<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\dy}{\frac{dy}{dx}}
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\begin{document}

{\bf Question}

For larger distances from the earth the gravitational potential is
more accurately represented by $-mMG/(x+R)$ where $G$ is constant,
$M$ is the mass of the earth, $m$ is the mass of the particle, $x$
is the distance from the surface of the earth and $R$ is the
earth's radius. Show that the force produced by this potential at
the surface of the earth is the same as the gravitational force
$-mg$ if the constants are such that $g=MG/R^2$ Write down the
total energy of a rocket of constant mass $m$ travelling
vertically. Assuming the rocket blasts off from the earth surface
(where $x=0$) with speed $v=V_0$ find the maximum height the
rocket reaches. Determine the critical velocity that ensures the
rocket never reaches a maximum height (this is the escape
velocity).


\vspace{0.25in}

{\bf Answer}

Kinetic energy =$\ds T=\frac{1}{2}mv^2 \hspace{0.5in}$ Potential
energy =$\ds V=-\frac{mHG}{x+R}$

Energy =$\ds T+V=\frac{1}{2}v^2-\frac{mHG}{x+R}$

Force =$\ds -\frac{dV}{dx}=-\frac{mHG}{(x+R)^2}$, near the earth's
surface $\ds x=c$

$\ds \Rightarrow \rm{Force}\approx -\frac{mHG}{R^2}$.  Compare
this with $\ds -mg$, and show these are the same if $\ds
g=\frac{HG}{R^2}.$

because the force only depends on $\ds x$, it is conservative

$\ds \Rightarrow$ energy remains constant.

Initially energy =$\ds \frac{1}{2}mv_0^2-\frac{mHG}{R}$

$\ds \Rightarrow
\frac{1}{2}mv^2-\frac{mHG}{x+R}=\frac{1}{2}mv_0^2-\frac{mHG}{R}$
always

so $\ds v^2=v_0^2-\frac{2HG}{R}+\frac{2HG}{x+R}$

The maximum height occurs when $\ds v=0$

$\ds \Rightarrow 0=v_0^2-\frac{2HG}{R}+\frac{2HG}{x+R}$

$\ds \Rightarrow
x=\frac{\frac{R^2v_0^2}{2HG}}{1-\frac{Rv_0^2}{2HG}}$ is the
maximum height.

\newpage
Note:

for $\ds 1-\frac{Rv_0^2}{2HG}&gt;0$ there is a maximum height that
has $\ds x&gt;0$.

for $\ds 1-\frac{Rv_0^2}{2HG}&lt;0$ there is no maximum height that
has $\ds x&gt;0$.

$\ds 1-\frac{Rv_0^2}{2HG}&lt;0 \Rightarrow 1&lt;\frac{Rv_0^2}{2HG}$

$\ds \Rightarrow v_0&gt;\sqrt{\frac{2HG}{R}}$

when $\ds v_0=\sqrt{\frac{2HG}{R}}$ this is the escape velocity.


\end{document}
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