho _e}$ and the mean density of the planet is ${\rho _p}$.

The earth and the planets are considered to be spherical so you can take their volume as the spherical volume of their radius.

So, calculate the mean density of the earth ${\rho _e}$

\[{\rho _e} = \dfrac{{{M_e}}}{{\dfrac{4}{3}\pi {R_e}^3}}\]

Similarly, calculate the mean density of the planet ${\rho _p}$

\[{\rho _p} = \dfrac{{{M_p}}}{{\dfrac{4}{3}\pi {R_p}^3}}\]

Step 4:

Now, express the escape velocities in terms of the mean densities of the earth and the planet.

The escape velocity at earth ${v_e}$, then becomes

\[

{v_e} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} \\

\Rightarrow {v_e} = \sqrt {\dfrac{{2G{M_e} \times \dfrac{4}{3}\pi {R_e}^2}}{{{R_e} \times \dfrac{4}{3}\pi {R_e}^2}}} \\

\Rightarrow {v_e} = \sqrt {\dfrac{{\dfrac{8}{3}\pi {R_e}^2G{M_e}}}{{\dfrac{4}{3}\pi {R_e}^3}}} \\

\Rightarrow {v_e} = \sqrt {\dfrac{8}{3}\pi G{R_e}^2{\rho _e}} \\

\]

Similarly rewrite the escape velocity at the planet ${v_p}$

\[

{v_p} = \sqrt {\dfrac{{2G{M_p}}}{{{R_p}}}} \\

\Rightarrow {v_p} = \sqrt {\dfrac{{2G{M_p} \times \dfrac{4}{3}\pi {R_p}^2}}{{{R_p} \times \dfrac{4}{3}\pi {R_p}^2}}} \\

\Rightarrow {v_p} = \sqrt {\dfrac{{\dfrac{8}{3}\pi {R_p}^2G{M_p}}}{{\dfrac{4}{3}\pi {R_p}^3}}} \\

\Rightarrow {v_p} = \sqrt {\dfrac{8}{3}\pi G{R_p}^2{\rho _p}} \\

\]

Step 5:

Now take the ratio of the escape velocity at the earth ${v_e}$ and the escape velocity at the planet ${v_p}$.

\[\dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt {\dfrac{8}{3}\pi G{R_e}^2{\rho _e}} }}{{\sqrt {\dfrac{8}{3}\pi G{R_p}^2{\rho _p}} }}\]

By the problem, you have the radius of the planet ${v_p}$ is twice that of the earth ${v_e}$ that is $\dfrac{{{R_p}}}{{{R_e}}} = 2$

By the problem, you have the mean density of the planet \[{\rho _p}\] is twice that of the earth ${\rho _e}$ that is $\dfrac{{{\rho _p}}}{{{\rho _e}}} = 2$

Hence, calculate the ratio

\[\dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt {\dfrac{8}{3}\pi G{R_e}^2{\rho _e}} }}{{\sqrt {\dfrac{8}{3}\pi G{R_p}^2{\rho _p}} }} = \dfrac{{\sqrt {\dfrac{8}{3}\pi G} }}{{\sqrt {\dfrac{8}{3}\pi G} }}\sqrt {{{\left( {\dfrac{{{R_e}}}{{{R_p}}}} \right)}^2}\left( {\dfrac{{{\rho _e}}}{{{\rho _p}}}} \right)} = \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2}\left( {\dfrac{1}{2}} \right)} = \dfrac{1}{{2\sqrt 2 }}\]

Final Answer:

The ratio of escape velocity at earth $\left( {{v_e}} \right)$ to the escape velocity at a planet $\left( {{v_p}} \right)$ whose radius and mean density are twice as that of earth is (B) $1$ : $2\sqrt 2 $.

**Note:**To calculate the mean density, you need to consider the planets as perfect spheres and take the volumes accordingly. Don’t mess up with the quantities of the earth and the planet. The kinetic energy should just balance the gravitational potential to possess the escape velocity.

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