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# The mass of a planet is half that of the earth and the radius of the planet is one fourth that of earth. If we plan to send an artificial satellite from the planet, the escape velocity will be (${V_e} = 11km{s^{ - 1}}$)A. $11km{s^{ - 1}}$B. $5.5km{s^{ - 1}}$C. $15.55km{s^{ - 1}}$D. $7.78km{s^{ - 1}}$

Last updated date: 17th Apr 2024
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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.

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 $v = \sqrt {\dfrac{{GM}}{R}}$