How do you derive the acceleration due to gravity at the Earth’s surface (g = GM/R²) with a labelled diagram and step-by-step proof?

What you are trying to show

You want a formula for the acceleration a freely falling object gets near Earth’s surface. The idea is to start with the gravitational force between Earth and the object, then connect that force to $F=ma$.

Labelled diagram (object at Earth’s surface)

 m (object)
 ●
 |
 | r = R (distance from Earth's center)
 |
 |
 (gravity force)
 ↓ F_g

 Earth (mass M)
 ___________
 / \
 | ● | ● = center of Earth
 \___________/
 R = Earth’s radius

Here, the object of mass $m$ is on the surface, so its distance from Earth’s center is $r=R$.

Using Newton’s law of gravitation

Newton’s law says the gravitational force between two masses $M$ and $m$ separated by distance $r$ is $$F_g = \frac{GMm}{r^2}$$ At the Earth’s surface, $r=R$, so $$F_g = \frac{GMm}{R^2}$$

Connecting gravitational force to acceleration (weight)

The same force is the object’s weight, and by Newton’s second law: $$F_g = ma$$ For free fall near Earth, the acceleration is $a=g$, so $$F_g = mg$$

Equate the two expressions and solve for g

Set the two formulas for the same force equal: $$mg = \frac{GMm}{R^2}$$ Cancel $m$ from both sides: $$g = \frac{GM}{R^2}$$ This is the required result for acceleration due to gravity at Earth’s surface.

Quick numerical check (optional)

Using $G=6.67\times 10^{-11}\,\text{N·m}^2\!\!/\text{kg}^2$, $M\approx 5.97\times 10^{24}\,\text{kg}$, $R\approx 6.37\times 10^6\,\text{m}$: $$g = \frac{(6.67\times 10^{-11})(5.97\times 10^{24})}{(6.37\times 10^6)^2} \approx 9.8\,\text{m/s}^2$$