(m = meters M = Mass)
6.67430 × 10^-11 N•m^2/kg^2 is obvious to intuitively decipher given the equation: F enclosed = G • M1 M2 / m^2 at the adjustment of multiplication and division between both equations to achieve units by themselves but why can gravitational constant also be adjusted to: m^3 / Kg / S^2
is it relevant to meter cubed volume per kg of density per acceleratory force? I'm just not sure..

Unit conversion can be done using the definition of Newtons. From Newton's second law of motion, we know that force is equal to mass times acceleration. One Newton is defined as the force needed to accelerate a mass of 1 kg by 1 m/s².

1 N = 1 kg m/s²

By substituting this, we can write the universal gravitational constant in terms of meters, kilograms, and seconds.

G = 6.67×10⁻¹¹ N m² / kg²

G = 6.67×10⁻¹¹ (kg m/s²) m² / kg²

G = 6.67×10⁻¹¹ m³ / kg / s²

To understand what these units mean, let's look at the gravitational equation:

F = G M m / r²

where

F is the gravitational force between two masses
G is the universal gravitational constant
M and m are the two masses
r is the distance between the masses

For illustration, let's say that M is the mass of a planet and m is the mass of an object on that planet. Thus, the gravitational force F is equal to the weight of the object, mg.

mg = G M m / r²

g = G M / r²

This equation tells us that the acceleration due to gravity, g, is a function of a planet's mass and radius. The universal gravitational constant G acts as a scale or multiplier between these measures. If we solve for G:

G = g r² / M

G = [m / s²] [m]² / [kg]

In the original equation, G related the mass of two objects and the distance between them to the force between those objects. By substituting Newton's second law, we can show that G also relates the acceleration that results from that force to the mass of the second object and the distance from it.