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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 \( \Pi=3.14 \).
Options:
of earth.
Given :
(1) The value of gravitational acceleration \( g=10 \mathrm{~ms}^{-2} \).
(2) Radius of earth \( R_{E}=6400 \mathrm{~km} \). Take \( \Pi=3.14 \).
Solution:
2893 Upvotes
Verified Answer
The correct answer is:
\( 83.73 \) minutes
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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