4 ms·

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

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 1y ago
> 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 1y ago
  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 1y 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
    - quantadev 1y ago
      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.

itishappy 1y ago

> 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 1y ago
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 1y 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
  - quantadev 1y ago
    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.