Answer:
The magnitude of the gravitational field strength near Earth's surface is represented by approximately [tex]9.82\,\frac{m}{s^{2}}[/tex].
Explanation:
Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:
[tex]F = G\cdot \frac{M\cdot m}{r^{2}}[/tex]
Where:
[tex]M[/tex] - Mass of the planet Earth, measured in kilograms.
[tex]m[/tex] - Mass of the person, measured in kilograms.
[tex]r[/tex] - Radius of the Earth, measured in meters.
[tex]G[/tex] - Gravitational constant, measured in [tex]\frac{m^{3}}{kg\cdot s^{2}}[/tex].
But also, the magnitude of the gravitational field is given by the definition of weight, that is:
[tex]F = m \cdot g[/tex]
Where:
[tex]m[/tex] - Mass of the person, measured in kilograms.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
After comparing this expression with the first one, the following equivalence is found:
[tex]g = \frac{G\cdot M}{r^{2}}[/tex]
Given that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.972 \times 10^{24}\,kg[/tex] and [tex]r = 6.371 \times 10^{6}\,m[/tex], the magnitude of the gravitational field near Earth's surface is:
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}[/tex]
[tex]g \approx 9.82\,\frac{m}{s^{2}}[/tex]
The magnitude of the gravitational field strength near Earth's surface is represented by approximately [tex]9.82\,\frac{m}{s^{2}}[/tex].
