A planet takes $T$ days to complete one orbit of the Sun. $T$ is known to be related to the planet's average distance $d$, in millions of kilometres, from the Sun. A graph of $ ext{log}_{10} T$ against $ ext{log}_{10} d$ is shown with data for Mercury and Uranus labelled. 7 (a) (i) Find the equation of the straight line in the form $ ext{log}_{10} T = a + b ext{log}_{10} d$ where $a$ and $b$ are constants to be found. 7 (a) (ii) Show that $T = K d^{n}$ where $K$ and $n$ are constants to be found. 7 (b) Neptune takes approximately 60,000 days to complete one orbit of the Sun. Use your answer to 7(a)(ii) to find an estimate for the average distance of Neptune from the Sun.

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A planet takes $T$ days to complete one orbit of the Sun - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 3

Question 7

A planet takes $T$ days to complete one orbit of the Sun. $T$ is known to be related to the planet's average distance $d$, in millions of kilometres, from the Sun. A... show full transcript

Worked Solution & Example Answer:A planet takes $T$ days to complete one orbit of the Sun - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 3

Step 1

Find the equation of the straight line in the form log_{10} T = a + b log_{10} d

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Answer

To find the equation, we need to calculate the slope of the line between the points for Mercury and Uranus.

Identify the coordinates of the points:
- Mercury: (1.76, 1.94)
- Uranus: (3.46, 4.49)
Use the formula for the slope (gradient) between two points: $b = \frac{y_2 - y_1}{x_2 - x_1}$ Substituting the values: $b = \frac{4.49 - 1.94}{3.46 - 1.76} = \frac{2.55}{1.70} \approx 1.5$
Then we can substitute into the linear equation. We know that:
- For Mercury, $1.94 = a + 1.5 \times 1.76$
- Solving for $a$ : $1.94 = a + 2.64\n$ $a = 1.94 - 2.64 = -0.7$
Thus, the equation of the line is: $\text{log}_{10} T = -0.7 + 1.5 \text{log}_{10} d$

Step 2

Show that T = K d^n

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Answer

To convert the equation from logarithmic to exponential form:

Start from the straight line equation: $\text{log}_{10} T = -0.7 + 1.5 \text{log}_{10} d$
Rearranging gives: $\text{log}_{10} T = \text{log}_{10} (d^{1.5}) - 0.7$
Using properties of logarithms, we can express it as: $\text{log}_{10} T = \text{log}_{10} \left( \frac{d^{1.5}}{10^{0.7}} \right)$
Exponentiating both sides leads to: $T = K d^{n},$ where $K = 10^{-0.7}$ and $n = 1.5$ . Thus, we have shown that $T = K d^{n}$ .

Step 3

Use your answer to 7(a)(ii) to find an estimate for the average distance of Neptune from the Sun.

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Answer

From the previous calculations, we have: $T = K d^{1.5}$ where $K = 10^{-0.7}$ . Given that Neptune takes approximately 60,000 days to complete one orbit:

Therefore, we can substitute $T = 60000$ : $60000 = 10^{-0.7} d^{1.5}$ Rearranging to find $d$ gives: $d^{1.5} = 60000 \times 10^{0.7}$ $d^{1.5} \approx 60000 \times 5.012 = 300720$ Taking both sides to the power of $rac{2}{3}$ : $d \approx (300720)^{\frac{2}{3}} \approx 4485.5$

Thus, the average distance of Neptune from the Sun is approximately 4500 million kilometres.

A planet takes $T$ days to complete one orbit of the Sun - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 3

Worked Solution & Example Answer:A planet takes $T$ days to complete one orbit of the Sun - AQA - A-Level Maths Pure - Question 7 - 2022 - Paper 3

Find the equation of the straight line in the form log_{10} T = a + b log_{10} d

Show that T = K d^n

Use your answer to 7(a)(ii) to find an estimate for the average distance of Neptune from the Sun.

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