What will be the escape velocity from a planet having radius thrice that of earth and the same mean density as that of earth?

Answer

What will be the escape velocity from a planet having radius thrice that of earth and the same mean density as that of earth?
Verified

Hint: The escape velocity of the planet is $\sqrt {\dfrac{{2GM}}{r}} $ where symbols have their usual meaning.As the density of the planet is the same as that of earth then we can find the relation between their masses then use these values in the formula.

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 earthR is the radius of earthNow, it is given that radius of planet is twice the radius of earth ie ${R_p} = 2{R_e}$$R_p$ is the radius of planet and $R_e$ is the radius of earthSince, the density of earth and planet are same, lets determine the ratio of their volumesLet volume of planet is ${V_p} = \dfrac{4}{3}\pi R_p^3$ ………………………… (2)And volume of earth is ${V_e} = \dfrac{4}{3}\pi R_e^3$ ……………………… (3)Now divide above two equations, we get⇒$\dfrac{{{V_p}}}{{{V_e}}} = \dfrac{{R_p^3}}{{R_e^3}} = \dfrac{{{{\left( {2R} \right)}^3}}}{{{R^3}}} = 8$Since, the density is same therefore,${M_p} = 8M$Now the escape velocity of planet is ${v_p} = \sqrt {\dfrac{{G{M_p}}}{{{R_p}}}} $On substituting the values in above equation, we get ⇒${v_p} = \sqrt {\dfrac{{8GM}}{{2R}}} $………………………….. (4)Dividing equation (4) by equation (1), we get⇒$\dfrac{{{v_p}}}{{{v_e}}} = \sqrt {\dfrac{8}{2}} = 2$⇒${v_p} = 2{v_e} = 2 \times 11 = 22km/s$Hence the escape velocity of planet is 22km/s

Therefore, the option (A) 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 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}} $