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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Circular Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Circular Motion

Determine the acceleration of a particle located on the surface of...

Determine the acceleration of a particle located on the surface of the earth at the equator due to the earth's rotation. The diameter of the earth is 12800 km12800 \text{ km} and it takes 2424 hours for the earth to complete one revolution around its axis.

Options

  1. A0.0168 m/s20.0168 \text{ m/s}^2
  2. B0.0336 m/s20.0336 \text{ m/s}^2
  3. C0.0672 m/s20.0672 \text{ m/s}^2
  4. D0.336 m/s20.336 \text{ m/s}^2

Correct answer

B. 0.0336 m/s20.0336 \text{ m/s}^2

Step-by-step solution

Radius of the earth, R=D2=128002=6400 km=6.4×106 mR = \dfrac{D}{2} = \dfrac{12800}{2} = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}

Time period of rotation, T=24 hours=24×3600 s=86400 sT = 24 \text{ hours} = 24 \times 3600 \text{ s} = 86400 \text{ s}

The acceleration of a particle at the equator due to the earth's rotation is the centripetal acceleration:

ac=ω2R=(2πT)2Ra_c = \omega^2 R = \left(\dfrac{2\pi}{T}\right)^2 R

Substituting the values:

ac=(2π86400)2×6.4×106a_c = \left(\dfrac{2\pi}{86400}\right)^2 \times 6.4 \times 10^6

ac=4π27464960000×6400000a_c = \dfrac{4\pi^2}{7464960000} \times 6400000

ac=25.6π27464.96a_c = \dfrac{25.6 \pi^2}{7464.96}

Taking π2≈9.8\pi^2 \approx 9.8:

ac≈25.6×9.87464.96=250.887464.96≈0.0336 m/s2a_c \approx \dfrac{25.6 \times 9.8}{7464.96} = \dfrac{250.88}{7464.96} \approx 0.0336 \text{ m/s}^2

Answer: 0.0336 m/s20.0336 \text{ m/s}^2
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