To calculate the acceleration due to gravity, [tex]\( g \)[/tex], on the surface of the Earth, we use the formula:
[tex]\[
g = G \frac{M}{R^2}
\][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.7 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \)[/tex],
- [tex]\( M \)[/tex] is the mass of the Earth, [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex],
- [tex]\( R \)[/tex] is the radius of the Earth, [tex]\( 6.4 \times 10^6 \, \text{m} \)[/tex].
The given formula can be broken down into several steps for clarity:
1. Calculate the square of the Earth's radius:
[tex]\[
R^2 = \left(6.4 \times 10^6 \, \text{m}\right)^2
\][/tex]
The calculation of [tex]\( R^2 \)[/tex] yields:
[tex]\[
R^2 = 6.4 \times 10^6 \times 6.4 \times 10^6 = 40.96 \times 10^{12} \, \text{m}^2
\][/tex]
2. Multiply the gravitational constant by the Earth's mass:
[tex]\[
G \times M = 6.7 \times 10^{-11} \, \text{N m}^2 \text{kg}^{-2} \times 6 \times 10^{24} \, \text{kg}
\][/tex]
The calculation of [tex]\( G \times M \)[/tex] gives us:
[tex]\[
G \times M = 40.2 \times 10^{13} \, \text{N m}^2 \text{kg}^{-1}
\][/tex]
3. Divide the result from step 2 by the result from step 1 to find [tex]\( g \)[/tex]:
[tex]\[
g = \frac{G \times M}{R^2} = \frac{40.2 \times 10^{13} \, \text{N m}^2 \text{kg}^{-1}}{40.96 \times 10^{12} \, \text{m}^2}
\][/tex]
4. Simplify the division:
[tex]\[
g = \frac{40.2 \times 10^{13}}{40.96 \times 10^{12}} \, \text{N kg}^{-1} \text{m}^{-1}
\][/tex]
5. Perform the division:
The value simplifies to approximately:
[tex]\[
g = 9.814 \, \text{m/s}^2
\][/tex]
So, the acceleration due to gravity [tex]\( g \)[/tex] on the surface of the Earth is approximately [tex]\( 9.814 \, \text{m/s}^2 \)[/tex].
