June 2025 Paper 1 Q9
9 The table below shows information about four of the moons of the planet Jupiter. The semi-major axis is the greatest distance that each moon reaches from the centre of its orbit around Jupiter. The orbital period is the time in Earth days that it takes to orbit the planet.
| Moon | Semi-major axis (km) | Orbital period (Earth days) |
|---|---|---|
| Io | 421 800 | 1.7627 |
| Europa | 671 100 | 3.5255 |
| Ganymede | 1 070 400 | 7.1556 |
| Callisto | 1 882 700 | 16.690 |
A student uses the equation \(T = kd^n\) to model the orbital period of Jupiter’s moons, where \(T\) is the orbital period in Earth days, \(d\) is the semi-major axis in km, and \(k\) and \(n\) are constants.
The student uses graph drawing software to plot \(\log_{10}T\) against \(\log_{10}d\). They find that the line of best fit for the data has gradient 1.504 and intercepts the \(\log_{10}T\) axis at \(-8.215\).
The student uses their equation to predict the orbital period for another moon of Jupiter called Thebe which has a semi-major axis of 221 900 km. They look up the value in an online encyclopedia and find it is 0.6761 Earth days.
| Scheme | Marks | AO |
|---|---|---|
| \(\log_{10}T = \log_{10}k + \log_{10}d^n\) \(\log_{10}T = \log_{10}k + n\log_{10}d\) | B1 | 2.1 |
| [1] |
Notes
B1: Uses laws of logs to split into two terms and then to simplify the second term. This line must be seen
AG
| Scheme | Marks | AO |
|---|---|---|
| Gradient \(= 1.504 = n\) | B1 | 3.3 |
| Intercept \(= -8.215 = \log_{10}k\) | M1 | 3.3 |
| So \(k = 10^{-8.215}\ [= 6.095\times 10^{-9}]\) | A1 | 1.1 |
| [3] |
Notes
B1: Allow \(n = 1.5\) or better seen
M1: soi
A1: Allow awrt \(6.1\times 10^{-9}\)
For using 2 data points from the table SC1 awrt \(n = 1.5\), SC1 \(5.159\times 10^{-9} \lt k \lt 7.076\times 10^{-9}\)
| Scheme | Marks | AO |
|---|---|---|
| For Thebe \(d = 221900\) Model predicts \(T = k\times d^{1.504} = 0.669\) | M1 | 3.4 |
| This is close to the given value 0.6761 which suggests that the model is suitable | A1FT | 3.5b |
| [2] |
Notes
M1: Uses the model to predict
A1FT: Appropriate comment based on similarity of their correct value and the given value.
FT their \(k\)
Do not accept a comment that implies the model should give an exact match
Alternative method
| Scheme | Marks |
|---|---|
| \(-8.215 + 1.504\log 22190 = -0.174\ldots\) | M1 |
| This is close to \(\log 0.6761 = -0.169989\ldots\) which suggests that the model is suitable | A1 |
M1: Uses the model to predict \(\log T\)
A1: Appropriate comment based on similarity of their correct values.
SC1 “Student equation would not be suitable as the data would have to be extrapolated” oe