Vulcan's fall

This is an exercise from "Newtonian Mechanics" by French, chapter 3, exercise 8.

Question

In Paradise Lost, Book I, John Milton describes the fall of Vulcan from Heaven to earth in the following words:

. . . from Morn
To Noon he fell, from Noon to dewy Eve,
A Summer's day ; and with the setting Sun
Dropt from the Zenith like a falling Star ....

1. Clearly air resistance can be ignored in this trip, which was mostly through outer space. If we assume that the acceleration had the value g (\(9.8 m/s^2\)) throughout, how high would Heaven be according to Milton's data? What would have been Vulcan's velocity upon entering the top of the atmosphere?
2. (Much harder) One really should take account of the fact that the acceleration varies inversely as the square of the distance from the earth's center. Obtain revised values for the altitude of Heaven and the atmospheric entry speed.

Answer

1. The time, \(t\), of Vulcan's fall is given as 17.5 hours in the answers on page 724, which is \begin{align*} t&=17.5 \mbox{ hours}\times60 \mbox{ minutes/hour}\times60 \mbox{ seconds/minute} \\ &=63,000 \mbox{ seconds}. \end{align*} Assuming constant acceleration, the distance to Heaven \begin{equation*} s=\frac12gt^2=0.5\times9.8\times63,000^2 \mbox{m} \approx 1.9\times10^{10}\mbox{m}, \end{equation*} about 19 million kilometers. (The book's answer on page 724 is out by a factor of ten.) Vulcan's velocity would be \(at=9.8\mbox{ms}^{-2}\times63,000\mbox{s}\approx 620\mbox{km/s}\), as given in the book.
2. The time of fall is as above, 63,000 seconds. The equation of motion in this case is \(F=ma=mMG/r^2\) which gives us a second-order non-linear differential equation \(r''(t)=k/r(t)^2\), where \(k=MG\). This equation's solution is fairly complex, and French doesn't discuss where he got the figure in the answers from. But we can relatively easily get the velocity using conservation of energy, although this wasn't covered in chapter 3. Vulcan's kinetic energy by the time he reaches the earth, at radius \(R\), is equal to \begin{equation*} \frac12mv^2=\int_{r_0}^R GmM/r^2 dr \end{equation*} so \(v=\sqrt{2GM\left(1/R-1/r_0\right)}\approx\sqrt{2GM/R}\) where, since \(r_0\) must be very large, since he fell for such a long time, that we can neglect the \(1/r_0\) term. Since \(GM/R^2=g\), the speed \(v=\sqrt{2gR}=\sqrt{2\times9.8ms^{-2}\times6.38\times10^{6}\mbox{m}}\approx 11\) km/s. (This answer is slightly different from the one given by French.)