Sure, let's prove that the acceleration due to the gravity of the Earth, [tex]\( g \)[/tex], is inversely proportional to the square of its radius, [tex]\( R \)[/tex].
To start, we use Newton's law of universal gravitation, which states that any two masses attract each other with a force [tex]\( F \)[/tex] that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is:
[tex]\[ F = G \frac{M \cdot m}{R^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force,
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( M \)[/tex] is the mass of the Earth,
- [tex]\( m \)[/tex] is the mass of an object,
- [tex]\( R \)[/tex] is the distance between the centers of mass of the two objects, which, in this context, is the radius of the Earth.
Next, we think about the force experienced by an object of mass [tex]\( m \)[/tex] due to the gravity of the Earth, which we call the weight of the object. According to Newton's second law of motion, the weight [tex]\( W \)[/tex] of the object is the product of the mass [tex]\( m \)[/tex] and the acceleration due to gravity [tex]\( g \)[/tex]:
[tex]\[ W = m \cdot g \][/tex]
Since the weight [tex]\( W \)[/tex] is due to the gravitational force, we set [tex]\( F = W \)[/tex]:
[tex]\[ G \frac{M \cdot m}{R^2} = m \cdot g \][/tex]
We can simplify this equation by canceling out the mass [tex]\( m \)[/tex] from both sides:
[tex]\[ G \frac{M}{R^2} = g \][/tex]
This rearranges to:
[tex]\[ g = G \frac{M}{R^2} \][/tex]
From this final equation, we see that [tex]\( g \)[/tex] (the acceleration due to gravity) is directly proportional to the gravitational constant [tex]\( G \)[/tex] and the mass of the Earth [tex]\( M \)[/tex], and inversely proportional to the square of the Earth's radius [tex]\( R \)[/tex].
Therefore, we have the relationship:
[tex]\[ g \propto \frac{1}{R^2} \][/tex]
Thus, we proved that the acceleration due to gravity is inversely proportional to the square of the Earth's radius.
