# Solving Kepler's 1st law as a function of time

• B
• jebez
In summary, the conversation discusses the poster's attempt to solve a differential equation using WolframAlpha and their question about purchasing Wolfram|Alpha Pro for better computation time. They also mention their goal of verifying Kepler's first law through solving the differential equation. Some solutions and equations are presented, but the conversation ultimately ends without a solution.

#### jebez

TL;DR Summary
p , ε , c constants :

θ'(t)(p/(1+ε cos(θ(t))))^2=c

θ(t)=?
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 ) post to Wolfram|Alpha Pro ? I don't want to buy Pro if it doesn't solve it ... Maybe someone who has Pro can post it ? Lend his/her account ?

In fact we have r(θ) = p/(1+ε cos(θ)) ( 1st Kepler's law ) , r²θ'(t)=c ( 2nd Kepler's law ) and I want r(t) so need to solve the differential equation .
https://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

Thanks .

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jebez said:
In fact we have r(θ) = p/{1 + ε cos(θ)) ( 1st Kepler law ) , r² θ'(t) = c ( 2nd Kepler law ) and I want r(t) so need to solve the differential equation .
https://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

Thanks .
Surprisingly, perhaps, there is no closed-form solution other than for circular orbits.

topsquark
jebez said:
TL;DR Summary: p , ε , c constants :

θ'(t) (p/(1 + ε cos(θ(t))))^2) = c

θ(t) = ?

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 ) post to Wolfram|Alpha Pro ? I don't want to buy Pro if it doesn't solve it ... Maybe someone who has Pro can post it ? Lend his/her account ?

In fact we have r(θ) = p/(1 + ε cos(θ)) ( 1st Kepler law ) , r² θ'(t) = c ( 2nd Kepler law ) and I want r(t) so need to solve the differential equation .
https://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion

Thanks .
Have a look at
https://www.wolframalpha.com/input?i=arccos(y(t))'=(a+b*+y(t))^2

topsquark
The equation is separable in $\theta$, so $$\int_{\theta(0)}^{\theta(t)} \frac{1}{(1 + \epsilon\cos\phi)^2}\,d\phi = \frac{ct}{p^2}.$$ But that's as far as you can go: there is no known antiderivative for the integrand when $\epsilon \neq 0$.

topsquark and PeroK
pasmith said:
The equation is separable in $\theta$, so $$\int_{\theta(0)}^{\theta(t)} \frac{1}{(1 + \epsilon\cos\phi)^2}\,d\phi = \frac{ct}{p^2}.$$ But that's as far as you can go: there is no known antiderivative for the integrand when $\epsilon \neq 0$.
What's wrong with my substitution ##y(t)=\cos\theta(t)##? Nasty, but it looks like it has an implicit solution.

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 M/r(t)²-r''(t)=0

r(t)=?
r(θ)=?

Having again r²θ'(t)=c ( 2nd Kepler's law ) , centrifugal force=m r θ'(t)² and gravity force=G M m/r² so
force=m r''(t)=centrifugal force - gravity force
m r''(t)=m c²/r(t)^3-G M m/r²
then above .

And same problem with WolframAlpha .

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The usual approach is to set $u = 1/r$ and $$\begin{split} \frac{dr}{dt} &= \frac{dr}{d\theta}\frac{d\theta}{dt} = \frac{L}{r^2}\frac{dr}{d\theta} = -L\frac{du}{d\theta} \\ \frac{d^2r}{dt^2} &= \frac{L}{r^2} \frac{d}{d\theta}\frac{dr}{dt} = -L^2u^2\frac{d^2 u}{d\theta^2} \end{split}$$ whence $$\frac{d^2u}{d\theta^2} + u = \frac{GM}{L^2}$$ which we can solve for $u(\theta)$.

fresh_42 said:
What's wrong with my substitution ##y(t)=\cos\theta(t)##? Nasty, but it looks like it has an implicit solution.

In fact WolframAlpha does give an antiderivative $$\int \frac{1}{(1 + \epsilon\cos\theta)^2}\,d\theta = \frac{\epsilon \sin \theta}{(\epsilon^2 - 1)(\epsilon \cos \theta + 1)} - \frac{2\operatorname{artanh}\left( \frac{(\epsilon-1)\tan(\frac12\theta)}{\sqrt{\epsilon^2-1}} \right)}{(\epsilon^2 - 1)^{3/2}} + C$$ but we then have the problem of solving that for $\theta$.

PeroK
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/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 cos(θ)(c(/1+ε cos(θ)))²/p^3=0

c²-p G M=0 as bonus .

Feel free to edit my posts to render equations , yes I can do it myself but it would be nice automatically ...

jebez said:
Feel free to edit my posts to render equations , yes I can do it myself but it would be nice automatically ...
Seriously?

Which word in the "LaTeX Guide" link below the Edit window did you have trouble understanding?

PhDeezNutz
jebez said:
I can't edit my previous messages to render equations , I'm just lazy to do that sorry and it isn't copyable ...
Please be sure to use LaTeX from now on here. Thank you.

- Copy your
jebez said:
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 M/r(t)²-r''(t)=0

As I suggested above, you should substitute $u = 1/r$ and eliminate $t$ as the independent variable in favour of $\theta$. This gives you $$u'' + u = \frac{GM}{L^2}$$ where $L = r^2 \dot \theta$ is constant. This is a linear ODE with constant coefficients which can be easily solved: $$r(\theta) =\frac{1}{u(\theta)} = \frac{L^2/GM}{1 + \epsilon\cos(\theta - \theta_0)}.$$

PeroK
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 to inverse t(θ) to θ(t) with WolframAlpha and Mathematica :

InverseFunction[(a^2 Integrate[(1/(1 + b Cos[x]))^2, x])/c]

but both didn't solve it , see https://community.wolfram.com/groups/-/m/t/2697668?p_p_auth=2VkMYTxA .

And ( maybe ) simplification :

t(θ)=(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 (b-1) tan(x/2) coth(x))/(b^2-1)^2+d))/c .

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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 θ'(t)=c((1+ε cos(θ(t)))/p)² aka y'(x)=c((1+b cos(y(x)))/a)² in Mathematica gives :

{{y(x)->InverseFunction[1/2 ((b sin(#1))/((b^2-1) (b cos(#1)+1))-(2 tanh^-1(((b-1) tan(#1/2))/Sqrt[b^2-1]))/(b^2-1)^(3/2))&][(c x)/(2 a^2)+Subscript[c, 1]]}}

$$\left\{\left\{y(x)\to \text{InverseFunction}\left[\frac{1}{2} \left(\frac{b (\text{\#1} \sin )}{\left(b^2-1\right) (b (\text{\#1} \cos )+1)}-\frac{2 ((b-1) (\text{\#1} \tan ))}{\left(b^2-1\right)^{3/2} \left(\left(2 \sqrt{b^2-1}\right) \tanh \right)}\right)\&\right]\left[\frac{c x}{2 a^2}+c_1\right]\right\}\right\}$$
What is # and & here ?

With Mathematica I can copy as LaTeX .

PeroK