The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is

The correct option is √3 Vₑ.

To find the escape velocity of the planet, we start with the formula for the escape velocity from the surface of a celestial body:
$V_e=\sqrt{\frac{2GM}{R}}$
where:

• $G$ is the universal gravitational constant,
• $M$ is the mass of the body, and
• $R$ is the radius of the body.

Let the mass of Earth be $M$ and its radius be $R$. Therefore, the escape velocity of Earth is:
$V_e=\sqrt{\frac{2GM}{R}}$

According to the problem, the planet has:

• Mass of the planet, $M_p=6M$
• Radius of the planet, $R_p=2R$

Substituting these values into the escape velocity formula for the planet ($V_p$):
$V_p=\sqrt{\frac{2G{M}_p}{R_p}}$
$V_p=\sqrt{\frac{2G\left(6M\right)}{2R}}$

Simplifying the expression under the square root:
$V_p=\sqrt{3\times \frac{2GM}{R}}$
$V_p=\sqrt{3}{V}_e$

Thus, the escape velocity of the planet is $\sqrt{3}$ times the escape velocity of Earth, which is written as √3 Vₑ.