Answer:
The magnitude of the gravitational field strength near Earth's surface is represented by approximately
.
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:
![F = G\cdot \frac{M\cdot m}{r^{2}}](https://tex.z-dn.net/?f=F%20%3D%20G%5Ccdot%20%5Cfrac%7BM%5Ccdot%20m%7D%7Br%5E%7B2%7D%7D)
Where:
- Mass of the planet Earth, measured in kilograms.
- Mass of the person, measured in kilograms.
- Radius of the Earth, measured in meters.
- Gravitational constant, measured in
.
But also, the magnitude of the gravitational field is given by the definition of weight, that is:
![F = m \cdot g](https://tex.z-dn.net/?f=F%20%3D%20m%20%5Ccdot%20g)
Where:
- Mass of the person, measured in kilograms.
- Gravity constant, measured in meters per square second.
After comparing this expression with the first one, the following equivalence is found:
![g = \frac{G\cdot M}{r^{2}}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7BG%5Ccdot%20M%7D%7Br%5E%7B2%7D%7D)
Given that
,
and
, the magnitude of the gravitational field near Earth's surface is:
![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}}](https://tex.z-dn.net/?f=g%20%3D%20%5Cfrac%7B%5Cleft%286.674%5Ctimes%2010%5E%7B-11%7D%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkg%5Ccdot%20s%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%285.972%5Ctimes%2010%5E%7B24%7D%5C%2Ckg%29%7D%7B%286.371%5Ctimes%2010%5E%7B6%7D%5C%2Cm%29%5E%7B2%7D%7D)
![g \approx 9.82\,\frac{m}{s^{2}}](https://tex.z-dn.net/?f=g%20%5Capprox%209.82%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
The magnitude of the gravitational field strength near Earth's surface is represented by approximately
.