Question:
A satellite P takes 2 years to make a circular orbit of radius R around the earth. The time period of another satellite Q if it moves in a circular orbit twice the orbital radius of satellite P around earth is:
A satellite P takes 2 years to make a circular orbit of radius R around the earth. The time period of another satellite Q if it moves in a circular orbit twice the orbital radius of satellite P around earth is:
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For satellites, time period increases as $r^{3/2}$, so even small increases in radius significantly increase orbital time.
Updated On: Jul 18, 2026
- 5.2 years
- 5.1 years
- 5.6 years
- 5.9 years
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The Correct Option is C
Solution and Explanation
Step 1: Applying Kepler’s Third Law.
For satellites orbiting the same central body: \[ T^2 \propto r^3 \] This gives: \[ \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2} \]
Step 2: Substituting given values.
Given: \[ T_1 = 2 \, \text{years}, \quad r_2 = 2R, \quad r_1 = R \] So: \[ \frac{T_2}{2} = (2)^{3/2} \]
Step 3: Evaluating power term.
\[ (2)^{3/2} = \sqrt{2^3} = \sqrt{8} = 2.828 \]
Step 4: Calculating time period of satellite Q.
\[ T_2 = 2 \times 2.828 = 5.656 \]
Step 5: Approximation.
\[ T_2 \approx 5.6 \, \text{years} \]
Step 6: Final conclusion.
\[ \boxed{5.6 \, \text{years}} \]
For satellites orbiting the same central body: \[ T^2 \propto r^3 \] This gives: \[ \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2} \]
Step 2: Substituting given values.
Given: \[ T_1 = 2 \, \text{years}, \quad r_2 = 2R, \quad r_1 = R \] So: \[ \frac{T_2}{2} = (2)^{3/2} \]
Step 3: Evaluating power term.
\[ (2)^{3/2} = \sqrt{2^3} = \sqrt{8} = 2.828 \]
Step 4: Calculating time period of satellite Q.
\[ T_2 = 2 \times 2.828 = 5.656 \]
Step 5: Approximation.
\[ T_2 \approx 5.6 \, \text{years} \]
Step 6: Final conclusion.
\[ \boxed{5.6 \, \text{years}} \]
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