Math Problem Statement

Jurgen is a scientist working for NASA. He maintains a telescope that is in orbit around the Earth. To test the accuracy of the telescope, he uses it to measure the radius of the Earth. The radius of the Earth is known to be 6378137 meters. (a) Jurgen measures the radius of the Earth to be 6,432,400 meters. What is the percentage error in his measurement? (b) (i) What value for the circumference of the Earth does Jurgen obtain? Round your answer to 4 significant figures. (ii) Write down the answer to part (b)(i) in the form a × 10^k, where 1 ≤ a < 10 and k ∈ Z.

Solution

The image presents a problem related to measuring the Earth's radius and calculating the percentage error and the Earth's circumference.

Problem Breakdown:

  1. Given Information:

    • The true radius of the Earth is R=6,378,137R = 6,378,137 meters.
    • Jurgen measures the radius to be Rmeasured=6,432,400R_{\text{measured}} = 6,432,400 meters.
  2. Questions:

    • (a) What is the percentage error in Jurgen's measurement?
    • (b)(i) Using his measurement, calculate the circumference of the Earth (assuming it's spherical). Round to 4 significant figures.
    • (b)(ii) Write the circumference in scientific notation a×10ka \times 10^k, where 1a<101 \leq a < 10 and kZk \in \mathbb{Z}.

(a) Percentage Error

The percentage error can be calculated using the formula:

Percentage Error=(RmeasuredRR)×100%\text{Percentage Error} = \left( \frac{|R_{\text{measured}} - R|}{R} \right) \times 100\%

(b)(i) Circumference of the Earth

The formula for the circumference CC of a sphere is:

C=2πRC = 2 \pi R

Using Jurgen's measured radius RmeasuredR_{\text{measured}}.

(b)(ii) Scientific Notation

Convert the answer from part (b)(i) into the form a×10ka \times 10^k, where 1a<101 \leq a < 10 and kk is an integer.

I'll proceed to calculate each of these now.### Solutions:

  1. (a) Percentage Error:

    • The percentage error in Jurgen's measurement is approximately 0.851%.
  2. (b)(i) Circumference of the Earth:

    • The circumference of the Earth, based on Jurgen's measurement, is approximately 40,415,961.1699 meters (rounded to 4 significant figures: 40,420,000 meters).
  3. (b)(ii) Scientific Notation:

    • The circumference in scientific notation is 4.0416×1074.0416 \times 10^7 meters.

Would you like more details on any of the steps, or do you have any further questions?

Related Questions:

  1. What is the actual circumference of the Earth using the true radius?
  2. How would the percentage error affect calculations related to satellite orbits?
  3. How does the Earth's non-spherical shape impact such measurements?
  4. What methods could Jurgen use to improve the accuracy of his radius measurement?
  5. How would a significant percentage error influence real-world applications, such as GPS?

Tip:

When performing calculations that involve real-world measurements, always consider the impact of measurement error, especially in critical fields like astronomy or engineering.

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Math Problem Analysis

Mathematical Concepts

Percentage error
Circumference calculation
Scientific notation

Formulas

Percentage Error = |Measured Value - True Value| / True Value * 100
Circumference of a sphere = 2πr
Scientific notation: a × 10^k

Theorems

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Suitable Grade Level

Grades 9-12