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The main text of the problem...
The rare NASA photo above demonstrates cows are in fact able to jump over the
moon.
The energy required to launch an object from the Earth is given by the following
equation:
$E_L = GM_Em\left(\frac{1}{r_0} - \frac{1}{r_1} \right)$
Where:
\begin{eqnarray*}
G & = & {\rm The \;Universal\; Gravitational\; Constant}\\
M_E & = & {\rm Earth's\; Mass}\\
m & = & {\rm The \;object's\; mass}\\
r_0 & = & {\rm Starting \;radius}\\
r_1 & = & {\rm Ending \;radius}
\end {eqnarray*}
listing different parts of the problem...
\begin{itemize}
\item[a. ] How much energy must a 450 kg cow expend in jumping over the moon?
(Don't worry about the moon's gravity or the Earth's atmosphere...)
\item[b. ] Assuming all the work is done during the jump itself, what would
her initial velocity have to be?
\item[c. ] What would her final velocity be upon her return to Earth?
\item[d. ] Again assuming all the work would be done during the initial jump, i.e.
for a distance approximately the length of her legs (1 meter), what would her
acceleration need to be?
\end{itemize}
Each section and subsection becomes a link to a seperate document.
(refer to the final product)
This way students would be able to look at hints (and solutions) individually.
\section* {Hints}
\subsection*{Hint 1}
The distance from the Earth to the moon is about $3.85 \times 10^8 \; m$
\subsection*{Hint 2}
$r_0$ is not zero.
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The solution for each part of the problem will appear on its own page
\section* {Solution}
\subsection*{Part a.}
$E_L = GM_Em\left(\frac{1}{r_0} - \frac{1}{r_1} \right)$
Where:
\begin{eqnarray*}
G & = & 6.672 \times 10^{-11}\; N \cdot m^2/kg^2\\
M_E & = & 5.98 \times 10^{24}\; kg\\
m & = & 450\; kg\\
r_0 & = & 6.38 \times 10^{6} \; m\\
r_1 & = & 3.85 \times 10^8 \; m
\end {eqnarray*}
So...
\begin{eqnarray*}
E_L & = & (6.672 \times 10^{-11}\; N \cdot m^2/kg^2)(5.98 \times 10^{24}\;kg)
(450\; kg)\left(\frac{1}{6.38 \times 10^{6} \; m} - \frac{1}{3.85 \times 10^8 \; m}
\right)\\
& = & {\bf 2.77 \times 10^{10} \; J}
\end{eqnarray*}
Which by the way is roughly equivalent to the energy given off by 6.5 tons of TNT...
Part b...
\subsection*{Part b.}
...So the jump gives her $2.77 \times 10^{10} \; J$ of kinetic energy...
\begin{eqnarray*}
KE & = & \frac{1}{2} m v^2\\
v & = & \sqrt{\frac{KE}{2m}}\\
& = & \sqrt{\frac{2.77 \times 10^{10} \; J}{2 \cdot 450\; kg}}\\
& = & {\bf 5550 \; m/s}
\end{eqnarray*}
... or about Mach 16
Part c...
\subsection*{Part c.}
Since gravity is a conservative force, the cow's initial velocity and final
velocity must be the same magnitude, only in the opposite direction...
{\bf 5550 m/s}
Part d...
\subsection*{Part d.}
So the cow goes from zero to 5550 m/s in 1 meter...
\begin{eqnarray*}
v^2 & = & {v_0}^2 + 2a\Delta x\\
2a\Delta x & = & v^2 - {v_0}^2\\
a & = & \frac{v^2 - {v_0}^2}{2\Delta x}\\
& = & \frac{(5550 \;m/s )^2 - (0 \; m/s )^2 }{2 \cdot 1 \; m}\\
& = & {\bf 2775 \; m/s^2}
\end {eqnarray*}
... or about 283 g's
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