The escape velocity from the surface of a planet is given by the formula:
\(v_e = \sqrt{\frac{2 G M}{R}}\)
where \(v_e\) is the escape velocity, \(G\) is the universal gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet.
For Earth, the escape velocity is given as \(11.2\ \text{km/s}\). Therefore,
\(11.2 = \sqrt{\frac{2 G M_{Earth}}{R_{Earth}}}\)
Now, consider a planet with the same mass as Earth, but with a radius \(\frac{1}{4}\) of Earth's radius:
The escape velocity for this planet is:
\(v_{planet} = \sqrt{\frac{2 G M_{planet}}{R_{planet}}}\)
Substitute the values:
\(v_{planet} = \sqrt{\frac{2 G M_{Earth}}{\frac{1}{4} R_{Earth}}}\)
Simplifying gives:
\(v_{planet} = \sqrt{\frac{4 \times 2 G M_{Earth}}{R_{Earth}}}\)
\(v_{planet} = \sqrt{4} \times \sqrt{\frac{2 G M_{Earth}}{R_{Earth}}}\)
\(v_{planet} = 2 \times \sqrt{\frac{2 G M_{Earth}}{R_{Earth}}}\)
Since \(\sqrt{\frac{2 G M_{Earth}}{R_{Earth}}}\) is the escape velocity of Earth:
\(v_{planet} = 2 \times 11.2 = 22.4\ \text{km/s}\)
Thus, the escape velocity from the new planet is \(22.4\ \text{km/s}\). Therefore, the correct answer is \(22.4\ \text{km/s}\).