6
\begin{enumerate}[label=(\alph*)]
\item Write down the dimensions of force.

The force $F$ of gravitational attraction between two objects with masses $m _ { 1 }$ and $m _ { 2 }$, at a distance $d$ apart, is given by

$$F = \frac { G m _ { 1 } m _ { 2 } } { d ^ { 2 } }$$

where $G$ is the universal gravitational constant.\\
In SI units the value of $G$ is $6.67 \times 10 ^ { - 11 } \mathrm {~kg} ^ { - 1 } \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 2 }$.
\item Write down the dimensions of $G$.
\item Determine the value of $G$ in imperial units based on pounds, feet, and seconds.

Use the facts that 1 pound $= 0.454 \mathrm {~kg}$ and 1 foot $= 0.305 \mathrm {~m}$.

For a planet of mass $M$ and radius $r$, it is suggested that the velocity $v$ needed for an object to escape the gravitational pull of the planet, the 'escape velocity', is given by the following formula.\\
$\mathrm { v } = \sqrt { \frac { \mathrm { kGM } } { \mathrm { r } } }$,\\
where $k$ is a dimensionless constant.
\item Show that this formula is dimensionally consistent.

Information regarding the planets Earth and Mars can be found in the table below.

\begin{center}
\begin{tabular}{|l|l|l|}
\hline
 & Earth & Mars \\
\hline
Radius (m) & 6371000 & 3389500 \\
\hline
Mass (kg) & $5.97 \times 10 ^ { 24 }$ & $6.39 \times 10 ^ { 23 }$ \\
\hline
Escape velocity ( $\mathrm { m } \mathrm { s } ^ { - 1 }$ ) & 11186 &  \\
\hline
\end{tabular}
\end{center}
\item Using the formula $\mathrm { v } = \sqrt { \frac { \mathrm { kGM } } { \mathrm { r } } }$, determine the escape velocity for planet Mars.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2021 Q6 [11]}}