Problem of the Week # 154 - March 9, 2015

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In summary, the conversation is about the different types of fruit and their nutritional benefits. They discuss how fruits can be sweet or sour, and how some fruits are high in vitamins and antioxidants. They also mention the importance of eating a variety of fruits for a well-balanced diet.
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Ackbach
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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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  • #2
No one answered this week's POTW. You can view my solution below:

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
$$\sqrt{\frac{r_0}{2 G M_e}}\frac{1}{\sqrt{r_0-r}}\left[\sqrt{r}(r-r_0)
+r_0\sqrt{r_0-r}\arctan\left(\sqrt{\frac{r}{r_0-r}}\,\right)\right]=
t+\frac{\pi r_0^{3/2}}{2\sqrt{2 G M_e}}.$$
 

FAQ: Problem of the Week # 154 - March 9, 2015

What is the "Problem of the Week # 154"?

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