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$\displaystyle t = \frac{ \frac{\pi}{2} - \arcsin \Big( \sqrt{ \frac{x}{r} }\Big) + \sqrt{ \frac{x}{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu } } \, r^{3/2}$

This is an equation from Equations for a falling body for falling from height r to height x with a standard gravitational parameter μ, solving for time t. What I need help with is how to rewrite the equation to solve for x instead if I have time t, standard gravitational parameter μ and height r already.

I have been able to multiply both sides by the square root of 2μ and divide by r to the 1.5 power. Then I subtract $\pi/2$, but I am stuck with the square roots that have $x/r$ in them.

Localth
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Pyrania
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3 Answers3

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Rearranging: $$ t = \frac{ \cos^{-1}\Big( \sqrt{ \frac{x}{r} }\Big) + \sqrt{ \frac{x}{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu } } \, r^\frac32\mathop \iff^{\frac x r=w} \cos^{-1}(\sqrt w)+\sqrt{w(1-w)}=\frac{\sqrt{2\mu}}{r^\frac32}t=b$$

Applying the regularized beta integral representation:

$$\cos^{-1}(\sqrt w)+\sqrt{w(1-w)}=\frac\pi2\int_0^{1-w}\frac{\sqrt u}{\sqrt{1-u}}du=\frac\pi 2\operatorname 1-I_w\left(\frac32,\frac12\right)=b$$

Therefore Mathematica’s inverse beta regularized applied:

$$x=r\operatorname I^{-1}_{1-\frac{2 \sqrt{2u}}{\pi r^{3/2}}}\left(\frac12,\frac32\right),0\le \frac{2 \sqrt{2u}}{\pi r^{3/2}} \le 1$$

shown here. If $\frac xr=\cos^2\left(\frac y2\right)$ instead, you would get $y+\sin(y)=2b$, a special case of Kepler’s equation for which there are series/integral representations here

Тyma Gaidash
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Let $y=\sqrt\frac{x}{r}=\sin\left(\frac{\phi}{2}\right)$

$$\begin{align}t &= \frac{ \frac{\pi}{2} - \sin^{-1} y + y\sqrt{ 1 - y^2 } }{ \sqrt{ 2 \mu } } \, r^{3/2}\\ \frac{t\sqrt{2\mu}}{r^{3/2}} -\frac{\pi}{2} &= - \sin^{-1} y + y\sqrt{ 1 - y^2 }\\ \frac{t\sqrt{2\mu}}{r^{3/2}} -\frac{\pi}{2} &= \frac{\sin\phi-\phi}{2}\\ \frac{2t\sqrt{2\mu}}{r^{3/2}} -\pi=g(t,r,\mu)&=\sin\phi-\phi \end{align}$$

If you could solve for $\phi$ you'd get $$x(\phi)=r\sin^2\left(\frac{\phi}{2}\right)$$

For $-\frac{\pi}{2}\le\phi\le \frac{\pi}{2}$, or $0\leq x\leq \frac{r}{2}$, a good approximation using the Taylor series becomes

$$g(t,r,\mu)\approx-\frac{\phi^3}{6}\implies x\approx r\sin^2\left(\sqrt[3]{\frac{3\pi r^{3/2}-6t\sqrt{2\mu}}{4r^{3/2}}}\right)\quad\text{if}\;0\le\frac{x}{r}\le\frac{1}{2}$$

Note that increasing the upper bound of $\phi$ to $\frac{3\pi}{4}$ which gives $\frac{x}{r} \leq \frac{2+\sqrt{2}}{4}=0.853\dots$, could still give a reasonable result depending on your application

Localth
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Let $$k=\frac{\pi }{2}-\frac{\sqrt{2\mu} t}{r^{3/2}}\qquad \text{and}\qquad x=r\,y^2$$

The equation becomes $$k=\sin ^{-1}\left(\sqrt{y^2}\right)-\sqrt{y^2-y^4}\sin ^{-1}\left(\sqrt{y^2}\right)-\sqrt{y^2-y^4}$$

Considering the positive solution $$k=\sin ^{-1}(y)-y \sqrt{1-y^2}$$

Using Taylor series, the rhs is $$\sin ^{-1}(y)-y \sqrt{1-y^2}=\sum_{n=1}^\infty a_n\, y^{2n+1}$$ where the first coefficients are

$$ \left\{\frac{2}{3},\frac{1}{5},\frac{3}{28},\frac{5}{72},\frac{35}{ 704},\frac{63}{1664},\frac{77}{2560},\frac{429}{17408},\frac{643 5}{311296},\frac{12155}{688128}\right\} $$

but this would require solving very high degree polynomial equations.

But, the simplest Padé approximant $$ \sin ^{-1}(y)-y \sqrt{1-y^2}\sim \frac{20 y^3}{30-9 y^2}$$ whose error is $\frac{33 y^7}{700}$ reduces, for an approximation, the problem to a cubic equation in $y$ $$20 y^3+ 9k y^2-30 k=0$$ whose discriminant is negative; so, only one real root. For $0 \leq k\leq \frac 12$ $$y=\frac{3}{20} k \left(2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{2000}{9 k^2}-1\right)\right)-1\right)$$ is a decent approximation of the solution.

For better, we can use more terms for the Taylor series and then use power series reversion to obtain

$$y=\sum_{n=0}^\infty b_n\, u^{2n+1}\qquad \text{where} \qquad u=\sqrt[3]{ \frac{3k}2}$$ the first coefficients being $$\left\{1,-\frac{1}{10},-\frac{19}{1400},-\frac{71}{25200},-\frac{17 753}{25872000},-\frac{1312063}{7207200000},-\frac{647915701}{127 13500800000}\right\}$$

A few numbers $$\left( \begin{array}{ccc} k & \text{estimate} &\text{solution} \\ 0.1 & 0.515718 & 0.515718 \\ 0.2 & 0.637418 & 0.637418 \\ 0.3 & 0.717212 & 0.717212 \\ 0.4 & 0.776602 & 0.776601 \\ 0.5 & 0.823347 & 0.823346 \\ 0.6 & 0.861235 & 0.861232 \\ 0.7 & 0.892424 & 0.892416 \\ 0.8 & 0.918267 & 0.918248 \\ 0.9 & 0.939666 & 0.939627 \\ 1.0 & 0.957253 & 0.957180\\ 1.1 & 0.971482 & 0.971350 \\ 1.2 & 0.982685 & 0.982458 \\ 1.3 & 0.991109 & 0.990734 \\ 1.4 & 0.996938 & 0.996338 \\ \end{array} \right) $$

Claude Leibovici
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