Question:
Two satellites A and B are orbiting the earth at heights of 2.5R and 7.5R from the centre. The ratio of their time periods is:
Two satellites A and B are orbiting the earth at heights of 2.5R and 7.5R from the centre. The ratio of their time periods is:
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$x^{3/2}$ is calculated as $x \cdot \sqrt{x}$. Here, $3^{3/2} = 3\sqrt{3}$.
Updated On: Apr 28, 2026
- $\sqrt{3}:1$
- $1:3\sqrt{3}$
- $1:\sqrt{3}$
- $1:2\sqrt{3}$
- $3\sqrt{3}:1$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Kepler's Third Law: $T^2 \propto r^3$ or $\frac{T_1}{T_2} = (\frac{r_1}{r_2})^{3/2}$.
Step 2: Analysis
Given $r_A = 2.5R$ and $r_B = 7.5R$. $\frac{r_A}{r_B} = \frac{2.5R}{7.5R} = \frac{1}{3}$.
Step 3: Calculation
$\frac{T_A}{T_B} = (\frac{1}{3})^{3/2} = \frac{1}{3\sqrt{3}}$. Final Answer: (B)
Kepler's Third Law: $T^2 \propto r^3$ or $\frac{T_1}{T_2} = (\frac{r_1}{r_2})^{3/2}$.
Step 2: Analysis
Given $r_A = 2.5R$ and $r_B = 7.5R$. $\frac{r_A}{r_B} = \frac{2.5R}{7.5R} = \frac{1}{3}$.
Step 3: Calculation
$\frac{T_A}{T_B} = (\frac{1}{3})^{3/2} = \frac{1}{3\sqrt{3}}$. Final Answer: (B)
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