Question:
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.
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.
Updated On: Mar 12, 2026
- 4 hours
- 6 hours
- 3 hours
- 12 hours
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The Correct Option is C
Solution and Explanation
Step 1: Relationship between time period and radius:
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:
\[ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3 \] \[ \left(\frac{T_2}{24}\right)^2 = \left(\frac{1}{4}\right)^3 = \frac{1}{64} \] \[ T_2 = 24 \times \frac{1}{8} = 3 \, \text{hours} \]
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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T2 ∝ a3