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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

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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 $ 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 ext{log}_{10} d = 1.76
    • extlog10T=1.94 ext{log}_{10} T = 1.94
  • Uranus:
    • extlog10d=3.46 ext{log}_{10} d = 3.46
    • extlog10T=4.49 ext{log}_{10} T = 4.49
  1. 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

  2. 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 ext{log}_{10} T = a + b ext{log}_{10} d becomes:

1.94=a+1.5(1.76)1.94 = a + 1.5(1.76)

Now, rearranging to find aa gives: a=1.94−1.5imes1.76=1.94−2.64=−0.7a = 1.94 - 1.5 imes 1.76 = 1.94 - 2.64 = -0.7

  1. Final equation: The equation of the line is: extlog10T=−0.7+1.5extlog10d ext{log}_{10} T = -0.7 + 1.5 ext{log}_{10} d

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 ext{log}_{10} T = -0.7 + 1.5 ext{log}_{10} d

We can convert this equation from logarithmic form to exponential form. Using the property of logarithms: extlog10a=bextimpliesa=10b ext{log}_{10} a = b ext{ implies } a = 10^b

Thus, we can express TT in terms of dd:

  1. Rearranging the equation: extlog10T+0.7=1.5extlog10d ext{log}_{10} T + 0.7 = 1.5 ext{log}_{10} d
  2. Exponentiate both sides: 10extlog10T=101.5extlog10d+0.710^{ ext{log}_{10} T} = 10^{1.5 ext{log}_{10} d + 0.7}
  3. This simplifies to: T=10−0.7imesd1.5T = 10^{-0.7} imes d^{1.5}
  4. We can define constants KK and nn as follows: K=10−0.7,extandn=1.5K = 10^{-0.7}, ext{ and } n = 1.5 Therefore, we have shown that: T=KdnT = 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

Given that Neptune takes approximately 60,000 days to complete one orbit of the Sun, we can substitute T=60000T = 60000 and n=1.5n = 1.5 into the equation: T=KdnT = K d^n

  1. First, calculate KK: Using K=10−0.7ightarrowKextisapproximately0.1995K = 10^{-0.7} ightarrow K ext{ is approximately } 0.1995

  2. Substituting values into the equation: 60000=0.1995imesd1.560000 = 0.1995 imes d^{1.5}

  3. 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.

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