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What is the free-fall time of a test mass on a mass $M$ from height $2R$ to $R$?

My Solution

The equation of motion of the test mass is $$\Delta{R}=-gt\Delta{t}$$ which on integration gives $$R={1\over2}gt^2$$ Then free-fall time $$\tau=\sqrt{2R\over g}$$ But acceleration due to gravity of a body of radius $R$ is $$g=\frac{GM}{R^2}$$ Therefore the free-fall time is $$\tau=\sqrt{2R^3\over GM}$$

However, since I don't know the radius of the mass $M$ , I'm using the expression $$\tau=\sqrt{y_0^3\over GM}\times\left[\sqrt{{y\over y_0}{\left(1-{y\over y_0}\right)}}+\cos^{-1}\sqrt{y\over y_0}\right]$$ With $y_0=2R,$ and $y=R$ the free-fall time I thus obtain is $$\tau=\left({{2\pi\over 3} +1}\right)\sqrt{R^3\over GM}$$

It does not feel right!

Sayontön Vöttacharjo
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1 Answers1

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The question you pose is actually much harder than you probably originally thought. The reason is you are trying to find the time elapsed in a time-varying gravitational field, remember the test mass (mass $m$) is not undergoing a constant acceleration, but it is varying, since $g = \frac{GM}{r^2}$ and $r$ is of course changing. However, you can still solve this problem. Since the gravitational force is conservative, we can apply the conservation of energy: $$\frac{1}{2} m v_f^2 -\frac{GMm}{r_f} = \frac{1}{2} m v_0 ^2 - \frac{GMm}{r_0}.$$ This is just the sum of the kinetic and potential energies initially and finally. Assume that the test mass (mass $m$) starts from rest, so that $v_0 = 0$, and our equation simplifies to $$ v_f^2 = 2GM\Big(\frac{1}{r_f}-\frac{1}{r_0}\Big).$$ Ok that is some progress but what we want is the time, not the velocity! Note however the definition of the velocity, which is $v \equiv \frac{dr}{dt}$, just the derivative of the position, so we get $$ \frac{dr}{dt} = \pm \sqrt{2GM \Big(\frac{1}{r_f}-\frac{1}{r_0}\Big)},$$ where I have taken the square root of both sides. Now you can notice, all we have to do is perform an integral to find the total time $\tau$ as you call it, but we need an integral variable, so instead of writing $r_f$, we write $r$ since we are integrating from our initial $r_0 = 2R$ to our final $r_f = R$. Ok, inverting the equation, we get $$\frac{dt}{dr} = \pm \frac{1}{\sqrt{2GM}}\frac{1}{\sqrt{\frac{1}{r}-\frac{1}{r_0}}}$$ By the way, I'm going to define $\mu \equiv GM$, which is something called the standard gravitational parameter, for convenience (and mainly because I am lazy and don't want to write it all the time). Keeping only the minus sign (so that the total time $\tau$ remains positive as you will see), we get the following integral: \begin{align} \tau &= \frac{1}{\sqrt{2\mu}}\int_{r_0}^{r_f} \frac{dr}{\sqrt{\frac{1}{r}-\frac{1}{r_0}}} \\ &= \frac{1}{\sqrt{2\mu}}\int_{2R}^{R} \frac{dr}{\sqrt{\frac{1}{r}-\frac{1}{2R}}}. \end{align} The rest of the problem is just performing this integration, which I will leave up to you. I hope this clarifies everything!

Josh Pilipovsky
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