Derivation of Kepler's first law



An adaptation/explanation of Stewart’s derivation of Kepler’s First Law from Newton’s Law of Gravitation

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Md.Rezaul Islam

First i see this proof it seems quite difficult now its clear,thanks.

Fenghua Liu

The first five minutes is essentially conservation of angular momentum right?

Efe Güleroğlu

By far the best!

Glen Check

Hi thanks for the clear explanation from S.K. !!

i have one question.
doesnt' c^ needs to be parallel to r^?
v x h // u ( because v⊥u and h⊥u)
if you plus u^ with c^ (whatever that isnt parellel to u^)
u^+c^ is not parellel to v^ x h^ anymore In your video graph,

John Williams

how do we know that c lies in the same plane as r and v?

Egor Timatkov

Wow, this is so awesome

Gonza2323

Great explanation! I just don't get what the c vector is. Is there anyway to calculate it in terms of the initial conditions?

Khushbu Raza

Hi! Quick question: I am a junior in high school currently taking Math Pre-Calc HL, so I havent learned calculus yet. I wanted to write an exploration/research paper on deriving and applying Kepler's 3 laws but bc i didnt learn calc, im having trouble understanding online proofs of it. do you think i could still learn these proofs, including yours, easily if i learn very simple calculus? or should I just try to find another topic on my level?

Joseph Baldwin

i love you

Samuel He

I don't understand at 3:55 why this term equals 0, can you explain me why we get that result ?

jdtaramona

thanks once again David. Despite being too tiny this extra quantity the derivation is a very interesting matter

jdtaramona

Thanks a lot once again David, maybe one of the most beautiful solutions in the history of physics/mathematics. Maybe I did not understand very well the physical meaning of vector C yielded during integration of motion equation (V x h)' = GM.U'. I suspect that C is the eccentricity vector. On the other hand, some books include an additional angle substracted from angle theta (true anomaly) as an argument of cosine function, and I do not know which angle is that.. r=p/1+ecos(theta-phi), where p is semi-lactus rectum and phi and C as well are determined by the initial conditions.
Finally, Have you thought to derive this Kepler law for a more general case as a trajectory of a massive particle in the schwarzschild geometry in the same way as yo have done here by means of vectors?

laurio jenah

that was difficult but you made it understandable 

Tomas Šiaulys

really clear… thank you! 🙂