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:

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:

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:

Given that
,
and
, the magnitude of the gravitational field near Earth's surface is:


The magnitude of the gravitational field strength near Earth's surface is represented by approximately
.