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.

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 $G = 6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}$.

The formula for escape velocity ($v_e$) is:

$$v_e = \sqrt{\frac{2GM}{R}}$$

The mass $M$ is already in kilograms (kg).
The gravitational constant $G$ is already in $\text{Nm}^2\text{kg}^{-2}$.
The radius $R$ needs to be converted from kilometers to meters:

$$R = 6,371 \text{ km} = 6,371 \times 10^3 \text{ m} = 6.371 \times 10^6 \text{ m}$$

$$v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}) \times (5.98 \times 10^{24} \text{ kg})}{6.371 \times 10^6 \text{ m}}}$$ $$v_e = \sqrt{\frac{2 \times 6.67 \times 5.98 \times 10^{-11+24}}{6.371 \times 10^6}} \text{ m/s}$$ $$v_e = \sqrt{\frac{79.7732 \times 10^{13}}{6.371 \times 10^6}} \text{ m/s}$$ $$v_e = \sqrt{\left(\frac{79.7732}{6.371}\right) \times 10^{13-6}} \text{ m/s}$$ $$v_e = \sqrt{12.5212 \times 10^7} \text{ m/s}$$

To take the square root, we can rewrite $10^7$ as $10 \times 10^6$:

$$v_e = \sqrt{125.212 \times 10^6} \text{ m/s}$$ $$v_e = \sqrt{125.212} \times \sqrt{10^6} \text{ m/s}$$ $$v_e \approx 11.189 \times 10^3 \text{ m/s}$$

$$v_e \approx 11.189 \times 10^3 \text{ m/s} = 11.189 \text{ km/s}$$

Rounding to one decimal place, $v_e \approx 11.2 \text{ km/s}$.

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