Question

The escape velocity from the earth is about 11 km/s. The escape speed from the planet, having twice the radius and the same mean density as earth is
A) 22 km/s
B) 11 km/s
C) 5.5 km/s
D) 15.5 km/s

Answer 1

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 earth
R is the radius of earth
Now, 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 earth
Since, the density of earth and planet are same, lets determine the ratio of their volumes
Let 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}} $