What is the gravitational field strength on the surface of the Earth? $\left(\mathrm{G}=6.67 \times 10-11 \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}\right.$, the mass of the Earth is $5.98 \times 10^{24} \mathrm{~kg}$, and the radius of the Earth is $6.38 \times 10^{6} \mathrm{~m}$.)
A) $10.1 \mathrm{~N} / \mathrm{kg}$.
B) $6.20 \mathrm{~N} / \mathrm{kg}$.
C) $20.3 \mathrm{~N} / \mathrm{kg}$.
D) $9.80 \mathrm{~N} / \mathrm{kg}$.
E) $3.40 \mathrm{~N} / \mathrm{kg}$.

Question

Quizplus · Accepted Answer

The gravitational field strength on the surface of the Earth can be calculated using the formula $g = \frac{GM}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the radius of the Earth. Substituting the given values, $g = \frac{(6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2)(5.98 \times 10^{24} \, \text{kg})}{(6.38 \times 10^6 \, \text{m})^2}$ results in approximately $9.80 \, \text{N/kg}$.

What Is the Gravitational Field Strength on the Surface of the Earth