[munchler]

In Step 2 (Gravity), why are we summing over the cube of the distance between the bodies in the denominator?

Edit: To answer myself, I think this is because one of the factors is to normalize the vector between the two bodies to length 1, and the other two factors are the standard inverse square relationship.

[itishappy]

You got it. The familiar inverse square formula uses a unit vector:

    a = G * m1 * m2 / |r|^2 * r_unit

    r_unit = r / |r|

    a = G * m1 * m2 / |r|^3 * r

[itishappy]

Oops, pretty big mistake on my part. I gave the formula for force and labeled it acceleration. Way too late to edit, but the correctly labeled formulas should have been:

    F = G * m1 * m2 / |r|^2 * r_unit
    F = G * m1 * m2 / |r|^3 * r

Or as acceleration, by dividing out an `m` using `F=ma`:

    a = G * m / |r|^2 * r_unit
    a = G * m / |r|^3 * r

[quantadev]

The force itself is `G * m1 * m2 / (r^2)`. That's a pure magnitude. The direction of the force is just the unit vector going from m1 to m2. You need it to be a unit vector or else you're multiplying up to something higher than that force. However, I don't get why you'd ever cube the 'r'. Never seen that. I don't think it's right, tbh.

[itishappy]

> I don't get why you'd ever cube the 'r'.

It's pulled out of the unit vector. Might be more clear if I notated the vector bits a bit:

    old    : new
    r      : r_vec
    |r|    : r_mag
    r_unit : r_dir

As you know, a vector is a magnitude and direction:

    r_dir = r_vec / r_mag

So the formulas from before become (also correctly labeled as `F` per my other comment):

    F = G * m1 * m2 / r_mag^2 * r_dir
    F = G * m1 * m2 / r_mag^2 * r_vec / r_mag
    F = G * m1 * m2 / r_mag^3 * r_vec

[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.