A celestial body has a mass of 5.98 x 1024 kg. An object is at a distance of 6,371 km from the center of the body. Calculate the escape velocity required for the object to break free from a celestial body's gravitational pull.

Step 1: Identify the given values and the formula for escape velocity. The mass of the celestial body is $M = 5.98 \times 10^{24} \text{ kg}$. The distance from the center of the body (radius) is $R = 6,371 \text{ km}$. The gravitational constant is $G = 6.67 \times 10^{-11} Nm^2kg^{-2}$.

The formula for escape velocity ($v_e$) is: $v_e = \sqrt{\frac{2GM}{R}}$

Step 2: Convert all values to SI units. The mass $M$ is already in kilograms (kg). The gravitational constant $G$ is already in $\text{Nm}^2kg^{-2}$. The radius $R$ needs to be converted from kilometers to meters: $R = 6,371 km = 6,371 \times 10^3 m = 6.371 \times 10^6 \text{ m}$

Step 3: Substitute the values into the escape velocity formula and calculate. $v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11} Nm^2kg^{-2}) \times (5.98 \times 10^{24} kg)}{6.371 \times 10^6 m}}$ $v_e = \sqrt{\frac{2 \times 6.67 \times 5.98 \times 10^{-11+24}}{6.371 \times 10^6}} \text{ m/s}$ $v_e = \sqrt{\frac{79.7732 \times 10^{13}}{6.371 \times 10^6}} \text{ m/s}$ $v_e = \sqrt{\left(\frac{79.7732}{6.371}\right) \times 10^{13-6}} \text{ m/s}$ $v_e = \sqrt{12.5212 \times 10^7} \text{ m/s}$ To take the square root, we can rewrite $10^7$ as $10 \times 10^6$: $v_e = \sqrt{125.212 \times 10^6} \text{ m/s}$ $v_e = \sqrt{125.212} \times \sqrt{10^6} \text{ m/s}$ $v_e \approx 11.189 \times 10^3 \text{ m/s}$

Step 4: Convert the result to kilometers per second (km/s). $v_e \approx 11.189 \times 10^3 m/s = 11.189 \text{ km/s}$ Rounding to one decimal place, $v_e \approx 11.2 \text{ km/s}$.

The final answer is $\boxed{11.2 kms^{-1}}$.