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Can you derive Newton's law of gravitation from Kepler's third law, assuming an elliptical orbit? Most of what I've seen have been people solving it with a circular orbit. However, I find it impossible for an elliptical orbit because the radius changes as one object orbits a bigger one.

I first tried work backward: deriving Kepler's law from Newton's. I ended with the fact there are two components to acceleration (the one working with or against the velocity and the centripetal one), which can be shown as this $$ \vec{a} = v'\hat{T}+v^2κ\hat{N} $$ v and v' stands for velocity and the derivative of velocity, T_hat is the unit tangent vector, κ is the curvature, and N_hat is the normal vector.

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I ended up with this $$ -\frac{GMm}{r^2}\hat{r} = m\vec{a} $$ $$ -\frac{GM}{r^2}\hat{r} = \vec{a} $$ $$ -\frac{GM}{r^2}\hat{r}=(v'\hat{T}+v^2κ\hat{N}) $$ Because the velocity does simply equal 2πr/T, like most examples use: $$ \frac{GMm}{r^2} = m\frac{v^2}{r} $$ I, again, don't see how one can derive Newton's law of gravitation from Kepler's 3rd law. I feel like I am missing something; are my understanding and reasoning wrong? or is there a way to derive it?

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    $\begingroup$ i think it should have been the other way round, to derive all kepler's three laws from F=GMm/r^2 is doable. $\endgroup$
    – James
    Commented Dec 5, 2022 at 2:40
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    $\begingroup$ Most of what I've seen have been people solving it with a circular orbit. There are an infinite number of force laws that allow a circular orbit. $\endgroup$
    – Ghoster
    Commented Dec 5, 2022 at 4:06
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    $\begingroup$ Yes you can, that’s how Newton figured it out. Check out Feynman’s lost lecture where he reproduces Newton’s geometrical arguments (with a slight modification) $\endgroup$
    – LPZ
    Commented Dec 5, 2022 at 10:29
  • $\begingroup$ @lpz Feyman's reconstruction is for a demonstration that Kepler's first law narrows down the possibilities to an inverse square law of gravity. However, this question enquires about starting from Kepler's third law. $\endgroup$
    – Cleonis
    Commented Dec 5, 2022 at 18:45
  • $\begingroup$ yes sorry got them mixed up $\endgroup$
    – LPZ
    Commented Dec 5, 2022 at 22:08

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I start with quoting Kepler's third law, as stated in the Wikipedia article about Keplers three laws

The square of a planet's orbital period is proportional to the cube of the length of the semi-major axis of its orbit.

Here's the thing: when the orbit is eccentric the length of the semi-major axis is a particular average of perihelion and aphelion distances.

That is: in the way Kepler's third law is stated the orbit of the planet is mathematically circularized.

The way that the third law is stated does not use the actual orbit, the actual orbit is substituted with a circular orbit that has the same total energy.


So: to show that Kepler's third law narrows down the possibilities to an inverse square law of gravity: there is only one way to do that: you have to use circular orbits.

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