inverse square law

[ydoaPs]

take the inverse square law for gravity(or electric charges for that matter) [math]\vec{F}=G\frac{m_1m_2}{r^2}[/math]. r² has to be a dot product(if it wasn't, it would make gravity undefined for rxr is 0). mass is a scalar and so is the gravitational constant. so, how do we get a vector from a bunch of scalars?

[timo]

The forumla you present above (the right hand side) actually gives the magnitude of the force, not the whole vector. You have to multiply this magnitude by a direction unit vector ([math] \vec r / r [/math]) to get the vector. Also, r is not a vector but a distance: The magnitude of the displacement vector. So it´s a simply product of to reals in the denominator.

[swansont]

You have to multiply this magnitude by a direction unit vector ([math] \vec r / r [/math']) to get the vector.

 

And a minus sign, since the force is attractive.

[timo]

I originally wrote [math] \pm \vec r /r [/math] because nowhere I see a convention about signs (r=r1-r2 or r2-r1 ?) or naming (the force excerted on object one by object two is the negative of the force excerted on object two by object one - which is F?). But then, I left it out to avoid confusion.