
The escape velocity on earth is $11.2km{s^{ - 1}}$. What is the escape velocity on the planet of radius four times the radius of earth and one fourth of the density of earth?
A.$11.2km{s^{ - 1}}$
B.$22.4km{s^{ - 1}}$
C.$44.8km{s^{ - 1}}$
D.$26.3km{s^{ - 1}}$
Answer
585.6k+ views
Hint: The escape velocity is the velocity needed for an object to escape the gravity field of the planet. The Escape Velocity depends on the radius and acceleration due to gravity. This should be changed into the terms of radius and density, as the data in the question is given in those terms.
Formula used:
$\dfrac{{{V_e}}}{{{V_x}}} = \sqrt {\dfrac{{{d_e}{R_e}^2}}{{{d_x}{R_x}^2}}} $
Complete answer:
Escape velocity is the velocity required to escape the Earth’s gravity field. It is mathematically given by
${V_e} = \sqrt {\dfrac{{2GM}}{R}} $
Where
${V_e}$ is the escape velocity of Earth
$G$ is the gravitational constant
$M$ is the mass of the Earth
$R$ is the radius of Earth
The mass of the earth can be written in terms of density and radius as
$M = d \times \dfrac{4}{3}\pi {R^3}$
Here, $d$ is density. Substituting this in the formula of escape velocity we get
${V_e} = \sqrt {\dfrac{{2G}}{R} \times d \times \dfrac{4}{3}\pi {R^3}} = \sqrt {\dfrac{{8\pi Gd{R^2}}}{3}} $
So, we see that
${V_e} \propto \sqrt {d{R^2}} $
Using this relation, assuming that another planet is X, we have
$\dfrac{{{V_e}}}{{{V_x}}} = \sqrt {\dfrac{{{d_e}{R_e}^2}}{{{d_x}{R_x}^2}}} $
It is given the question, that ${R_x} = {\text{ }}4{R_e}$ and ${d_x} = {\text{ }}\dfrac{{{d_e}}}{4}$. Substituting these in the formula
$\eqalign{
& \dfrac{{{V_e}}}{{{V_x}}} = \sqrt {\dfrac{{{d_e} \times {R_e}^2}}{{\dfrac{{{d_e}}}{4} \times {{\left( {4{R_e}} \right)}^2}}}} \cr
& \Rightarrow \dfrac{{{V_e}}}{{{V_x}}} = \sqrt {\dfrac{1}{4}} = \dfrac{1}{2} \cr
& \Rightarrow {V_x} = {V_e} \times 2 = 11.2km{s^{ - 1}} \times 2 = 22.4m{s^{ - 1}} \cr
& \therefore {V_x} = 22.4m{s^{ - 1}} \cr} $
So, the correct answer is “Option B”.
Note:
In the above problem we are generalizing the formula of escape velocity for the Earth. But the formula applies to every planet. Don’t try to memorize the formula we’ve used in the above problem. Simply remember the escape velocity formula and change it according to the data given in the question. An alternative formula for escape velocity is ${V_e} = \sqrt {2gR} $.
Formula used:
$\dfrac{{{V_e}}}{{{V_x}}} = \sqrt {\dfrac{{{d_e}{R_e}^2}}{{{d_x}{R_x}^2}}} $
Complete answer:
Escape velocity is the velocity required to escape the Earth’s gravity field. It is mathematically given by
${V_e} = \sqrt {\dfrac{{2GM}}{R}} $
Where
${V_e}$ is the escape velocity of Earth
$G$ is the gravitational constant
$M$ is the mass of the Earth
$R$ is the radius of Earth
The mass of the earth can be written in terms of density and radius as
$M = d \times \dfrac{4}{3}\pi {R^3}$
Here, $d$ is density. Substituting this in the formula of escape velocity we get
${V_e} = \sqrt {\dfrac{{2G}}{R} \times d \times \dfrac{4}{3}\pi {R^3}} = \sqrt {\dfrac{{8\pi Gd{R^2}}}{3}} $
So, we see that
${V_e} \propto \sqrt {d{R^2}} $
Using this relation, assuming that another planet is X, we have
$\dfrac{{{V_e}}}{{{V_x}}} = \sqrt {\dfrac{{{d_e}{R_e}^2}}{{{d_x}{R_x}^2}}} $
It is given the question, that ${R_x} = {\text{ }}4{R_e}$ and ${d_x} = {\text{ }}\dfrac{{{d_e}}}{4}$. Substituting these in the formula
$\eqalign{
& \dfrac{{{V_e}}}{{{V_x}}} = \sqrt {\dfrac{{{d_e} \times {R_e}^2}}{{\dfrac{{{d_e}}}{4} \times {{\left( {4{R_e}} \right)}^2}}}} \cr
& \Rightarrow \dfrac{{{V_e}}}{{{V_x}}} = \sqrt {\dfrac{1}{4}} = \dfrac{1}{2} \cr
& \Rightarrow {V_x} = {V_e} \times 2 = 11.2km{s^{ - 1}} \times 2 = 22.4m{s^{ - 1}} \cr
& \therefore {V_x} = 22.4m{s^{ - 1}} \cr} $
So, the correct answer is “Option B”.
Note:
In the above problem we are generalizing the formula of escape velocity for the Earth. But the formula applies to every planet. Don’t try to memorize the formula we’ve used in the above problem. Simply remember the escape velocity formula and change it according to the data given in the question. An alternative formula for escape velocity is ${V_e} = \sqrt {2gR} $.
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