Deducing Kepler's second law from Newton's laws?

[vinicius0197]

I've searched a little bit and found that I can derive kepler's third law from Newton's law of gravitation. That's okay. But I want to deduce kepler's second law too: "An imaginary line joining a planet and the sun sweeps out an equal area of space in equal amounts of time".
I know it's possible to do that using angular momentum, and from there proving that angular momentum in this case is constant. But how would I do that?

 

[brainpushups]

How do you feel about going straight to the source? See propositions 1 and 2:

https://en.wikisource.org/wiki/The_Mathematical_Principles_of_Natural_Philosophy_(1846)/BookI-II

 

[Korak Biswas]

I've searched a little bit and found that I can derive kepler's third law from Newton's law of gravitation. That's okay. But I want to deduce kepler's second law too: "An imaginary line joining a planet and the sun sweeps out an equal area of space in equal amounts of time".
I know it's possible to do that using angular momentum, and from there proving that angular momentum in this case is constant. But how would I do that?

I've searched a little bit and found that I can derive kepler's third law from Newton's law of gravitation.
That's okay. But I want to deduce kepler's second law too: "An imaginary line joining a planet and the sun sweeps out an equal area of space in equal amounts of time".
I know it's possible to do that using angular momentum, and from there proving that angular momentum in this case is constant. But how would I do that?

Keep the sun at the origin. At any time t let $\vec { r(t)}$ be the position vector of a planet. At the moment t+dt the position vector will be $\vec {r (t)}+\vec{dr (t)}$. So within the interval of time dt the planet sweeps out an area $\vec{da}= \frac {1}{2} \vec{r (t)}\times\vec{dr (t)} \Rightarrow \vec{\frac{da}{dt}}= \frac {1}{2} \vec{r (t)}\times\vec{\frac{dr}{dt}}$ Now you know that Newton's gravitational force is central in nature or mathematically $\frac {d^2\vec{r}}{dt^2} = f (r)\vec{r} \Rightarrow \vec{r}\times \frac {d^2\vec{r}}{dt^2} = \vec{0}\Rightarrow \frac {d}{dt}(\vec{r (t)}\times\vec{\frac{dr}{dt}})=\vec{0}\Rightarrow \frac{d}{dt}(\vec{\frac{da}{dt}})=\vec{0}$ So, $\vec{\frac{da}{dt}}$ is a constant vector. This is Kepler's second law.