The approximate mean distance $x$ (in millions of kilometres) from the centre of the Sun and the period of the orbit $T$ (in Earth years) are recorded in the table.

	Mercury	Venus	Mars	Uranus
$x$	58	108	228	2871
$T$	0.24	0.62	1.88	84.11

It is believed that the planets orbiting around the Sun obey a law of the form $T = kx^n$, where $k$ and $n$ are constants.

1. Transform the non-linear equation into a linear form using logarithms.

$T = kx^n$
Take $\lg$ on both sides:
$\lg T = \lg(kx^n)$
$\lg T = \lg k + \lg x^n$
$\lg T = n \lg x + \lg k$

2. Calculate the corresponding values for the linear variables $\lg x$ and $\lg T$.

Let $X = \lg x$ and $Y = \lg T$.

| $x$ | 58 | 108 | 228 | 2871 |
| $T$ | 0.24 | 0.62 | 1.88 | 84.11 |
| $X = \lg x$ | 1.76 | 2.03 | 2.36 | 3.46 |
| $Y = \lg T$ | -0.62 | -0.21 | 0.27 | 1.93 |

3. Plot the straight line graph.

Plot the points $(1.76, -0.62)$, $(2.03, -0.21)$, $(2.36, 0.27)$, and $(3.46, 1.93)$ on a Cartesian plane with $X$-axis as $\lg x$ and $Y$-axis as $\lg T$. Draw the line of best fit passing through these points.

1. Calculate the gradient of the line to find $n$

From part (a), the linearised equation is $\lg T = n \lg x + \lg k$. The gradient of the line is $n$.
Using the points for Mercury $(1.763, -0.620)$ and Uranus $(3.458, 1.925)$:
$n = \frac{1.925 - (-0.620)}{3.458 - 1.763} = \frac{2.545}{1.695} \approx 1.50$

2. Find the $y$-intercept to determine $k$

The $y$-intercept of the graph is $\lg k$.
Substituting $n = 1.5$ and the point $(3.458, 1.925)$ into the equation:
$1.925 = 1.5(3.458) + \lg k$
$\lg k = 1.925 - 5.187 = -3.262$

3. Solve for $k$ and state the final answers to two significant figures

$k = 10^{-3.262} \approx 5.47 \times 10^{-4}$
To two significant figures, $n = 1.5$ and $k = 5.5 \times 10^{-4}$.

1. Identify the value of $x$ and calculate $\lg x$

The distance is $149.6 \times 10^6\text{ km}$, so $x = 149.6$.
$\lg x = \lg 149.6 \approx 2.175$

2. Read the corresponding value of $\lg T$ from the graph and solve for $T$

From the graph, when $\lg x = 2.175$, $\lg T \approx 0$.
$T = 10^0 = 1$

1. Calculate the value on the vertical axis for the given $T$

Given $T = 11.86$, the value on the vertical axis is:
$\lg T = \lg 11.86 \approx 1.074$

2. Read the corresponding value of $\lg x$ from the graph and calculate $x$ in km

From the graph of $\lg T$ against $\lg x$, when $\lg T = 1.074$, we read the corresponding value on the horizontal axis:
$\lg x \approx 2.89$
$x = 10^{2.89} \approx 776$
Since $x$ is in millions of kilometres, the distance is:
$776 \times 10^6 = 7.76 \times 10^8\text{ km}$ (Accept values around $7.8 \times 10^8\text{ km}$ depending on the graph reading)