The earth's gravitational field intensity at its surface is about \(G = 6.7 \times 10^{-11}\,\mathrm{Nm^2/kg^2}\), mass of the earth is \(6 \times 10^{24}\,...
Question 1 Report
The earth's gravitational field intensity at its surface is about
\(G = 6.7 \times 10^{-11}\,\mathrm{Nm^2/kg^2}\), mass of the earth is \(6 \times 10^{24}\,\mathrm{kg}\), radius of the earth is \(6.4 \times 10^6\,\mathrm{m}\), g on the earth = \(9.8\mathrm{m/s^2}\)
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.
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