Question 6 - A-Level Maths

Exam Board	OCR MEI
Module	Further Mechanics Major (Further Mechanics Major)
Year	2021
Session	November
Topic	Dimensional Analysis

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 $6.67 \times 10 ^ { - 11 } \mathrm {~kg} ^ { - 1 } \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 2 }$.
Write down the dimensions of $G$.
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.
Show that this formula is dimensionally consistent. Information regarding the planets Earth and Mars can be found in the table below.
Earth Mars
Radius (m) 6371000 3389500
Mass (kg) $5.97 \times 10 ^ { 24 }$ $6.39 \times 10 ^ { 23 }$
Escape velocity ( $\mathrm { m } \mathrm { s } ^ { - 1 }$ ) 11186
Using the formula $\mathrm { v } = \sqrt { \frac { \mathrm { kGM } } { \mathrm { r } } }$, determine the escape velocity for planet Mars.

	Earth	Mars
Radius (m)	6371000	3389500
Mass (kg)	\(5.97 \times 10 ^ { 24 }\)	\(6.39 \times 10 ^ { 23 }\)
Escape velocity ( \(\mathrm { m } \mathrm { s } ^ { - 1 }\) )	11186