Does this differential equation has analytic solution?

Question

I'm not competent in solving differential equations, so I've stuck in solving next equation that I've derieved:

$ Gm_\bigoplus v''(t)^2 = 4v(t)^2v'(t)^3 $, where $ v(0) = 0 $ and $ v'(0) = \frac{Gm_\bigoplus}{(R_\bigoplus - h_0)^2} $

I've asked Wolfram Alpha, but it understands me badly and gives no solutions. (Probably, I'm not a master of promts.)

Maybe some introduction of derivation of equation can help. So, I've thought about what if appreciate height about Earth when body falling down freely. I'm about use Newton's law of universal gravitation to get gravity acceleration at some height $h$. I've started with that $ h(t) = h_0 - \int_0^t v(x) \, dx $ and $ g(h) = \frac{Gm_\bigoplus}{(R_\bigoplus + h)^2} $. Next $ v(t) = \int_0^t g(h(x)) \, dx $. Differentiate once: $ v'(t) = g(h(t)) $. Expand: $ v'(t) = Gm_\bigoplus (R_\bigoplus + h_0 - \int_0^t v(x) \, dx)^{-2} $. Some movements to get: $ v'(t)^{-\frac{1}{2}} = \frac{R_\bigoplus + h_0 - \int_0^t v(x) \, dx}{\sqrt{Gm_\bigoplus}} $. Differentiate once: $ -\frac{1}{2}v''(t)v'(t)^{-\frac{3}{2}} = -\frac{v(t)}{\sqrt{Gm_\bigoplus}} $. Square both sides: $ \frac{v''(t)^2}{4v'(t)^3} = \frac{v(t)^2}{Gm_\bigoplus} $. Rearrange: $ Gm_\bigoplus v''(t)^2 = 4v(t)^2v'(t)^3 $. Initial conditions same as above.

P.S. I'm not a physician, so it's mostly for math understanding and less for practical applications. $ v(t) = gt $ is quite close approximation, though. (Ofc, it's standard free fall motion.)

Answer 1

Let's start with the equation before it was squared, namely $$ \frac{1}{2}(v')^{-3/2}v''=Kv, \tag{1} $$ where $K:=\frac{1}{\sqrt{Gm_\bigoplus}}$. Since $v''=\frac{dv'}{dt}=\frac{dv'}{dv}\frac{dv}{dt}=v'\frac{dv'}{dv}$, we can rewrite $(1)$ as $$ \frac{1}{2}(v')^{-1/2}\frac{dv'}{dv}=Kv. \tag{2} $$ Integrating $(2)$, and using the initial conditions $v(0)=0$, $v'(0)=a:=\frac{Gm_\bigoplus}{(R_\bigoplus - h_0)^2}$, we obtain $$ (v')^{1/2}=\frac{1}{2}Kv^2+a^{1/2} \implies v'=\frac{dv}{dt}=\left(\frac{1}{2}Kv^2+a^{1/2}\right)^2. \tag{3} $$ Integrating $(3)$, and again using the initial condition $v(0)=0$, we obtain $$ t=\int_0^v\frac{ds}{\left(\frac{1}{2}Ks^2+a^{1/2}\right)^2}. \tag{4} $$ Making the substitution $s=\beta\tan\theta$, where $\beta:=\sqrt{\frac{2a^{1/2}}{K}}$, we can rewrite $(4)$ as $$ t=\int_0^{\theta_v}\frac{\beta\sec^2\theta\,d\theta}{a(\tan^2\theta+1)^2} =\frac{\beta}{a}\int_0^{\theta_v}\cos^2\theta\,d\theta, \tag{5} $$ where $\theta_v:=\arctan(v/\beta)$. Going ahead with the integration, we obtain \begin{align} \frac{at}{\beta}&=\int_0^{\theta_v}\frac{1}{2}(\cos 2\theta+1)\,d\theta \\ &=\frac{1}{2}\left(\frac{1}{2}\sin 2\theta_v+\theta_v\right) \\ &=\frac{1}{2}(\sin\theta_v\cos\theta_v+\theta_v) \\ &=\frac{1}{2}\left(\frac{\tan\theta_v}{\sec^2\theta_v}+\theta_v\right) \\ &=\frac{1}{2}\left(\frac{\tan\theta_v}{1+\tan^2\theta_v}+\theta_v\right) \\ &=\frac{1}{2}\left(\frac{\beta v}{\beta^2+v^2}+\arctan\left(\frac{v}{\beta}\right)\right). \tag{6} \end{align} Equation $(6)$ gives $v$ as an implicit function of $t$. However, one can find approximate expressions for $v(t)$ in two limits:

• if $v\ll\beta$, the RHS of $(6)$ can be approximated by $\frac{v}{\beta}$, so $$ v\sim at\qquad(v\ll\beta); \tag{7} $$
• if $v\gg\beta$, the RHS of $(6)$ can be approximated by $\frac{\pi}{4}-\frac{1}{3}\left(\frac{\beta}{v}\right)^3$, so $$ v\sim\beta\left[3\left(\frac{\pi}{4}-\frac{at}{\beta}\right)\right]^{-1/3}\qquad(v\gg\beta). \tag{8} $$

It follows from $(8)$ that $v(t)$ blows up at $t=\frac{\pi\beta}{4a}$.