Quick question about the gravitational law

[Kevin VD]

Aloha fellow forumdwellers,

“Long” time reader, first time poster reporting for duty! Got a small question which I hope you peeps can help me understand (since I'm someone who wants to understand how things work, not just blindly copy it from a notebook). Gonna start my first year of college physics in 4 months, and I've done some reading pure out of interest, but I can't seem to find an answer to my problem:

Why is “r” squared in Newton's Gravitation law (or Coulombs law for that matter)? I've been staring myself almost insane on the representation of 2 attracting particles, and can't find a logical answer to this! Can someone explain me why this is?

With the representation of 2 attracting particles I mean the picture on Wikipedia under Newton's law of universal gravitation.

Cheers,

Kevin


PS: sorry for my bad English, still working on it!

 

[Muphrid]

You can accept that masses or charged particles create fields. The relationship between the fields they create and the particles themselves is encoded in a simple equation:

\nabla F = \rho

That is, for an arbitrary source of field \rho, it creates a field F whose derivative is the soruce itself. In other words, the field is just the integral of the source.

When that source is a point particle, the solution in 3D follows a specific form.

F_{\text{point}} = \frac{1}{4\pi r^2} \hat r

You've seen this before, of course; it's possible to derive this solution as the point particle solution based on the properties of integrals and point sources. It's interesting to note that the 1/4\pi r^2 part is specific to 3D. In 2D, you get 1/2 \pi r instead. This actually appears in, for example, the electric field of a line charge.

So, really the only thing I've done is tell you that point particles' create very specific fields because that's what the differential equation says, but in doing so, another question arises: why do these laws of physics obey this differential equation at all? That's something I can't answer, and I don't think anyone can.