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Hint the escape velocity of the planet is $\sqrt {\dfrac{{2GM}}{r}} $where symbols have their usual meaning.
As mass and radius of the planet are given and escape velocity of the earth is also given then we will get a relation between all these and then we will get the answer.
Complete step-by-step answer:
The escape velocity of earth is given by:
${v_e} = \sqrt {\dfrac{{GM}}{R}} = 11km/s$……………………… (1)
Where, G is the universal gravitational constant i.e. $G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$
M is the mass of earth
R is the radius of earth
Now, let escape velocity of planet is ${v_p} = \sqrt {\dfrac{{G{M_p}}}{{{R_p}}}} $………………… (2)
Where $M_P$ and $R_P$ is the mass and radius of the planet.
Now, it is given that the mass of planet is half that of the earth i.e. ${M_P} = \dfrac{M}{2}$
And radius of planet is one fourth of the radius of earth i.e. ${R_P} = \dfrac{R}{4}$
Substitute these values in equation (2), we get
$
\Rightarrow {V_P} = \sqrt {\dfrac{{2 \times 4 \times GM}}{{2R}}} = \sqrt 2 \times \sqrt {\dfrac{{2GM}}{R}} \\
$
Using equation (1), we get
$ \Rightarrow {V_P} = \sqrt 2 \times 11 = 1.414 \times 11 = 15.55km{s^{ - 1}}$
Thus, the escape speed of planet is ${V_P} = 15.55km{s^{ - 1}}$
Hence, C option is correct.
Note Escape velocity of an object of mass m for a planet of mass M and radius R is given by the sum of potential energy and kinetic energy and equating to zero
⇒$\dfrac{{m{v^2}}}{2} - \dfrac{{GmM}}{R} = 0$, m will cancel out and therefore escape velocity is $v = \sqrt {\dfrac{{GM}}{R}} $
As mass and radius of the planet are given and escape velocity of the earth is also given then we will get a relation between all these and then we will get the answer.
Complete step-by-step answer:
The escape velocity of earth is given by:
${v_e} = \sqrt {\dfrac{{GM}}{R}} = 11km/s$……………………… (1)
Where, G is the universal gravitational constant i.e. $G = 6.67 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}$
M is the mass of earth
R is the radius of earth
Now, let escape velocity of planet is ${v_p} = \sqrt {\dfrac{{G{M_p}}}{{{R_p}}}} $………………… (2)
Where $M_P$ and $R_P$ is the mass and radius of the planet.
Now, it is given that the mass of planet is half that of the earth i.e. ${M_P} = \dfrac{M}{2}$
And radius of planet is one fourth of the radius of earth i.e. ${R_P} = \dfrac{R}{4}$
Substitute these values in equation (2), we get
$
\Rightarrow {V_P} = \sqrt {\dfrac{{2 \times 4 \times GM}}{{2R}}} = \sqrt 2 \times \sqrt {\dfrac{{2GM}}{R}} \\
$
Using equation (1), we get
$ \Rightarrow {V_P} = \sqrt 2 \times 11 = 1.414 \times 11 = 15.55km{s^{ - 1}}$
Thus, the escape speed of planet is ${V_P} = 15.55km{s^{ - 1}}$
Hence, C option is correct.
Note Escape velocity of an object of mass m for a planet of mass M and radius R is given by the sum of potential energy and kinetic energy and equating to zero
⇒$\dfrac{{m{v^2}}}{2} - \dfrac{{GmM}}{R} = 0$, m will cancel out and therefore escape velocity is $v = \sqrt {\dfrac{{GM}}{R}} $
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