homework and exercises - Free fall in non-uniform field

Imagine I'm a space-diver, with mass $m_1 $, 500 miles above the Earth's surface at $x_i$. I want to calculate my position, velocity, and acceleration as a function of time, accounting for the Earth's non-uniform gravitational field, and neglecting air resistance. I've done some basic calculations, and am confused by the answer I get; it seems to imply no acceleration as a function of time, if I start off at rest. Purely Newtonian regime. Here, I imagine I'm falling purely along the x-axis:

Conservation of energy: (Earth mass $m_2$, diver mass $ m_1$)

$$ \frac{1}{2}m_1 \dot{x}^2 = \frac{G m_2 m_1}{x} $$

Square root and integrating:

$$ \int_{x_i}^{x(t)} x^{1/2} dx = \int_0^t (2Gm_2)^{1/2} dt $$

This gives the solution

$$ x(t) = ( \frac{3}{2} (2Gm_2)^{1/2} t + x_i^{3/2} )^{2/3} $$

However, my problem with this is that it seems to imply that the velocity scales:

$$ \dot{x} \sim t^{-1/3} ,$$

but if I start off at rest at t = 0, it seems that my velocity will not increase, but decrease with time? What am I doing wrong here? Is there something wrong with my assumption that I can simply place myself at $ x_i $ with zero velocity? One would expect the diver's velocity to increase as a function of time, and for the gravitational field (since it will be stronger as I get closer to the surface of the Earth).

This should be straightforward, but I'm missing something?

Answer

Your mistake is in your conservation energy equation. The way you wrote it is valid only when falling from infinity, from rest. The correct is: $$dE=dK+dU=0,$$ that is $$mvdv=-\frac{K}{x^2}dx,$$ where $K\equiv Gm_1m_2$. Integrating from $(x_i,v_i)$ to $(x,v)$ we get $$\frac 12m(v^2-v_i^2)=K\left(\frac{1}{x}-\frac{1}{x_i} \right).$$ This is the correct equation which you have to start with. Now $$v=\frac{dx}{dt}=\pm\sqrt{v_{i}^2+\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{x_i} \right)}.$$ Assuming $v_i=0$ and integrating again from $(t=0,x_i)$ to $(t,x)$ we obtain $$t=-\int_{x_i}^{x(t)}\frac{dx}{\sqrt{\frac{2K}{m}\left(\frac{1}{x}-\frac{1}{x_i} \right)}},$$ where I am using the minus sign because the axis is oriented upwards. To solve this integral you use the substitution $x=x_i\sin^2{\theta}$, $$t=-\sqrt{\frac{2mx_i^3}{K}}\int_{\frac{\pi}{2}}^{\theta(x)}\sin^2{\theta}d\theta,$$ where $\theta(x)=\arcsin{\sqrt{\frac{x}{x_i}}}$. Therefore, $$t=\sqrt{\frac{mx_i^3}{2K}}\left[\frac{\pi}{2}-\arcsin{\sqrt{\frac{x}{x_i}}}-\frac 12 \sin\left(2\arcsin{\sqrt{\frac{x}{x_i}}}\right)\right].$$ However this equation cannot be solved for $x$.