Recent content by jebez

Earth's elliptic orbit : assume(ε>0,ε<1); plot2d(2456084229048821270250000000/489238659320312440009*subst(ε=0.0167086,integrate(1/(1+ε*cos(θ))^2,θ)),[θ,-π,π]);

Erratum : https://en.wikipedia.org/wiki/Kepler_problem instead of https://en.wikipedia.org/wiki/Kepler's_equation in my #21 post . In Maxima : assume(ε>0,ε<1); integrate(1/(1+ε*cos(θ))^2,θ); assume(ε>0,ε<1,θ>-π,θ<π); solve(t=integrate(1/(1+ε*cos(θ))^2,θ),θ);

https://en.wikipedia.org/wiki/Kepler's_equation just gives at the end the Kepler's 1st law of my #17 post in general ( ##-\frac{k}{m}##instead of ##GM## ) . The problem is solved , we've r(θ) & t(θ) , the summary with " the conversation ultimately ends without a solution " needs to be edited ...

In : var('θ,ε,t') assume(0<ε<1) assume(-π<θ<π) from sage.symbolic.integration.integral import indefinite_integral latex(solve(t==indefinite_integral(1/(1+ε*cos(θ))^2,θ),θ)) Out : ##\left[\sin\left(θ\right) = \frac{\sqrt{-ε^{2} + 1} {\left(ε \cos\left(θ\right) + ε\right)} - \sqrt{-t^{2} ε^{8} +...

I can't edit my previous post ... There're (cos(θ)+1) as denominators , so we must precise : -π<θ<π

I'm back , sorry for long time . In short : ##L=r^2ω=constant## ( Kepler's 2nd law ) ##r=\frac{L^2}{GM(1+εcos(θ))}## ( Kepler's 1st law ) ⇒##\frac{1}{ω}=\frac{dt}{dθ}=\frac{L^3}{(GM(1+εcos(θ)))^2}## ⇔##t(θ)=\frac{L^3}{(GM)^2}∫\frac{1}{(1+εcos(θ))^2}dθ## ##∫\frac{1}{(1+εcos(θ))^2}dθ## ...

t(θ)=$$\frac{a^2 \left(\frac{b (x \sin )}{\left(b^2-1\right) (b (x \cos )+1)}-\frac{2 (b-1) (x \tan ) (x \coth )}{2 \left(b^2-1\right)^2}+d\right)}{c}$$ ( a=p , b=ε , c=c , d=t0 ) given by Mathematica ( I've 15 days trial ) , not WolframAlpha . There's also the differential equation...

I can't edit my messages #6 and #9 but r=c²/(G M(1+ε cos(θ))) , c²/r(t)³-G M/r(t)²-r''(t)=0 can be simplified : c²/r(t)-r(t)²r''(t)=G M c²G M(1+ε cos(θ))/c²-c⁴/(G M(1+ε cos(θ)))²ε cos(θ)(G M)³(1+ε cos(θ))²/c⁴=G M . Sorry pasmith but I don't understand from -L du/dθ in #7 ... Well I found how...

I found t(θ) : 1/θ'(t)=dt/dθ=t'(θ)=(p/(1+ε cos(θ)))²/c t(θ)=see https://www.wolframalpha.com/input?i2d=true&i=Divide[a²*antiderivative+\(40)Divide[1,1+b*cos\(40)x\(41)]\(41)²,c] . Now we need to from t(θ) to θ(t) , how to do this ? With WolframAlpha ? I can't edit my previous messages to...

I found r=p/(1+ε cos(θ(t)) is solution to c²/r(t)^3-G M/r(t)²-r''(t)=0 : r²θ'(t)=c θ'(t)=c/r² (dr/dθ)(dθ/dt)=dr/dt=v=r'(θ)c/r²=c ε sin(θ)/p v'(θ)=c ε cos(θ)/p (dv/dθ)(dθ/dt)=dv/dt=r''(t)=ε cos(θ)(c(/1+ε cos(θ)))²/p^3 c²/r(t)^3-G M/r(t)²-r''(t)=0 c²(1+ε cos(θ)/p)^3-G M(1+ε cos(θ)/p)²-ε...

Well sorry I thought equations will be rendered automatically like WolframAlpha , it seems to need LaTeX but I never used this and it seems tiresome ... In fact I want to verify if r(t) is solution to this differential equation to prove the Kepler's 1st law : c , G , M constants : c²/r(t)^3-G...

Hi I posted this differential equation to WolframAlpha https://www.wolframalpha.com/input?i2d=true&i=Power[\(40)Divide[a,1+b*cos\(40)y\(40)x\(41)\(41)]\(41),2]*y'\(40)x\(41)=c but no solution , " Standard computation time exceeded... Try again with Pro computation time " Should I ( buy and )...