[quantadev]

Ok, I see what you're doing. Your multiplying the force vector by a non unit-vector, and then dividing back out the linear amount to correct for it. You never see this in a physics book because it's a computational hack, probably because it saves you the CPU cost of not having to do the 3 division operations it takes to get each component (X,Y,Z) of the unit vector.

This makes sense to do in computer code also because if you were going to raise r_mag to a power, you might as well raise it to 3 instead of 2, because it's not extra cost, but you do avoid the three divisions, by never calculating a unit vector. Back when I was doing this work, was decades ago and I had no idea about cost of floating points. Thanks for explaining!

[itishappy]

Glad I could help!

Also fun is that taking the magnitude involves a square root that can sometimes be avoided, but that doesn't really help us here because of the power of three. If the denominator were squared we could just use `r_mag^2 = r_x^2 + r_y^2`, but we still need the root to get the direction. It is kinda interesting though that in 2d it expands to a power of `3/2`:

    F_vec = G * m1 * m2 / (r_x^2 + r_y^2) ^ (3/2) * r_vec

[quantadev]

Yeah, on paper (or mathematical symbolics) it comes down to what's more clear and representing reality. That's why I initially said I know there's no cubic relations in the physics of this, which was correct.

But that doesn't mean that therefore there's no correct physics equations (for gravity) involving the cube of a distance, even when there's only squares in these "laws" of physics.

In both cases the power of 2, as well as 3/2, is there merely to "cancel out" the fact that you didn't use a unit vector (in the numerator) and therefore need to divide that out in the denominator, to end up scaling the force magnitude against a unit vector.