Free fall in non-uniform field

Question

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 1

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 $x$.