\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

{\bf Question}

A particle of mass $m$ moves in a straight line form a fixed
particle of mass $M$ under their influence of their mutual
gravitational attraction.  Show that the potential of the
gravitational force is $\ds -\frac{Gmm}{r},$ where $R$ is the
distance between two masses.  A rocket departs from the earth in a
straight line.  Shortly after lift-off when the engines have
stopped firing the rocket has speed $u$.

\begin{description}
\item[(a)] Show that in order that the rocket can escape the
earth's gravitational field $u > u_E,$ where $u_E$ is the so
called escape velocity for a particle.  (treat the gravitational
field of the earth as that of a particle with the mass of the
earth and write down conservation of energy for the rocket.)

\item[(b)] Calculate $u_E$ given that $G = 6.6726 \times 10^{-11}$
Nm$^2$kg$^{-2}$, the mass of the earth is $5.976 \times 10^{24}$kg
and the radius of the earth is $6.38 \times 10^6$m.


\end{description}

\vspace{.25in}

{\bf Answer}

$$$$
\begin{center}
\setlength{\unitlength}{.5in}
\begin{picture}(5,1)
\put(3,1){\makebox(0,0){$r$}}

\put(3.1,1){\vector(1,0){0.8}}

\put(2.9,1){\vector(-1,0){0.8}}

\put(2,.9){$\bullet$}

\put(3.9,.9){$\bullet$}

\put(2,0.8){\makebox(0,0){m}}

\put(5,0.9){\vector(-1,0){0.5}}

\put(5.6,0.9){\makebox(0,0){$\ds \frac{GMm}{r^2}$}}

\put(0,1){\vector(1,0){.3}}

\put(0,1){\vector(0,1){0.3}}

\put(0,1.5){\makebox(0,0){\bf j}}

\put(0.4,.9){\makebox(0,0){\bf i}}

\end{picture}
\end{center}

Gravitational Force $ \ds {\bf F} = -\frac{GMm}{r^2} \bf i$

Potential $\ds = - \int F \, dr = - \frac{GMm}{r}$

\begin{description}
\item[(a)]

Conservation of energy: Kinetic Energy + Potential Energy =
Constant

Therefore $\ds \frac{1}{2} m u^2 - \frac{GMm}{r} = $ constant

Initially $u = v$, and $r = R$ (the radius of the earth)

Therefore $\ds \frac{1}{2}mv^2 - \frac{GMm}{r} = \frac{1}{2} m v^2
- \frac{GMm}{R}$

\item[(b)]
For a rocket to escape the Earth's gravitational field requires
$v\geq$ 0 as  $r\rightarrow \infty,$ hence

$$ \frac{1}{2}m^2 \geq \frac{GMm}{R} \Rightarrow v \geq
\sqrt{\frac{2GM}{R}}$$

Using the Earth's data gives an escape velocity of 11.2kms$^{-1}$
\end{description}



\end{document}
