Question:
The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The satellite in an orbit at a distance of three times earth radii from 2 jts surface will be
The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The satellite in an orbit at a distance of three times earth radii from 2 jts surface will be
Updated On: Apr 8, 2024
- 83 minutes
- $ 83\times \sqrt{8} $ minutes
- 664 minutes
- 249 minutes.
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The Correct Option is C
Solution and Explanation
: $ {{T}_{2}}={{T}_{1}}{{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{3/2}}=83{{\left( \frac{R+3R}{R} \right)}^{3/2}} $ $ =83\times 8=664 $ minutes.
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Concepts Used:
Keplers Laws
Kepler’s laws of planetary motion are three laws describing the motion of planets around the sun.
Kepler First law – The Law of Orbits
All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
Kepler’s Second Law – The Law of Equal Areas
It states that the radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
Kepler’s Third Law – The Law of Periods
It states that the square of the time period of revolution of a planet is directly proportional to the cube of its semi-major axis.
T2 ∝ a3