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Q.
The escape velocity from the earth is $11.2\, km / s$. The escape velocity from a planet having twice the radius and the same mean density as the earth, is:
At a certain velocity of projection, the body will go out of the gravitational field to the earth and will never return, this initial velocity is called escape velocity
$v_e=\sqrt{\frac{2 G M_e}{R_e}}$
where $G$ is gravitational constant, $M_e$ is mass of earth and $R_e$ is radius.
For planet $R_p=2 R_e, d_p=d_e$
Also since earth is assumed spherical in shape its mass is given by
$M =\frac{4}{3} \pi R^3 \times d$
$\therefore v_e =\sqrt{\frac{2 G}{R_e} \times \frac{4}{3} \pi R^3 d}$ (1)
$v_p =\sqrt{\frac{2 G}{2 R_e} \times \frac{4}{3} \pi\left(2 R_e\right)^3 d}$(2)
Dividing Eq. (1) by Eq. (2), we get
$\frac{v_e}{v_p}=\frac{1}{2} $
$\Rightarrow v_p =2 v_e=2 \times 11.2=22.4\, km / s$