Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The ratio of escape velocity at the Earth ( $v_e$ ) to the escape velocity at a planet ( $v_p$ ) whose radius and mean density are twice that of the Earth is:

1 : 2\sqrt{2}

1 : 4

1 : \sqrt{2}

1 : 2

Step-by-Step Solution

Escape Velocity Formula: The escape velocity $v_e$ from a spherical body of mass $M$ and radius $R$ is given by $v_e = \sqrt{\frac{2GM}{R}}$ .
In Terms of Density: To relate velocity to density ( $\rho$ ), express mass as $M = \text{Volume} \times \text{Density} = \frac{4}{3}\pi R^3 \rho$ . Substituting this into the escape velocity formula: $v = \sqrt{\frac{2G}{R} \left( \frac{4}{3}\pi R^3 \rho \right)} = \sqrt{\frac{8\pi G \rho R^2}{3}}$ From this, we see that $v \propto R\sqrt{\rho}$ .
Ratio Calculation: Let Earth's radius be $R$ and density be $\rho$ . Then the planet has radius $R' = 2R$ and density $\rho' = 2\rho$ . $\frac{v_e}{v_p} = \frac{R\sqrt{\rho}}{R'\sqrt{\rho'}} = \frac{R\sqrt{\rho}}{(2R)\sqrt{2\rho}}$ $\frac{v_e}{v_p} = \frac{1}{2\sqrt{2}}$ The ratio is $1 : 2\sqrt{2}$ .

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