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# The ratio of escape velocity at earth ( ${v_e}$ ) to the escape velocity at a planet (${v_p}$ ) whose radius and mean density are twice as that of earth is:A) $1:2$B) $1:2\sqrt 2$C) $1:4$D) $1:\sqrt 2$

Last updated date: 10th Aug 2024
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Hint: To solve this question we need to know what is meant by escape velocity and the equation which defines the escape velocity in terms of radius and density. Also know about the factors that escape velocity depends upon.

Complete step by step solution:
The minimum velocity with which an object is needed to be projected so that it would escape the gravitational attraction or pull of any massive body is known as the escape velocity of the body (massive bodies like planets or celestial bodies).
Mathematically escape velocity is expressed as
${v_{esc}} = \sqrt {\dfrac{{2GM}}{r}}$
Where ${v_{esc}}$ is the escape velocity of the body
$G$ is universal gravitational constant
$M$ is the mass of the body whose gravitational pull is to be escaped
$r$ is the distance from the centre of the mass
Now, in the question we are given that radius and mean density of a planet are twice as that of the earth.
Let us denote the radius and density of earth as ${r_e}$ and ${\rho _e}$ respectively. Also we denote the radius and density of the other planet as ${r_p}$ and ${\rho _p}$
$\Rightarrow {r_p} = 2{r_e}$ and ${\rho _p} = 2{\rho _e}$ --- ( $1$ )
Now, we find the mass of earth and the planet in terms of their radius and density.
$\Rightarrow {M_e} = \dfrac{4}{3}\pi r_e^3 \times {\rho _e}$ is the mass of earth ---( $2$ )
$\Rightarrow {M_p} = \dfrac{4}{3}\pi r_p^3 \times {\rho _p}$ is the mass of the planet
Substituting the values from equation ( $1$ ) in the equation of mass of the planet, we get
$\Rightarrow {M_p} = \dfrac{4}{3}\pi {(2{r_e})^3} \times 2{\rho _e}$
$\Rightarrow {M_p} = \dfrac{{64}}{3}\pi r_e^3 \times {\rho _e}$ ---( $3$ )
Now that we have the equations of masses in terms of the given data from the question, we substitute these equations of masses in the equation of escape velocity.
Thus, substituting equation ( $2$ ) and equation ( $3$ ) in the equation of the escape velocity, we get:
Escape velocity of earth is equal to ${v_{earth}} = \sqrt {\dfrac{{2G{M_e}}}{{{r_e}}}}$
$\Rightarrow {v_{earth}} = \sqrt {\dfrac{{2G\dfrac{4}{3}\pi r_e^3 \times {\rho _e}}}{{{r_e}}}}$
Simplifying this to get
$\Rightarrow {v_{earth}} = \sqrt {\dfrac{8}{3}G\pi r_e^2{\rho _e}}$ --- ( $4$ )
Escape velocity of the planet is equal to ${v_{planet}} = \sqrt {\dfrac{{2G{M_p}}}{{{r_p}}}}$
$\Rightarrow {v_{planet}} = \sqrt {\dfrac{{2G\dfrac{{64}}{3}\pi r_e^3 \times {\rho _e}}}{{2{r_e}}}}$
Simplifying this to get
$\Rightarrow {v_{planet}} = \sqrt {\dfrac{{64}}{3}G\pi r_e^2 \times {\rho _e}}$ --- ( $5$ )
As the question says, we have to find the ratio of the escape velocity of earth to the escape velocity of the planet. So, to find this we divide equation ( $4$ ) by equation ( $5$ )
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt {\dfrac{8}{3}G\pi r_e^2{\rho _e}} }}{{\sqrt {\dfrac{{64}}{3}G\pi r_e^2 \times {\rho _e}} }}$
Cancelling out all the common terms, we get:
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt {\dfrac{8}{3}} }}{{\sqrt {\dfrac{{64}}{3}} }}$
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt 8 }}{{\sqrt {64} }}$
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{{\sqrt 8 }}{8}$
$\Rightarrow \dfrac{{{v_e}}}{{{v_p}}} = \dfrac{1}{{\sqrt 8 }} = \dfrac{1}{{2\sqrt 2 }}$
Thus, the ratio ${v_e}:{v_p} = 1:2\sqrt 2$

Therefore, option (B) is the correct option.

Note: Remember that the escape velocity depends on the mass and radius of the body whose gravitational pull is to be escaped. Escape velocity does not depend on the mass and radius of the smaller object which is propelled.