MHB Problem of the Week # 154 - March 9, 2015

[Ackbach]

Here is this week's POTW:

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In freshman physics, or high school physics, you analyzed free-fall motion by assuming that the gravitational force was a constant, $mg$, from which you can use the constant-acceleration kinematic equations. Let's improve on this approximation one notch by introducing a more accurate gravitational force:
$$F=\frac{GM_{e}m}{r^2},$$
Newton's Universal Law of Gravitation. Here $G$ is the gravitational constant, $M_e$ is the mass of the earth, $m$ is the mass of the object in the next sentence, and $r$ is the distance from the center-of-mass of the Earth to the center-of-mass of the object. Assuming that the object starts from rest at some initial distance $r_0$ from the center of the earth, find its position as a function of time. You may neglect air resistance, the dependence of pressure on height, the Coriolis effect, etc. That is, for this model, the only force present is gravity.

[EDIT] You may obtain an implicit result.

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[Ackbach]

No one answered this week's POTW. You can view my solution below:

Spoiler

We have that
$$F=-\frac{G M_e m}{r^2}=m \ddot{r}.$$
Multiplying both sides by $\dot{r}$ and canceling the $m$'s yields
$$-G M_e r^{-2} \dot{r}=\ddot{r} \, \dot{r}.$$
Integrating once yields
\begin{align*}
-GM_e \int r^{-2} \dot{r} \, dt &=\int \dot{r} \ddot{r} \, dt \\
\frac{GM_e}{r}+C&=\frac{\dot{r}^2}{2}.
\end{align*}
Plugging in the initial conditions yields
$$\frac{GM_e}{r_0}+C=0 \quad \implies \quad
C=-\frac{GM_e}{r_0}.$$
Hence, our DE is now
$$\frac{GM_e}{r}-\frac{GM_e}{r_0}=\frac{\dot{r}^2}{2},$$
or
$$\dot{r}=\sqrt{2GM_e\left(\frac{1}{r}-\frac{1}{r_0}\right)}.$$
We can separate variables as follows:
$$\frac{dr}{\sqrt{2GM_e\left(\frac{1}{r}-\frac{1}{r_0}\right)}}=dt.$$
When we perform these integrals, we obtain
$$\sqrt{\frac{r_0}{2 G M_e}}\frac{1}{\sqrt{r_0-r}}\left[\sqrt{r}(r-r_0)
+r_0\sqrt{r_0-r}\arctan\left(\sqrt{\frac{r}{r_0-r}}\,\right)\right]=t+C.$$
The LHS is not one in which we can simply plug in $r_0$ for $r$, but we can
do the limit as $r\to r_0$. This yields
$$\frac{\pi r_0^{3/2}}{2\sqrt{2 G M_e}}=C,$$
and thus our implicit expression for $r$ as a function of $t$ is given by
$$\sqrt{\frac{r_0}{2 G M_e}}\frac{1}{\sqrt{r_0-r}}\left[\sqrt{r}(r-r_0)
+r_0\sqrt{r_0-r}\arctan\left(\sqrt{\frac{r}{r_0-r}}\,\right)\right]=
t+\frac{\pi r_0^{3/2}}{2\sqrt{2 G M_e}}.$$