A

Step 1: We start with the formula for gravitational potential due to a mass \( M \):
\( V = -\frac{GM}{r} \)
where \( G \) is the gravitational constant (approximately \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \text{s}^{-2} \)), \( M \) is the mass of the Earth, and \( r \) is the radius of the Earth.
Step 2: Calculate the mass of the Earth using its volume and density.
The volume of the Earth is given by \( V = \frac{4}{3} \pi r^3 \), where \( r = 6.4 \times 10^6 \, \text{m} \).
Thus, \( V = \frac{4}{3} \pi (6.4 \times 10^6)^3 \approx 1.083 \times 10^{21} \, \text{m}^3 \).
Step 3: Now calculate the mass \( M \) using density \( \rho \) where \( \rho = 5.5 \times 10^{3} \, \text{kg/m}^3 \):
\( M = \rho V = 5.5 \times 10^3 imes 1.083 \times 10^{21} \approx 5.95 \times 10^{24} \, \text{kg} \).
Step 4: Substitute \( M \) and \( r \) into the gravitational potential formula:
\( V = -\frac{(6.674 \times 10^{-11})(5.95 \times 10^{24})}{6.4 \times 10^6} \).
Step 5: Calculate \( V \): \( V \approx -9.81 \, \text{J/kg} \) (considering the negative sign since it is gravitational potential).
Therefore, the gravitational potential at the Earth's surface is approximately -9.81 J/kg.