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Q.
The escape velocity for the earth is $v_e$. The escape velocity for a planet whose radius is $\frac{1}{4}th$, the radius of the earth and mass half that of the earth is
Escape velocity of a body from the surface of the earth is given by
$v_{e}=\sqrt{\frac{2 G M_{e}}{R_{e}}}$ ...(i)
So, according to the question,
$M_{P}=\frac{M_{e}}{2} \text { and } R_{P}=\frac{R_{e}}{4}$
Now, escape velocity of a body from the surface of the planet is given by
$v_{P}=\sqrt{\frac{2 G M_{P}}{R_{P}}}$
or $V_{{P}} =\sqrt{\frac{2 G \times M_{e} \times 4}{2 R_{e}}}=\sqrt{2} \sqrt{\frac{2 G M_{e}}{R_{e}}}$
$=\sqrt{2 v_{e}}$ [from Eq.(i)]