The earth's gravitational field intensity at its surface is about (G = 6.7 × 10−11 − 11 Nm2 2 /kg2 2 , mass of the earth is 6 × 1024 24 kg, radius of the ea...

The earth's gravitational field intensity at its surface is about

(G = 6.7 × 10${}^{-11}$Nm${}^2$/kg${}^2$, mass of the earth is 6 × 10${}^{24}$kg, radius of the earth is 6.4 × 10${}^6$m, g on the earth = 9.8m/s${}^2$)

Answer Details

The earth's gravitational field intensity at its surface can be calculated using the formula: g = G * M / r^2 where G is the gravitational constant, M is the mass of the earth, r is the radius of the earth, and g is the gravitational field intensity at the surface of the earth. Substituting the given values, we get: g = (6.7 × 10^-11 Nm^2/kg^2) * (6 × 10^24 kg) / (6.4 × 10^6 m)^2 g = 9.8 N/kg (approx.) Therefore, the answer is 9.8N/kg.