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by itishappy 398 days ago
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
1 comments

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.