The escape velocity of an object from a planet is $16 \mathrm{kms}^{-1}$. If the escape velocity of the…
The escape velocity of an object from a planet is $16 \mathrm{kms}^{-1}$. If the escape velocity of the object from another planet having twice the density and three times the radius of the planet is $v \sqrt{2} \mathrm{~ms}^{-1}$, then the value of $v$ is
$12$
$48$
$18$
$36$
Solution
Escape speed is given by
$\begin{aligned} v & =\sqrt{\frac{2 G M}{R}} \\ & =\sqrt{\frac{2 G \times \frac{4}{3} \pi R^3 \times d}{R}}=\quad(\because M=d \times V) .\end{aligned}$
$\Rightarrow \quad v=\sqrt{\frac{8}{3} G \pi R^2 d}$
Now given for a planet (lets say $A$ )
$v_1=16=\sqrt{\frac{8}{3} G \pi R^2 d}$
and for planet $B$,
$v_2=\sqrt{\frac{8}{3} G \pi(3 R)^2 \times 2 d}$.
$\begin{aligned} & \Rightarrow \quad \frac{v_1}{v_2}=\frac{16}{v_2}=\sqrt{\frac{R^2 \times d}{(3 R)^2 \times 2 d}} \\ & \Rightarrow \quad \frac{16}{v_2}=\sqrt{\frac{1}{18}}=\frac{1}{3 \sqrt{2}}\end{aligned}$
Hence, $v_2=48 \sqrt{2} \mathrm{~m} / \mathrm{s}=v \sqrt{2} \mathrm{~ms}^{-1}$(given)
$\therefore \quad v=48$