The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
- 4 hours
- 6 hours
- 3 hours
- 12 hours
The Correct Option is C
Solution and Explanation
The orbital time period (\(T\)) is related to the orbital radius (\(R\)) as: \[ T^2 \propto R^3 \] Step 2: Given change in radius:
The new radius is: \[ R_2 = \frac{R_1}{4} \] Step 3: Relate the time periods:
\[ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3 \] \[ \left(\frac{T_2}{24}\right)^2 = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \] \[ T_2 = 24 \times \frac{1}{8} = 3 \, \text{hours} \]
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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