KCET · Physics · Waves and Sound
What is a period of revolution of earth satellite ? Ignore the height of satellite above the surface of earth.
Given :
(1) The value of gravitational acceleration \( g=10 \mathrm{~ms}^{-2} \).
(2) Radius of earth \( R_{E}=6400 \mathrm{~km} \). Take \( \Pi=3.14 \).
- A \( 85 \) minutes
- B \( 156 \) minutes
- C \( 83.73 \) minutes
- D \( 90 \) minutes
Answer & Solution
Correct Answer
(C) \( 83.73 \) minutes
Step-by-step Solution
Detailed explanation
Given, gravitational acceleration, \(g=10 \mathrm{~ms}^{-1}\)
radius of Earth, \(R_{E}=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}\)
We know, period of revolution of Earth satellite is given as
\(T=2 \pi \sqrt{\frac{R_{E}}{g}}=2 \times 3.14 \times \sqrt{\frac{6400 \times 10^{3}}{10}}\)
\(=2 \times 3.14 \times 80 \times 10\)
Therefore
\(T=5024 \mathrm{~s}=\frac{5024}{60} \mathrm{~min}=83.733 \mathrm{~min}\)
Thus, period of revolution of Earth satellite is \(83.73\) minutes.
radius of Earth, \(R_{E}=6400 \mathrm{~km}=6400 \times 10^{3} \mathrm{~m}\)
We know, period of revolution of Earth satellite is given as
\(T=2 \pi \sqrt{\frac{R_{E}}{g}}=2 \times 3.14 \times \sqrt{\frac{6400 \times 10^{3}}{10}}\)
\(=2 \times 3.14 \times 80 \times 10\)
Therefore
\(T=5024 \mathrm{~s}=\frac{5024}{60} \mathrm{~min}=83.733 \mathrm{~min}\)
Thus, period of revolution of Earth satellite is \(83.73\) minutes.
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