Question

Which of Kepler's laws gave Newton the idea about gravitational force and how it is related to distance?and how it is related to distance?

Answer 1

So if we say that there is some radial force f=f(r,....) pulling a planet in our system back to the Sun and so keeping it in orbit, and we balance the forces along that radial direction, then:

f=mω2r△

We know from Kepler's 3rd third law : T2∝r3 or T2=αr3.

We also know that period, T, and angular speed, ω, are related for circular motion as: T=2πω, so △ becomes:

f=m.4π2T2r=m.4π2αr3r=βr2, ie f∝1r2

That of course completely ignores Kepler's 1st Law , namely that planetary orbits are elliptical . The math becomes more turgid with ellipses, and certainly would have been for Newton as he was making it all up it as he went along.

Using post-Newton maths, and actually assuming (naughty) an inverse law F=−μmr2 er, with μ=GM, the Lagrangian in a polar co-ordinate plane tells us that:

..r−.θ2r=−μr2    ⋆ddt(mr2.θ)= const.    □

□ is just a statement that angular momentum L is preserved, and so we can say that |L|=mr2.θ⇒.θ=Lmr2= const.; and we then pop that into ⋆ to get:

..r=−μr2+L2m2r3    ∘

That's a non-linear DE; but the strategy I have borrowed - namely, we want a polar co-ordinate expression r=r(θ) which describes an ellipse - drives the solution, and that involves eliminating t from the equations. (I found this approach on a pretty ancient site that I will link at the bottom, I needed to translate it into more modern language.)

First we introduce a new fictional variable, u on terms that:

r=r(u(t))=1u(t) and u=u(θ(t))

We can do that, this is physics not maths, and we assume everything is sufficiently smooth, as it is in the physical world. In maths terms, we are, I think, assuming that these inter-related functions are invertible, so the chain rule is operating on full-power.

What's truly mind-blowing is that, thanks to Newton et al, we are now just pushing letters about according to certain rules, because it follows from the chain rule that:

drdt=drdu⋅dudt

=−1u2⋅dudθdθdt

=−1u2⋅dudθLmr2=−Lmdudθ

It then follows that:

d2rdt2=−Lmddt(dudθ)

=−Lmd2udθ2dθdt

=−Lmd2udθ2Lmr2

=−L2u2m2d2udθ2

Then ∘ becomes:

−L2u2m2d2udθ2=−μu2+L2m2u3

⇒d2udθ2+u=m2μL2

That's now trivial, it solves as:

u(θ)=Acos(θ−ϕ)+m2μL2

Or

r(θ)=1Acos(θ−ϕ)+m2μL2

hope it helps