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.
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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 $ ext{log}_{10} T = a + b ext{log}_{10} d$
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Answer
To find the equation of the straight line, we will use the coordinates of the two known points from the graph:
Mercury:
extlog10d=1.76
extlog10T=1.94
Uranus:
extlog10d=3.46
extlog10T=4.49
Calculate the slope (b) using the formula:
b = rac{y_2 - y_1}{x_2 - x_1} = rac{4.49 - 1.94}{3.46 - 1.76} = rac{2.55}{1.70} = 1.5
Use point-slope form to find a:
Select one point, let's use the coordinates for Mercury (1.76, 1.94). The equation extlog10T=a+bextlog10d becomes:
1.94=a+1.5(1.76)
Now, rearranging to find a gives:
a=1.94−1.5imes1.76=1.94−2.64=−0.7
Final equation:
The equation of the line is:
extlog10T=−0.7+1.5extlog10d
Step 2
Show that $T = K d^n$ where K and n are constants to be found.
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Answer
Using the equation we just derived:
extlog10T=−0.7+1.5extlog10d
We can convert this equation from logarithmic form to exponential form. Using the property of logarithms:
extlog10a=bextimpliesa=10b
Thus, we can express T in terms of d:
Rearranging the equation:extlog10T+0.7=1.5extlog10d
Exponentiate both sides:10extlog10T=101.5extlog10d+0.7
This simplifies to:
T=10−0.7imesd1.5
We can define constants K and n as follows:
K=10−0.7,extandn=1.5
Therefore, we have shown that:
T=Kdn
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
Given that Neptune takes approximately 60,000 days to complete one orbit of the Sun, we can substitute T=60000 and n=1.5 into the equation:
T=Kdn
First, calculate K:
Using K=10−0.7ightarrowKextisapproximately0.1995
Substituting values into the equation:60000=0.1995imesd1.5
Rearranging to solve for d:
ightarrow d^{1.5} ext{ is approximately } 300250.176$$
4. **Now, take both sides to the power of $ rac{2}{3}$ to isolate $d$:**
$$d ext{ is approximately } igg(300250.176 igg)^{ rac{2}{3}} ext{ which is approximately } 4485.5 ext{ million kilometers.}$$
Thus, the average distance of Neptune from the Sun is approximately 4485 million kilometers.
