Standard +0.3 This is a straightforward dimensional analysis problem requiring students to rearrange the formula and equate dimensions. While it's a Further Maths mechanics question, it only requires basic algebraic manipulation and knowledge of standard dimensions (time, length, mass), making it easier than average for A-level standard.
Kepler's Third Law of planetary motion for the period of a circular orbit around the Earth is given by the formula,
$$t = 2\pi\sqrt{\frac{r^3}{Gm}}$$
where,
\(t\) is the time taken for one orbit
\(r\) is the radius of the circular orbit
\(m\) is the mass of the Earth
\(G\) is a gravitational constant.
Use dimensional analysis to determine the dimensions of \(G\) [4 marks]
Question 6:
6 | [ ]
States or clearly uses 2π =1
or is dimensionless.
Condone no units. | 1.2 | B1 | [ ]
2π =1
r3
t =2π
Gm
4π2r3
G =
mt2
4π2 r3
[ ]
G =
[ ]t2
m
1×L3
[ ]
G =
MT2
[ G ] = L3M–1T–2
Uses dimensional analysis to
form an equation to find the
dimensions of G
Must use dimensions not units. | 1.1a | M1
Obtains L3 or M–1 or T–2
Condone use of units for this
mark. | 1.1a | M1
Obtains L3M–1T–2for the
dimensions of G | 1.1b | A1
Question total | 4
Q | Marking instructions | AO | Marks | Typical solution
Kepler's Third Law of planetary motion for the period of a circular orbit around the Earth is given by the formula,
$$t = 2\pi\sqrt{\frac{r^3}{Gm}}$$
where,
$t$ is the time taken for one orbit
$r$ is the radius of the circular orbit
$m$ is the mass of the Earth
$G$ is a gravitational constant.
Use dimensional analysis to determine the dimensions of $G$ [4 marks]
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2024 Q6 [4]}}