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. 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. Show that $T = K d^n$ where $K$ and $n$ are constants to be found. 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. ... 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 log10 T = a + b log10 d

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Answer

To find the equation of the straight line, we need to determine the slope and intercept using the data points for Mercury and Uranus. The coordinates given are:

Mercury: (1.76, 1.94)
Uranus: (3.46, 4.49)

Calculate the slope (b):

Using the formula for the slope between two points: $b = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4.49 - 1.94}{3.46 - 1.76} = \frac{2.55}{1.70} \approx 1.5$
Calculate the y-intercept (a):

We can use one of the points, say Mercury's:

$1.94 = a + 1.5(1.76)$

Rearranging gives:

$a = 1.94 - 1.5(1.76) = 1.94 - 2.64 = -0.70$

Thus, the equation of the line is: $\text{log}_{10} T = -0.70 + 1.5 \text{log}_{10} d$

Step 2

Show that T = K d^n

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Answer

Starting from the equation found in the previous step:

$\text{log}_{10} T = -0.70 + 1.5 \text{log}_{10} d$

We can use properties of logarithms to rewrite the equation in exponential form. The equation can be expressed as:

$\text{log}_{10} T - \text{log}_{10} d^{1.5} = -0.70$

This implies:

$\text{log}_{10} \left( \frac{T}{d^{1.5}} \right) = -0.70$

Taking the antilogarithm (base 10) of both sides gives:

$\frac{T}{d^{1.5}} = 10^{-0.70} \implies T = K d^{1.5}$

where $K = 10^{-0.70}$ .

Step 3

Neptune takes approximately 60,000 days to complete one orbit of the Sun.

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Answer

Given that Neptune takes 60,000 days, we can substitute into the equation:

$60000 = K d^{1.5}$

Where we found earlier that:

$K = 10^{-0.70} \approx 0.1995$

Substituting for K:

$60000 = 0.1995 \cdot d^{1.5}$

To find $d^{1.5}$ :

$d^{1.5} = \frac{60000}{0.1995} \approx 300,301.505$

Now, we find $d$ :

$d = (300301.505)^{\frac{2}{3}} \approx 4485.5$

Hence, the average distance of Neptune from the Sun is approximately 4485.5 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 log10 T = a + b log10 d

Show that T = K d^n

Neptune takes approximately 60,000 days to complete one orbit of the Sun.

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